111Present address: Department of Physics, Winona State University, Winona, MN
55987, USA

# How many acres of potatoes does a society need?

N T Moore<EMAIL_ADDRESS>
###### Abstract

One of the main difficulties in a class on Sources of Energy and Social Policy
is the wide variety of units used by different technologists (BTU’s, Barrels
of oil, Quads, kWh, etc). As every student eats, I think some of this
confusion can be resolved by starting and grounding the class with a
discussion of food and food production. A general outline for this
introduction is provided and two interesting historical cultural examples,
Tenochtitlan and the Irish Potato Famine, are provided. Science and Social
Policy classes are full of bespoke units and involve many different contexts.
Starting the class with a discussion of food energy is a nice way for everyone
to start with the same context. In addition, discussion of Food Energy can
lead to interesting historical claims.

Keywords: Energy, Social Policy, kcals, Tenochtitlan, Irish Potato Famine,
History, self-reliance

##  1 Introduction

When the United States entered World War One one of the problems they faced
was logistics. How much food do you need to ship overseas to Europe to feed a
million soldiers? That early work in nutrition led to the $3000$ Calorie diet
many people remember from secondary Health Education class. A bit about
“Calorie” (uppercase) vs “calorie” (lowercase) units you might remember:
$1~{}Calorie=1~{}kilocalorie~{}(kcal)$, and a dietitian might build a
$3000kcal$ diet for a 20 year old basketball player. A calorie is the amount
of energy it takes to heat a gram of water by a degree Celsius. There are
about $4.2$ Joules in a single calorie, and a Joule occurs all over
introductory physics. If you need to buy a new home furnace, the sales
brochure might advertise that it is capable of delivering $100,000$ BTU’s of
heat each hour. What’s a BTU? Heat a pound of water by $1^{\circ}F$. Of course
Heat Pumps are far more efficient than simply burning methane or propane, but
they consume kilo-watt-hours (kWh) of electricity, not BTU’s. What’s a kWh?
Run a $1000$ Watt toaster for an hour and you’ll have pulled one kWh off the
grid, it will cost you about $\$0.13$ in Minnesota. If you decide to put solar
panels in your backyard, they will probably collect about $10\%$ of the
$3.5kWh$ the the sun delivers to each square meter of your lawn (in Minnesota)
each day.

As the previous paragraph illustrates, there are a frustratingly large number
of different units in an “Energy” class. At Winona State, this 3 credit class
fulfills a “Science and Social Policy” general education requirement and is
taken by students from across the university. Lots of college majors don’t
require a math class beyond algebra or introductory statistics and the
population is largely math-averse. You could jokingly say that one of the main
things students learn in the class is unit conversion, but it isn’t far off.
Nearly every field finds energy a useful representation, and every profession
has their own set of units and terminology most well suited for quick
calculation. Would a medical lab scientist talk about the fractional acre-foot
of urine needed test kidney function? No, but someone in the central valley of
California would certainly care about the acre-feet of water necessary to grow
almonds! Does a gas station price their gasoline in dollars per kWh? Given the
growing electrification of cars, they might soon.

Everyone eats, maybe not $3000kcals$ per day, but at least something every
day. When I teach our energy class, [1, 2], I spend a few weeks talking about
food energy before all other types. While food production is not central to
climate change and wars over oil, food is essential in a way that diesel and
gasoline are not. Vehicle fuel makes modern life possible, but we could live,
unpleasantly, without it. We can’t live without fats and protein.

##  2 Food Energy

To introduce Food Energy, I ask the students to work through a few questions:

Figure 1:  A proto-college-student at Winona’s China King Buffet, dreaming
about visiting the steam tables every day.

###  2.1 Converting food into body heat

Planning to save money, one college student decides to go to an all-you-can-
eat buffet each day at 11am, eg figure 1. If he brings homework and stretches
the meal out for a few hours he can get all $3000kcals$ with only one bill.
Food is fuel for the human body – could too much fuel make his body feel sick?
If his body burned all this food at once, how much warmer would he get? Useful
information: the student has a mass of $80kg$ and is made mostly of water. A
Calorie heats $1kg$ of water $1^{\circ}C$.

Here’s a possible answer: equate food energy with calorimetric heating and
assume human bodies have the same heat capacity as water, about
$1\frac{kcal}{kg\cdot^{\circ}C}$. This allows us to calculate the body’s
temperature increase.

$\displaystyle 3000kcals$ $\displaystyle=$ $\displaystyle 80kg\cdot
1\frac{kcal}{kg\cdot^{\circ}C}\cdot\Delta T$ $\displaystyle\Delta T$
$\displaystyle\approx$ $\displaystyle+37.5^{\circ}C$

Students are normally quite surprised at this number. Although wildly
unrealistic, $\Delta T\approx+6^{\circ}C$ is typically fatal, there is a
related phenomena of diet-induced thermogenesis[3] known informally as “the
meat sweats”. Some students connect this calculation to feeling quite hungry
after a cold swim in the pool (a similar effect). On a larger scale,
discussing what’s wrong with this estimate is useful. The main storage
mechanism for storing food energy is fat tissue, which the calculation
completely ignores. Infants are generally born with little fat, and an infant
sleeping through the night often coincides with the baby developing enough fat
tissue to store sufficient kcals to make it though a night without waking up
ravenously hungry. A related follow-up is that if a person is stranded in the
wilderness, they should immediately start walking downstream (ie, towards
civilization) as they likely won’t be able to harvest an amount of kcals
equivalent to what they already have stored on their hips and abdomen.[4] The
contrast of bear hibernation [5] and songbirds constantly eating through the
winter are related connections to investigate.

###  2.2 Biophysical Power

A more realistic question to follow up with relates to the average power given
off by a person over a day. Again, assuming $3000kcal$ is burned over
$24hours$, with useful information: $1kcal\approx 4200J$ and $1J/s=1W$.

$\frac{3000kcal}{24hours}\cdot\frac{4200J}{1kcal}\cdot\frac{1hour}{3600sec}\approx
145W$ (1)

Most students still remember $75Watt$ lightbulbs, but given the spread of LED
lighting, “A person’s body heat is two $75W$ light bulbs” will probably only
make sense for a few more years. Desert or cold-weather camping, alone versus
with friends, and survival swimming are also examples for students to make
sense of this answer. If you can take advantage of other people’s waste body
heat, you’ll sleep more pleasantly and survive longer in cold water.

Another application to discuss is that of “brown fat,” a sort of biological
space heater that humans and other mammals develop in response to cold
weather. This tissue’s mitochondria can burn lipids and carbohydrates in a
useless proton pumping scheme, which produces metabolic heat [6, 7, 8, 9].
Most common in rodents and infants, this mechanism can be stimulated by
extended exposure to cold temperatures – the original work was done on
lumberjacks in Finland [10] . The idea of a biological space heater that takes
a month to turn on and a month to turn off matches the lived experience of
college students in Minnesota, who wear down jackets in $4^{\circ}C$ weather
in November, and beachwear in the same $4^{\circ}C$ weather in March.
Additionally, transplants to northern climates often take a few years to “get
used to” the colder weather up north. It seems just as easy to say that
transplants’ bodies take a few years to develop the brown fat cells which
allow them to be comfortable in cold weather.

One other distinction to emphasize is the difference between power and energy.
A graph of a human body’s “kcal content” over the course of a day can be a
useful illustration. When sedentary, this graph probably has the slope of
$-150W\approx-125\frac{kcals}{hour}$. If the $3000kcal$ meal at the buffet
takes an hour, this period corresponds to an energy-time slope of
$+3000\frac{kcal}{hour}\approx+3500W$.

In medicine, these slopes are effectively equivalent to “Metabolic Equivalent
of Task” (METS), a common measure in cardiology and exercise physiology. METS
is power normalized by mass, $1METS=1\frac{kcal}{kg\cdot hour}$, and METS
levels are available for many different physical activities. [11]

###  2.3 Burning off food energy

Imagine that after eating a $600kcal$ bacon-maple long-john (donut), you
decide to go for a hike to “work off” the Calories. Winona State is in a river
valley bounded by $200m$ tall bluffs. How high up the bluff would you have to
hike to burn off the donut? Useful information: human muscle is about $1/3$
efficient, and on Earth’s surface, gravitational energy has a slope of about
$10~{}\frac{Joules}{kg\cdot m}$.

Figure 2: An Energy Bar Chart to illustrate the $1/3$ efficient student hiking
up a bluff to burn off the morning’s donut. The initial state (left) is the
hiker at the bottom of the hill, with donut in stomach. The final state
(right) is the hiker at the top of the bluff with $2/3$ of the energy removed
to the atmosphere by sweat and exhalation of warm air. $1/3$ of the donut’s
energy is stored in elevation. The system for this diagram includes the earth,
the hiker, and the donut. The system does not include the atmosphere around
the hiker.

One way to approach this problem is by using Energy Bar Charts [12] to
illustrate how the energy held in food changes form as it is used. An
approximation for this question is shown in figure 2. In this story, the
“system” is taken to be the earth, food, and hiker. The hiker’s body is
assumed to be $1/3$ efficient, which means one of the food energy blocks of
energy is transformed into gravitational energy (elevation) at the end of the
hike. The other $2$ blocks of energy are transformed into heat and leave the
hiker’s body, most likely by mechanisms of respiration and sweat evaporation.
The purpose of a bar chart like this is to provide a pictorial and
mathematical representation of the energy conservation equation given in 2.

$\displaystyle\frac{1}{3}\cdot 600kcal\cdot\frac{4200J}{1kcal}$
$\displaystyle=$ $\displaystyle 80kg\cdot 10\frac{Joules}{kg\cdot m}\cdot
height$ (2) $\displaystyle height$ $\displaystyle\approx$ $\displaystyle
1000m$ (3)

This estimate is again surprising to students. Five trips up the bluff to burn
off $\$2$ of saturated fat, sugar, and flour! A nice followup calculation is
to imagine a car that can burn a $100kcal$ piece of toast in the engine: from
rest, what speed will the toast propel it to? If (again) the engine converts
$1/3$ of the energy into motion (kinetic energy), a $1300kg$ Honda Civic will
reach a speed of about $15\frac{m}{s}\approx 33mph$!

The point of these energy calculations is not to give students an eating
disorder. Rather, the numbers show food’s amazing power. A single slice of
toast will bring a car up to the residential speed limit! A day’s food,
$3000kcal$, will power you up an $5000m$ mountain peak! The body-work food
allows us to do is astonishing, and increases in food production have made
modern comforts, unimaginable $150$ years ago, possible to the point of being
taken for granted.

###  2.4 Where does food energy come from?

One feature of the aught’s “homesteading” culture [13] is the idea that a
person should probably be able to move to the country, eat a lot of peaches,
and grow all their own food. Learning that farming labor is skilled labor can
be brutal and disheartening. Eating $3000kcals$ each day means planting,
weeding, harvesting, and storing more than a million kcals each year [14].
Where will those Calories come from? Is your backyard enough to homestead in
the suburbs [15]?

At some point between 1920 and 1950, US chemical manufacturers realized that
in the post-war period, they could repurpose processes developed for
manufacturing munitions and chemical warfare agents, to produce chemicals that
would kill insects and increase the nitrogen levels in the soil. As figures 3
and 4 show, the epoch of “Better Living Through Chemistry” produced a dramatic
increase in per-acre yields across all comodity food crops, particularly corn
and potatoes.

Figure 3:  USDA per acre Corn and Potato production figures, plotted over
time. Data is given in harvet units, $56lbs$ bushels per acre for field corn
and hundred-weight (CWT) for potatoes. By mass, corn is about $4.5$ times more
calorie dense than potato which results in a nearly equal $kcal/acre$ values
for both crops in figure 4. Details on the data source and conversions are
given in A.  Figure 4:  USDA per acre crop production figures, plotted over
time. Production data is scaled by estimated dietary kcal content to show
that, over all crops, there has been a dramatic increase in kcal production
since about 1940. Details of the data source and conversions are given in A.
The idea for this plot came from an online blog, [17]. It would be interesting
to know if there are patterns of scaling among vegetable families (grains,
legumes, tubers, etc) in the same way that there are family classifications
for the minimal energy required for transport [18].

However, if you’re discussing backyard Calorie production it isn’t reasonable
to use modern yield estimates for planning. “Roundup Ready” Corn, Soybean, and
Sugar Beet seeds are not available to the public, nobody wants to put on a
respirator to apply Atrazine ten feet from the back door, and the edge effects
from deer and insects are much smaller on a $600$ acre field than they are in
an community garden allotment. As mentioned in the introduction, in 1917 the
USDA published a pamphlet [16] giving detailed Calorie estimates a farmer
might expect from a given acre of a crop. A table from this pamphlet is shown
in Figure 5. The pamphlet data came from pre-war, pre-chemical agriculture,
and the yields cited were produced with horses, manure, lime, and large
families full of children. If you want to be self sufficient, these yield
numbers are probably a good upper bound on what’s realistically possible by a
dedicated Luddite.

Figure 5:  A table from a USDA booklet giving 1917 yields for various farm
products.

So, another question using this data. If you want to feed your family of four
people potatoes, how much land will you need to cultivate? Here’s an estimate:
a family of 4 requires $3000kcal/person$ each day[19]. If we over-estimate and
produce food for the entire year, the family will need about $4.4$ million
kcals.

$4~{}people\cdot\frac{3000kcal}{person\cdot
day}\cdot\frac{365~{}days}{year}\approx 4.4Mkcal$ (4)

A brief aside for those bored by the simplistic unit conversion: when I ask
students to solve problems like these, one undercurrent of conversation is
“Should I divide by 365 or multiply?” Particularly with online homework
systems, checking your answer for reasonability isn’t typically graded. Asking
the students to reason proportionally with units is a skill that can give
meaning to numbers.

From figure 5 we can estimate $1.9~{}million~{}kcals$ per acre of potato
production. Again the students might ask, should I multiple $4.4$ and $1.9$ or
should I divide them? It can be useful in a class discussion to have the
students discuss and vote which of the following two forms will give the
meaningful answer.

$\displaystyle\frac{4.4Mkcal}{family}\cdot\frac{1acre}{1.9Mkcal}$ or
$\displaystyle\frac{4.4Mkcal}{family}\cdot\frac{1.9Mkcal}{1acre}$ (5)

The choice of operation is difficult to make without seeing the units present,
which is again a learning opportunity for the students.

What does the answer of $2.3$ acres mean? The university’s $91m\times 49m$
football field has an area of about $1.1$ acres, so you could say that a
football field planted in potatoes will probably feed a family through the
winter [20]. Can a person enjoy the benefits of urban living and grow all
their own food? The population density of New Jersey is
$1,263~{}people/mile^{2}\approx 1.97~{}people/acre$ and our 4 person family
needs $2.3$ acres for their potatoes. Unless the social model is one of a
country Dacha or an endless suburb with no duplexes or apartment buildings,
urban living and food self-sufficiency seem mutually exclusive.

More emotionally charged conversations can be had about converting the United
States to all organic agriculture, which, for corn, typically has a yield
penalty of about $20-40bu/acre$ when compared to conventional production. The
1917 data isn’t directly applicable, but it relates. At $180bu/acre$
conventional corn requires $\approx 24~{}million~{}acres$ (half of Wisconsin,
or all of Indiana) to feed the US population ($350$ million people) corn for a
year. The remainder of the corn belt can be devoted to animal feed, ethanol,
and export. If the corn belt was devoted to producing organic corn at lower
yield [21], we probably wouldn’t starve, but cheap meat and ethanol vehicle
fuel would likely disappear.

##  3 Example: How big could Tenochtitlan have been?

The questions described thus far have largely been centered within a physics
context. The paper closes with two more examples that leverage this food
energy picture to make historical claims. The first example relates to the
pre-Columbian capital of the Aztec Empire, Tenochtitlan, now known as Mexico
City. Tenochtitlan was built on and around a endorheic lake, Texcoco. Crops
were grown in shallow parts of the lake via chinampas [22], floating patches
of decaying vegetation and soil. Given the proximity to water and decaying
vegetation, these fields were very fertile [23, 24] and some continue to be
used in the present day [25].

Estimates of Tenochtitlan’s population in 1500CE vary widely, from 40,000 [26]
to more than 400,000 [27] inhabitants, comparable in size to Paris at that
time. These estimates come from oral and written records and estimates of
archaeological building density and land area. While cannibalism was part of
Aztec religious ritual and practice [28], the staple Calorie sources for the
Aztecs were corn and beans.

Few if any Native American cultures made use of draft animals for food or
power before the Columbian Exchange. This means that the food that fed
Tenochtitlan must have been brought to the city center by foot or canoe. How
much land must have been devoted to chinampas to feed the population, or
conversely, how many people could be supported by the land within walking or
paddling distance from the city center?

A 1964 paper in Scientific American [24] gives a general outline of the
chinampas in the area of Tenochtitlan in 1500CE. This map seems to be the
basis for the similar figure in Wikipedia [29]. Descriptions of chinampas
agriculture indicate that as many as $7$ successive crops could be grown and
harvested from the same plot of soil each year, two of which could be maize
(corn). This is truly amazing productivity, given that in the midwest United
States corn is normally grown, at most, every other year because of it’s
extreme nutrient demands on the soil.

There are many ways to approach this estimation problem. We could assume a
Tenochtitlan population of $100,000$ people has a $3000kcal/day$ diet that
comes completely from corn. Assuming that corn’s density and nutritional
content haven’t changed in the $4$ centuries preceding the 1917 data in figure
5, we could assume $1lbs$ of corn contains $\approx 1594kcal$ of food energy.
Looking at the map with ImageJ [30], it seems like the recorded area devoted
to chinampas might be about $16,000~{}acres$ – details are given in B. With
these assumptions, we could equate the corn energy production from chinampas
with the population’s yearly food need. Note, in this version of the story,
the corn productivity, $P\frac{bu}{acre}$ is treated as an unknown variable.

$\displaystyle Food~{}production$ $\displaystyle=$ $\displaystyle
16,000acres\cdot\frac{2~{}corn~{}crops}{year}\cdot
P\frac{bu~{}of~{}corn}{acre}$ $\displaystyle Population~{}requires$
$\displaystyle=$ $\displaystyle
100,000~{}people\cdot\frac{3000kcal}{person\cdot
day}\cdot\frac{365days}{year}\cdot\frac{1lbs~{}corn}{1594kcal}\cdot\frac{1bu}{56lbs}$
$\displaystyle P\approx 38\frac{bu}{acre}$ (6)

This crop productivity is in remarkable agreement with the 1917 USDA yields,
$35bu/acre$, which seems to validate the assumed $100,000$ person population
of Tenochtitlan. Some references [24] describe an extensive tribute system
that Aztec government required of it’s subjects, which certainly would have
been necessary to support populations on the upper end of historical estimates
[27].

##  4 Example: Was the Irish Potato Famine a Natural Disaster?

In contrast to native cultures of the Americas, Ireland’s population boomed
with the Columbian Exchange and the introduction of the potato. [31, 32].
Figure 6 shows that from about 1700 onward there was a dramatic growth in the
island’s population. There’s never just one reason for historical events, but
unlike grains, potatoes thrived in Ireland’s cool damp climate and potatoes,
kale, and milk form a nutritionally complete diet that greatly reduced hunger-
related mortality among the poor working-class in Ireland. If you look closely
at the data in figure 6 you might believe that there were two weather and
potato related famines, the most obvious 1845-49 and the second, with much
smaller effect on population in 1740-1. Both famines were precipitated by poor
weather, but an important difference is that in 1740, Ireland was a sovereign
state but by 1845 the island was effectively an economic colony of the British
Empire [32].

As the story goes, the two main commodity crops in Ireland were potatoes (for
humans), and oats, which as horse feed, were something like gasoline in
today’s economy. A sovereign government can halt the export of food to feed
English horses, which is what happened in 1741 (and 1782). The grain was
diverted back as relief to starving people in Ireland, reducing the famine’s
mortality. However, by 1845 most of Irish farmland was economically controlled
by foreign (English) markets, and grain traders typically refused to divert
oats (horse feed) as famine relief for the sake of their investment income.

This inflammatory claim, which is certainly a simplified version of history,
serves as a useful evaluation example for students. Specifically, in years
that the potato crop failed because of weather or late blight, could the
amount of oats produced (and exported) have fed the Irish population? More
broadly, was the Great Famine due to weather and disease, natural causes “we
can’t do anything about,” or was the depth of the tragedy a result of
political choices?

Some estimates follow: Ireland’s population in 1845 was about $8.5$ million
people. The island has an area of about $84,400km^{2}$ [33] and you might
estimate that $64\%$ of the land ($54,000km^{2}$) is arable for agriculture
[37]. It seems reasonable to use the 1917 productivity, figure 5, to make
calculations for Ireland in 1845. Reminder, in 1917, potatoes produced
$1.908\times 10^{6}kcal/acre$ and oats $1.254\times 10^{6}kcal/acre$. With
students, evaluation of the claim could be approached as a series of
questions:

How much food does the island need?

$\displaystyle food~{}needed~{}per~{}year$ $\displaystyle=$ $\displaystyle
8.5\times 10^{6}~{}people\cdot\frac{3000}{person\cdot
day}\cdot\frac{365days}{year}$ $\displaystyle\approx$ $\displaystyle 9.3\times
10^{12}kcals$

How much land area, sown in potatoes, would produce this food?

$\displaystyle 9.3\times 10^{12}kcals/\left(1.908\times
10^{6}\frac{kcal}{acre}\right)$ $\displaystyle=$ $\displaystyle 4.87\times
10^{6}acres$ $\displaystyle\approx$ $\displaystyle 19,700km^{2}$

How much land area, sown in oats, would produce this food?

$\displaystyle 9.3\times 10^{12}kcals/\left(1.254\times
10^{6}\frac{kcal}{acre}\right)$ $\displaystyle=$ $\displaystyle 7.41\times
10^{6}acres$ $\displaystyle\approx$ $\displaystyle 30,000km^{2}$

Summed, $49,700km^{2}$, these two areas devoted to oats and potatoes are
roughly equivalent to the amount of arable land estimated above for Ireland,
$54,000km^{2}$ [37]. What do the numbers mean? Did there have to be a famine?
If all of the potato crop failed because of late blight, there would likely
have been enough oats to feed the population a $2000kcal$ ration of oats with
leftover to spare. Like the Holodomor or the Great Leap Forward, the numbers
suggest that large-scale suffering wasn’t a natural disaster, but rather a
human disaster resulting from poor government policy insensitive to the value
of human life.

Figure 6:  The population of Ireland over time, file from Wikipedia [35], data
sources [36]. The humble potato, kale, and milk were part of an amazing
population boom. Note that there were two weather-related “potato” famines in
Ireland, in about 1740 and 1850. government policy response to the famines
could explain the drastic difference in subsequent population following each
of the two famines. The population of Ireland finally re-reached it’s 1851
peak in 2021 [34].

##  5 Conclusion

A class about Energy and Social Policy and the author hasn’t mentioned climate
change, coal, or solar panels even once! What is he thinking?

How many tons of carbon does your car release in a year? How many shiploads of
iron oxide will we have to dump into the ocean for phytoplankton to eat up the
equivalent about of carbon? Every question in a class like this is, to at
least some extent, informed by numerical calculation and it’s pretty arrogant
to assume that “those students” don’t need to (or can’t) do the math. If
you’re going to have success talking about numerical calculations, you might
as well start with examples that everyone can relate to, and everyone eats!
Along the way you might find fascinating historical questions to investigate.

The work was influenced and improved by discussions with Diane Dahle-Koch,
Larry Moore, John Deming, Carl Ferkinhoff, and Sarah Taber.

##  Appendix A Creating the historical kcal/acre figure from USDA data

The United States Department of Agriculture (USDA) provides historical crop
information via the National Agricultureal Statistics Service [38] . Data was
downloaded in spreadsheet csv format and then combined and plotted via a
Python Jupyter notebook.

Each crop has its own bespoke units, for example potatoes are sold by
hundredweight (CWT) but sugar beets are measured by the ton. Every imaginable
agricultural product seems to be tracked in the NASS site, for example Maple
Syrup production is tracked and given in gallons of syrup per tap! Conversion
factors used are summarized in Table 1. Calorie (kcal) density for each crop
was taken from the USDA’s Food Data Central [39]. Within this database, foods
are identified by an FDC ID.

An example calculation (implemented in the Jupyter notebook) follows for Corn.
In 2022 the USDA reported an average production of $172.3$ bushels of corn per
acre of farmland.

$172.3\frac{bu}{acre}\cdot\frac{56lbs~{}corn}{bu}\cdot\frac{453.6~{}grams}{lbs}\cdot\frac{365~{}kcal}{100~{}grams}=15,974,657\frac{kcal}{acre}$
(7)

Obviously the result is only reasonable to two significant figures!

Table 1:  A summary of units and conversions used to create figure 4 from USDA
NASS data. $1cwt$ is a hundred pounds of potatoes. A bushel, $1bu$, is a
volume unit of about 35liters and corresponds to about 60lbs of grain. Calorie
content per 100 gram (mass) of food is taken from the USDA’s “Food Data
Central” database. For context, typical serving sizes are included. It isn’t
clear from any of these resources if lb is pound-force (lbf) or pound-mass
(lbm) and so I am treating them as “grocery store units” where $1lbs\approx
453.6grams$.

  * Crop | per acre unit | production unit | kcals per 100gram | typical portion | FDC ID
---|---|---|---|---|---
Corn | bu/acre | $1bu=56lbs$ | 365 | 1 cup is 166g | 170288
Potatoes | cwt/acre | $1CWT=100lbs$ | 77 | 0.5 cup is 75g | 170026
Soybeans | bu/acre | $1bu=60lbs$ | 446 | 1 cup is 186g | 174270
Sunflowers | lbs/acre |  | 584 | 1 cup is 140g | 170562
Wheat | bu/acre | $1bu=60lbs$ | 327 | 1 cup is 192g | 168890

Raw data from the USDA NASS is plotted in figure 7. The scaling described in
equation 7 produces figure 4 earlier in the paper.

Figure 7:  Average USDA per acre yields for a number of commodity crops over
time. This “raw” data (in bespoke harvest units) was scaled to produce the
data in figure 4 earlier in the paper.

##  Appendix B Estimating land area devoted to chinampas with ImageJ

ImageJ is a free software program developed by the National Institutes of
Health for photo analysis, [30]. I used the program to measure a calibration
scale in a map and I also used the program to measure the area of two polygons
that I drew on the map. The length and both areas are shown in figure 8.

Specifically, to find the area of the two large chinampas areas near
Tenochtitlan, I took a screenshot from the 1964 paper, [24], and saved it in
jpg format. Then, I opened the image in the Windows-Java edition of ImageJ
[30]. The length of the 10 mile distance scale was 213 pixels. The long
chinampas area at the south end of the lake was measured with a Polygon
selection via the Measure tool to have an area of $9940~{}pixel^{2}\approx
21.9miles^{2}$. The smaller region near Chalco had an area of about
$1439~{}pixel^{2}\approx 3.2miles^{2}$. While there were certainly other
regions devoted to chimanpas agriculture, the portion visible near the Aztec
capital seems to be about $25.1miles^{2}$ or $16,000acres$.

Figure 8:  Three screen captures showing chinampa areas and the calibration
stick used to convert pixel-squared area into $miles^{2}$. The image being
analyzed is available in [24].

## References

## References

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  * [34] Carroll R. Ireland’s population passes 5m for first time since C19th famine. The Guardian [Internet]. 2021 Aug 31 [cited 2023 Jan 16]. Available from: https://www.theguardian.com/world/2021/aug/31/ireland-population-surpasses-5m-for-first-time-since-1851
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  * [36] Specific census data sources for the Ireland population plot are given at the “talk” page of [35], specifically https://commons.wikimedia.org/wiki/File_talk:Population_of_Ireland_since_1600.png
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# Unconventional Superfluidity in a model of Fermi-Bose Mixtures

K Sheshadri1<EMAIL_ADDRESS>A Chainani2<EMAIL_ADDRESS>1226, Bagalur,
Bangalore North, Karnataka State, India 562149  2Condensed Matter Physics
Group, National Synchrotron Radiation Research Center, Hsinchu 30076, Taiwan

###### Abstract

A finite-temperature ($T>0$) study of a model of a mixture of spin-zero
hardcore bosons and spinless fermions, with filling fractions $\rho_{B}$ and
$\rho_{F}$, respectively, on a two-dimensional square lattice with composite
hopping $t$ is presented. The composite hopping swaps the locations of a
fermion and a boson that occupy nearest-neighbor sites of the lattice. The
superfluid order parameter $\psi$, the femion hopping amplitude $\phi$, the
chemical potential $\mu$, the free energy minimum $\tilde{F}$ and entropy $S$
are calculated in the limit $\rho_{B}+\rho_{F}=1$ within a mean-field
approximation, and lead to a phase diagram in the $\rho_{F}-T$ plane. This
phase diagram consists of a metallic superfluid phase under a dome-shaped
$T(\rho_{F})$, and insulating normal liquid and insulating normal gas phases
outside the dome. These phases are separated by coupled discontinuous
transitions as indicated by jumps in $\psi$ and $\phi$. The maximum critical
transition temperature $T_{c}$ is observed very close to $\rho_{F}=1/2$. While
$\tilde{F}(T)$ is continuous with a derivative discontinuity at
$T=T_{c}(\rho_{F})$ for $0<\rho_{F}\leq 1/2$ (first-order transition), it
becomes discontinuous for $\rho_{F}>1/2$ (zeroth-order transition), where the
entropy becomes negative for a range of temperatures below $T_{c}$. The ratio
of $T_{c}$ to Fermi band width agrees remarkably with the ratio of
$T_{c}$/$T_{F}$ (where $T_{F}$ is the Fermi temperature) of unconventional
superfluids and superconductors like Fermi-Bose mixtures, the high-$T_{c}$
cuprates, iron-based and hydride superconductors, that exhibit experimental
values of $T_{c}$ spread over nine orders of magnitude from $\sim 200$nK to
$\sim 260$K.

##  I Introduction

Fermi-Bose mixtures (FBMs) constitute an unusual and important state of
matter, including well known examples like He3-He4 mixtures Ebner , the mixed
phase of type-II superconductors, ultracold atom systemsTruscott ; Schreck ;
Hadzibabic , unconventional superconductors which exhibit Bardeen-Cooper-
Schrieffer to Bose-Einstein condensation (BCS-BEC) crossoverLubashevsky ;
Okazaki ; Kasahara ; Rinott , and so on. Experimental and theoretical studies
of FBMs have shown remarkable results, particularly in terms of the BCS-BEC
crossover across a Feshbach resonance Regal , that have revealed their
distinct aspects compared to the limiting cases of BCS superconductivity and
BEC superfluidity Randeria ; Ketterle .

The BCS-BEC crossover was originally predicted to occur for excitons in
semiconductors Keldysh  and quarks in high energy physicsKerbikov . However,
it was first reported experimentally in ultracold fermionic atoms with s-wave
interactions Bartenstein . Unusual and unexpected results include formation of
a Feshbach moleculeCumby , and the role of three body physicsBloom  in FBMs.
On the other hand, the role of BCS-BEC crossover in condensed matter involves
experimental results on iron-based superconductorsLubashevsky ; Okazaki ;
Kasahara ; Rinott  and its relation to well-accepted theoretical
resultsRanderia ; Randeria2 ; quick . While interactions between fermions
mediated by phonons define the BCS theory of superconductivity, several
studies have also considered their importance in mixtures of ultracold atoms
Bijlsma ; Viverit ; Capuzzi ; Albus . The Boson-Fermion (BF) model Schafroth ,
which preceded the BCS theory, discusses itinerant fermions hybridizing with
bosons composed of bound pairs of fermions of opposite spins. The BF model was
subsequently used to study electrons interacting with local lattice
deformations Ranninger  as well as high temperature superconductivity
Ranninger2 ; Friedberg ; Gershkenbein ; Domanski . Recent studies have applied
it for describing resonance superfluids in the BCS-BEC crossover regime Shin ,
as well as a temperature driven crossover in an FBM Maska2 . These studies
have shown the importance and interplay of bosonic and fermionic degrees of
freedom in various physical systems.

Early studies on mixtures investigated the role of an attractive interaction
between fermions and bosons. It was shown that an FBM with attractive
interactions undergoes a collapse when the fermion number exceeded a critical
valueUfrecht . The breakthrough in controlling a Feshbach resonance in FBMs
allowed researchers to effectively tune the boson-fermion interaction and
control the system from collapsing at high densities Ospelkaus . In contrast,
theoretical studies employing repulsive interactions could describe various
stable density configurations of FBMs. The role of finite temperatures and
going beyond the mean-field approximation was also investigated Viverit . In
the case of a strongly repulsive quasi one-dimensional FBM,Guan  it was shown
that the phase diagram as a function of applied magnetic field $H$ displays a
pure boson phase for $H=0$, polarized fermions and bosons coexisting for
$0<H<H_{c}$, and a fully polarized fermion phase for $H>H_{c}$. More
interestingly, for an FBM on a 2D optical lattice in the framework of an
extended single band Hubbard model with Coulomb interaction terms between
bosons ($U_{BB}$), between fermions ($U_{FF}$) and an additional Coulomb
interaction between bosons and fermions ($U_{BF}$), it was shown that the
bosons can mediate an attractive interaction between fermions, leading to
fermion paired states with different $s,~{}p$ and $d$ orbital symmetries Wang
. Further, the phase diagram as a function of $U_{BF}$ versus fermion number
also revealed the existence of spin density wave and charge density wave
phases. The authors also predicted that for experimentally accessible regime
of parameters, the 2D FBM would exhibit superfluidity with an unconventional
fermion pairing having a transition temperature around a percent of the Fermi
energy. On the other hand, the role of interaction-dependent temperature
effects in an FBM were investigated by CramerCramer . It was shown that
adiabatic temperature changes of the FBM occur which depend on the interaction
between fermions and bosonsCramer .

In addition, the dynamics of FBMs has also been investigated and it was shown
that long range density wave phases can be obtained for fermions and bosons
hopping independently in the presence of on-site boson-boson $U_{BB}$ and
boson-fermion $U_{BF}$ Coulomb interactions Lewenstein ; Pollet . However, a
composite hopping that exchanges a fermion with a boson, when they occupy
neighboring sites, was not considered in earlier work. This form of hopping
was proposed recently by us zeroTarxiv ; zeroTPRO  and distinguishes our work
from earlier work on FBMs. In this work, we calculate the thermodynamic
properties of a model of FBM on a two-dimensional square lattice with
composite hopping between neighboring spinless fermions and hardcore bosons,
extending our earlier study of $T=0$ properties zeroTarxiv ; zeroTPRO . As in
the previous work, we use a mean-field approximation and restrict ourselves to
the case

$\rho_{F}+\rho_{B}=1,$ (1)

where $\rho_{F}$ and $\rho_{B}$ are the filling fractions of the fermions and
bosons, respectively. To recall, at $T=0$, the model displays two distinct
phases separated by coupled first-order transitions at Fermi filling fraction
$\rho_{F}\simeq 0.3$: for $\rho_{F}<0.3$ the Fermi sector is insulating and
the Bose sector is a normal liquid, while for $\rho_{F}>0.3$ the Fermi sector
is metallic and the Bose sector is a superfluid. In the present work, we find
that thermal fluctuations suppress superfluidity, and at a certain
$T=T_{c}(\rho_{F})$ there is a discontinuous transition to an insulating non-
superfluid phase as shown by the superfluid amplitude $\psi(T)$ and the
fermion hopping amplitude $\phi(T)$. We further find that the transition
occurring at $T_{c}(\rho_{F})$ is first order for $0.3<\rho_{F}\leq 1/2$ (the
minimum free energy $\tilde{F}(T)$ is continuous with a discontinuity in its
first derivative), but is zeroth order for $1/2<\rho_{F}<1$ (the minimum free
energy $\tilde{F}(T)$ is discontinuous). In the latter regime, the entropy
becomes negative for a range of temperatures below $T_{c}$. We compute the
ratio of $T_{c}$ to the Fermi band width, and find remarkable agreement with
measured values in the range of $0.02$ to $0.20$ for a wide variety of
unconventional superfluids and superconductors, including Fermi-Bose mixtures,
the high-$T_{c}$ cuprates, iron-based superconductors and hydrides, that have
their $T_{c}$ spread over nine orders of magnitude from a few hundred
nanokelvins to a few hundred kelvinszeroTPRO . Our estimate for the
superconducting $T_{c}$ in the solid-state context with known experimental
band widths or the Fermi temperature are consistent with observed $T_{c}$’s of
the cuprates as well as iron-based superconductors.

##  II The Composite-Hopping Model and Its Mean-Field Thermodynamics

We consider the composite-hopping model with Hamiltonian

$\displaystyle H$ $\displaystyle=$
$\displaystyle-\alpha\sum_{i}\left[b_{i}^{\dagger}b_{i}+f_{i}^{\dagger}f_{i}-1\right]-\mu\sum_{i}f_{i}^{\dagger}f_{i}$
(2) $\displaystyle-t\sum_{<ij>}f_{i}^{\dagger}f_{j}b_{j}^{\dagger}b_{i}$

that was proposed in a recent study for $T=0$zeroTarxiv ; zeroTPRO , where the
notation used above is also explained. In this work also, we consider a FBM on
a two-dimensional square lattice. The composite hopping term, the last term
above, results in swapping of a hardcore boson and a spinless fermion when
they occupy nearest-neighbor sites. Using the mean-field approximation, this
term is transformed according to

$\displaystyle f_{i}^{\dagger}f_{j}b_{j}^{\dagger}b_{i}$ $\displaystyle\simeq$
$\displaystyle\langle f_{i}^{\dagger}f_{j}\rangle~{}(\langle
b_{j}^{\dagger}\rangle b_{i}+\langle b_{i}\rangle b_{j}^{\dagger}-\langle
b_{j}^{\dagger}\rangle\langle b_{i}\rangle)$ (3) $\displaystyle+\langle
b_{j}^{\dagger}\rangle\langle b_{i}\rangle f_{i}^{\dagger}f_{j}-\langle
f_{i}^{\dagger}f_{j}\rangle\langle b_{j}^{\dagger}\rangle\langle
b_{i}\rangle,$

so $H$ is approximated by a mean-field Hamiltonian

$\displaystyle H^{MF}$ $\displaystyle=$ $\displaystyle
H_{0}+H_{1}+H_{2},~{}~{}\mathrm{where}$ $\displaystyle H_{0}$ $\displaystyle=$
$\displaystyle N(2\phi\psi^{2}+\alpha),$ $\displaystyle H_{1}$
$\displaystyle=$
$\displaystyle-(\alpha+\mu)\sum_{i}f_{i}^{\dagger}f_{i}-\frac{1}{z}\psi^{2}\sum_{<ij>}f_{i}^{\dagger}f_{j},~{}\mathrm{and}$
$\displaystyle H_{2}$ $\displaystyle=$ $\displaystyle\sum_{i}\left[-\alpha
b_{i}^{\dagger}b_{i}-\phi\psi(b_{i}+b_{i}^{\dagger})\right].$ (4)

We have taken $zt=1$ ($z$ is the coordination number of the lattice), and
introduced the thermodynamic expectation values

$\phi=\langle f_{i}^{\dagger}f_{j}\rangle,~{}~{}\psi=\langle
b_{i}\rangle=\langle b_{j}^{\dagger}\rangle.$ (5)

We assume $\phi~{}\mathrm{and}~{}\psi$ to be real and homogeneous and consider
$\psi$ to be the superfluid order parameter FisherWeichman89 ; shesh93 ;
shesh95 ; gutz ; spin1_1 ; spin1_2 ; spin1_3 . For the hardcore bosons, we use
the single-site boson occupation number basis $\\{|0\rangle,~{}|1\rangle\\}$
for diagonalizing the $2\times 2$ matrix $h_{2}$ of $H_{2}/N$, i.e.,

$h_{2}=\begin{bmatrix}0&-\phi\psi\\\ -\phi\psi&-\alpha\end{bmatrix}$ (6)

that has the eigenvalues

$\lambda_{\pm}=\frac{1}{2}\left[-\alpha\pm
R\right],~{}~{}\mathrm{where}~{}~{}R=\sqrt{\alpha^{2}+4\phi^{2}\psi^{2}}.$ (7)

Using the Fourier transform

$f_{i}=\frac{1}{\sqrt{N}}\sum_{\bf k}e^{i{\bf k.r_{i}}}f_{\bf k},$ (8)

the Hamiltonian $H_{1}$ of the fermion sector becomes

$H_{1}=\sum_{\bf k}(\varepsilon_{\bf k}-\mu)f_{\bf k}^{\dagger}f_{\bf k},$ (9)

where

$\displaystyle\varepsilon_{\bf k}$ $\displaystyle=$
$\displaystyle-\alpha-\psi^{2}\gamma_{\bf k},~{}~{}\mathrm{and}$
$\displaystyle\gamma_{\bf k}$ $\displaystyle=$ $\displaystyle\frac{2}{z}(\cos
k_{x}+\cos k_{y}).$ (10)

The free energy per lattice site $F=F(\alpha,\psi)$ is now

$\displaystyle F$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\alpha-R)+2\phi\psi^{2}-T\ln(1+e^{-R/T})$ (11)
$\displaystyle-T\frac{1}{N}\sum_{k}\ln\left[1+e^{(\mu-\varepsilon_{\bf
k})/T}\right].$

We take $k_{B}=1$ here and in the following. To calculate
$\phi=\phi(\alpha,\psi)$, we use its definition in (5) and go over to
$k$-space to get

$\phi=\frac{1}{Nz}\sum_{<ij>}\langle
f_{i}^{\dagger}f_{j}\rangle=\frac{1}{N}\sum_{\bf k}\gamma_{\bf k}\langle
f_{\bf k}^{\dagger}f_{\bf k}\rangle.$ (12)

By definition,

$\langle f_{\bf k}^{\dagger}f_{\bf k}\rangle=\frac{\mathrm{Tr}(f_{\bf
k}^{\dagger}f_{\bf
k}e^{-H_{1}/T})}{\mathrm{Tr}(e^{-H_{1}/T})}=\frac{1}{1+e^{(\varepsilon_{\bf
k}-\mu)/T}},$ (13)

and so

$\phi=\frac{1}{N}\sum_{\bf k}\frac{\gamma_{\bf k}}{1+e^{(\varepsilon_{\bf
k}-\mu)/T}}.$ (14)

Using $\gamma_{\bf k}=-(\alpha+\varepsilon_{\bf k})/\psi^{2}$ (the first
equation in (II)) we obtain

$\phi=-\frac{1}{N}\sum_{\bf k}\frac{\alpha+\varepsilon_{\bf
k}}{\psi^{2}}\frac{1}{1+e^{(\varepsilon_{\bf k}-\mu)/T}}.$ (15)

Introducing the density of states

$\rho(E)=\frac{1}{N}\sum_{\bf k}\delta(E-\varepsilon_{\bf k}),$ (16)

we can write

$\phi=-\frac{1}{\psi^{2}}\int_{E_{0}}^{\mu}dE~{}\frac{\alpha+E}{1+e^{(E-\mu)/T}}\rho(E),$
(17)

where $\mu$ is chosen such that the fermion filling fraction

$\rho_{F}(\alpha,\psi)=\int_{E_{0}}^{\mu}dE\frac{1}{1+e^{(E-\mu)/T}}\rho(E)$
(18)

has a desired value. Here, $E_{0}=-\alpha-\psi^{2}$ is the minimum value of
fermion energy. To calculate the density of states (16), we convert the
$k$-sum in to an integral according to $(1/N)\sum_{\bf k}\to(1/4\pi^{2})\int
d{\bf k}$. Since $\varepsilon_{\bf-k}=\varepsilon_{\bf k}$, the $k$-space
integral is four times the integral over the first quadrant of the Brillouin
zone, and so we have

$\rho(E)=\frac{1}{\pi^{2}}\int_{0}^{\pi}dk_{x}\int_{0}^{\pi}dk_{y}~{}\delta(E+\alpha+\psi^{2}\gamma_{\bf
k}).$ (19)

The integral over $k_{y}$ can be easily evaluated, and we get

$\displaystyle\rho(E)$ $\displaystyle=$
$\displaystyle\frac{2}{\pi^{2}\psi^{2}}f\left(\frac{\alpha+E}{\psi^{2}}\right),~{}\mathrm{where}$
$\displaystyle f(u)$ $\displaystyle=$
$\displaystyle\int_{0}^{\pi}\frac{dk_{x}}{\sqrt{1-(2u+\cos k_{x})^{2}}}.$ (20)

We can readily see that the function $f(u)$ is real only when $-1\leq u\leq
1$, and is non-negative. Therefore we have the inequality
$-\alpha-\psi^{2}\leq E\leq-\alpha+\psi^{2}$ for the fermion energy $E$. We
substitute the above expression for $\rho(E)$ in to equations (17) and (18)
and transform the integrals to obtain

$\displaystyle\rho_{F}$ $\displaystyle=$
$\displaystyle\frac{2}{\pi^{2}}\int_{-1}^{u_{F}}du\frac{f(u)}{1+e^{(u-u_{F})\psi^{2}/T}}~{}\mathrm{and}$
$\displaystyle\phi$ $\displaystyle=$
$\displaystyle-\frac{2}{\pi^{2}}\int_{-1}^{u_{F}}du\frac{uf(u)}{1+e^{(u-u_{F})\psi^{2}/T}},$
(21)

where $u_{F}=(\alpha+\mu)/\psi^{2}$. This helps us choose the value of $\mu$
for a desired $\rho_{F}$, given the values of $(\alpha,\psi)$. For any
$(\rho_{F},T)$, we determine $\psi$ and $\alpha$ by solving $\partial
F/\partial\psi=0~{}\mathrm{and}~{}\partial F/\partial\alpha=0$ simultaneously.
These two equations, and the two equations in (II) above, are solved
iteratively to obtain $(\alpha,\psi)$ as well as $(\phi,\mu)$ for any chosen
$(\rho_{F},T)$. In general, there could be multiple solutions $(\alpha,\psi)$.
We substitute each solution in $F(\alpha,\psi)$ and denote the resulting free
energy minimum for the solution by $\tilde{F}$; the correct solution is the
one that corresponds to the lowest $\tilde{F}$. From equation (11), we obtain

$\displaystyle\frac{\partial F}{\partial\psi}$ $\displaystyle=$ $\displaystyle
2\psi(\phi+\phi_{\psi}\psi)\left[1-\frac{\phi}{R}\chi\right],~{}\mathrm{and}$
$\displaystyle\frac{\partial F}{\partial\alpha}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[(1-2\rho_{F})-\frac{\chi}{R}\alpha\right]+2\phi_{\alpha}\psi^{2}\left[1-\frac{\chi}{R}\phi\right],$

where
$\phi_{\psi}=\partial\phi/\partial\psi,~{}\phi_{\alpha}=\partial\phi/\partial\alpha$,
and

$\chi(R,T)=\frac{e^{R/T}-1}{e^{R/T}+1}.$ (23)

Since $\phi~{}\mathrm{and}~{}\phi_{\psi}$ are positive (see equations (14) and
(II)), we always have $\phi+\phi_{\psi}\psi>0$, so $\partial F/\partial\psi=0$
gives

$\psi=0~{}~{}\mathrm{or}~{}~{}R=\phi\chi(R,T).$ (24)

Using this in equation (II), $\partial F/\partial\alpha=0$ gives

$\alpha=(1-2\rho_{F})R/\chi.$ (25)

##  III Tractable Limits and Some General Observations

Now that we have derived the implicit equations for $\psi$ and $\alpha$, we
first analyze two simple limits: the $\psi=0$ (disordered phase) and the $T=0$
limits. We obtain closed-form solutions for $\alpha,~{}\psi$ and $\tilde{F}$
in these limits. For the general case, we make some observations before moving
on to a discussion of the numerical results in the next section.

From equation (25), we get
$\alpha^{2}[\chi^{2}-(1-2\rho_{F})^{2}]=4\phi^{2}\psi^{2}(1-2\rho_{F})^{2}$.
The solution $\psi=0$ corresponds to the disordered state, i.e., the Bose
sector is in a non-superfluid, normal phase. We then get $\alpha=0$ or
$\chi=|1-2\rho_{F}|$. In this state, $R=|\alpha|$, so using the definition of
$\chi$ in equation (23) we get

$\alpha=0,~{}T[\ln(1-\rho_{F})-\ln\rho_{F}]$ (26)

in the disordered state ($\psi=0$). Of these two solutions for $\alpha$, we
must pick the one corresponding to the lower $\tilde{F}$. Substituting the
above in equation (11) for $F$, we obtain $\tilde{F}_{I}=-T(2\ln 2)$ for the
first solution $\alpha=0$, and $\tilde{F}_{II}=-T[\ln 2-\ln(1-\rho_{F})]$ for
the second solution $\alpha=T[\ln(1-\rho_{F})-\ln\rho_{F}]$. We can see that
$\tilde{F}_{I}<\tilde{F}_{II}$ for $\rho_{F}<1/2$ and
$\tilde{F}_{II}<\tilde{F}_{I}$ for $\rho_{F}>1/2$. This shows that in the
disordered phase, $\alpha=0$ for $\rho_{F}\leq 1/2$ and
$\alpha=T[\ln(1-\rho_{F})-\ln\rho_{F}]$ (which is the same as
$\chi=|1-2\rho_{F}|$) for $\rho_{F}>1/2$. In Fig. (1) we show the behavior of
$-\tilde{F}_{D}/T$ (which is the disordered-phase entropy $S_{D}$) as a
function of $\rho_{F}$.

At $T=0$, we get $\tilde{F}=0$ for the disordered state (since both
$\tilde{F}_{I}$ and $\tilde{F}_{II}$ vanish). For the ordered state, $\chi=1$
so that

$\psi_{0}^{2}=\rho_{F}(1-\rho_{F}),~{}~{}\alpha_{0}=(1-2\rho_{F})\phi_{0},$
(27)

and the minimum free energy is
$\tilde{F}_{0}=\rho_{F}^{2}[-\phi_{0}+u_{F}(\rho_{F}-1)]$ by taking the
$T\rightarrow 0$ limit in equation (11) and substituting the above expressions
for $\psi_{0},~{}\alpha_{0}$. Here $\psi_{0},~{}\phi_{0},~{}\alpha_{0}$ and
$\tilde{F}_{0}$ are $T=0$ values. This gives coupled first-order transitions
at $\rho_{F}\simeq 0.3$ at $T=0$ between a metallic superfluid for
$\rho_{F}>0.3$ and an insulating normal liquid for $\rho_{F}\leq 0.3$. We
referred to the latter as insulating normal gas in our earlier workzeroTarxiv
; zeroTPRO . However, since the destruction of superfluidity of the Bose
sector in this regime is due to an interplay between correlation and quantum
effects and not due to temperature, this unusual phase should be more
appropriately called an insulating normal liquid rather than an insulating
normal gas.

As we increase the temperature at a fixed $\rho_{F}$, thermal fluctuations
suppress superfluidity, reducing $\psi$. To see how this happens, we rewrite
the ordered-phase self consistency equation $R=\phi\chi$ in the form
$T=J(\psi)$ where $J(\psi)=\phi x/\ln[(1+x)/(1-x)]$; here
$x=\sqrt{(1-2\rho_{F})^{2}+4\psi^{2}}$. The function $J(\psi)$ has zeros at
$\psi=0$ (when $\phi=0$) and $\psi=\psi_{0}$ (when $x=1$). Since it is
positive, it must have a maximum in the interval $(0,\psi_{0})$. This is
graphically illustrated in Fig. (2). As $T$ increases from zero, the line
$y=T$ intersects the curve $y=J(\psi)$ at two points $\psi=\psi_{1},\psi_{2}$.
Since $J(\psi)$ is a single-valued function, we have
$\psi_{2}<\psi_{1}<\psi_{0}$ and further, $\psi_{1}$ decreases and $\psi_{2}$
increases as $T$ is increased. The solution $\psi=\psi_{1}$ corresponds to the
ordered minimum of the free energy. As the temperature increases further,
$y=T$ increases, while the maximum of $J(\psi)$ decreases, and at a certain
temperature $T_{2}(\rho_{F})$, the line $y=T$ becomes a tangent to $y=J(\psi)$
at the maximum, when $\psi_{1}=\psi_{2}$. For $T>T_{2}(\rho_{F})$, the self
consistency equation $T=J(\psi)$ has no solution, and therefore $T_{c}\leq
T_{2}$.

For $T\leq T_{2}$, when the ordered solution exists, there are two
possibilities. (1) The ordered free energy minimum ($\tilde{F}_{O}$) remains
lower than the disordered free energy minimum ($\tilde{F}_{D}$) for all $T\leq
T_{2}$. In this case the transition temperature is $T_{c}=T_{2}$, and occurs
when the ordered minimum ceases to exist. The free energy minimum value is
discontinuous at the transition, since the system switches to the only
solution that exists, namely $\psi=0$. This corresponds to a zeroth-order
transition. A zeroth-order phase transition has previously been considered in
the theory of superfluidity and superconductivity Maslov2004 . More recently,
the reentrant phase transition in black holes has been discussed as a zeroth-
order transition.Gunasekaran ; Altamirano01 ; Zou ; Hennigar ; Altamirano02 ;
Amin .

Our numerical computations obtain this scenario for $\rho_{F}>1/2$. (2) There
exists a certain temperature $T_{1}<T_{2}$ such that
$\tilde{F}_{O}<\tilde{F}_{D}$ for $T<T_{1}$, and $\tilde{F}_{O}>\tilde{F}_{D}$
for $T>T_{1}$, with the two phases coexisting at $T_{c}=T_{1}$, which is a
point of first-order transition. Our numerical results obtain this result for
$\rho_{F}\leq 1/2$.

We can show that the transition is indeed zeroth order for $\rho_{F}>1/2$
based on the disordered-phase behavior of $\alpha$ that we derived above.
Formally, the solution of the self-consistency equation $R=\phi\chi$ is
$\psi^{2}=(1/4)[\chi^{2}-(1-2\rho_{F})^{2}]$. As we saw above,
$\chi=|1-2\rho_{F}|$ in the disordered phase when $\rho_{F}>1/2$, so the
ordered solution $\psi>0$ does not exist in the disordred phase. If
$T_{c}<T_{2}$, then we would have the ordered solution existing in the
disordered phase for $T_{c}<T<T_{2}$, which is a contradiction. So we must
have $T_{c}=T_{2}$ in this case, resulting in a zeroth-order transition.

For $\rho_{F}\leq 1/2$, however, the ordered-phase self-consistency equation
in the disordered phase (where $\alpha=0$) becomes
$2\psi=(e^{2\phi\psi/T}-1)/(e^{2\phi\psi/T}+1)$, that has a finite $\psi$
solution for $T<T_{2}$. If $T_{c}=T_{2}$, then we would have an ordered
solution existing in the disordered phase for $T>T_{c}=T_{2}$, a
contradiction. We therefore have $T_{c}<T_{2}$ in this case. The transition
criterion in this case is clearly $\tilde{F}_{O}=\tilde{F}_{D}$ since the
ordered and disordered minima both exist at the transition. The transition for
$\rho_{F}\leq 1/2$ is therefore first order.

We now turn to the correlation function $C({\bf
r})=(1/N)\sum_{i}\langle\delta\rho_{{\bf r}_{i}}\delta\rho_{{\bf r}_{i}+{\bf
r}}\rangle$ where $\delta\rho_{{\bf r}_{i}}=f_{i}^{\dagger}f_{i}-\rho_{F}$. We
obtain $C({\bf r})=-n^{2}({\bf r})$ where $n({\bf r})=(1/N)\sum_{\bf k}\langle
f_{\bf k}^{\dagger}f_{\bf k}\rangle e^{i{\bf k}.{\bf r}}$, and $C_{\bf
q}=-(1/N)\sum_{\bf k}\langle f_{\bf k}^{\dagger}f_{\bf k}\rangle\langle f_{\bf
k+q}^{\dagger}f_{\bf k+q}\rangle$ for the Fourier transform of $C({\bf
r})$zeroTarxiv ; zeroTPRO . In our earlier work, we could show that $n({\bf
r})$ shows a periodicity of twice the lattice spacing for $\rho_{F}=1/2$,
while no apparent periodicity was obtained for other values of $\rho_{F}$.
Using the expression for the Fermi function in Eq. (13), we can readily see
that in the disordered phase ($\psi=0$), we obtain $C_{\bf q}=-1/4$. This
shows that there is no density wave (DW) order in the insulating normal gas
phase, and the DW order obtained for the $T=0$ metallic superfluidzeroTarxiv ;
zeroTPRO  is stabilized by the nesting of the Fermi surface and the
superfluidity of the Bose sector.

##  IV Discussion

Figure 1: The disordered-phase entropy $S_{D}=-\tilde{F}_{D}/T$ (equal to
$-\tilde{F}_{I}/T$ for $\rho_{F}\leq 1/2$ and $-\tilde{F}_{II}/T$ for
$\rho_{F}>1/2$) plotted as a function of $\rho_{F}$. The high-temperature
entropy is $2\ln 2$, as expected, for $0<\rho_{F}\leq 1/2$. However, it is
anomalously high for $1/2<\rho_{F}\leq 1$, the regime of zeroth-order
transition. Figure 2: A figure illustrating the graphical solution of the
self-consistency equation $T=J(\psi)$. The function $J(\psi)$ vanishes at
$\psi=0,\psi_{0}$ and has a maximum in between. The horizontal lines show
plots of $y=T$ (for $0<T<T_{2},~{}T=T_{2},~{}T>T_{2}$). For $0<T<T_{2}$, the
line $y=T$ intersects the curve $y=J(\psi)$ (red curve) at two points
$\psi=\psi_{1},\psi_{2}$, with $\psi_{0}\geq\psi_{1}>\psi_{2}$, that
correspond, respectively, to a minimum and a maximum of $F$. For $y=T_{2}$,
the two points merge and the line is a tangent to the curve $J(\psi)$ at its
maximum. For $T>T_{2}$, the self-consistency equation has no solution for any
real value of $\psi$. Figure 3: Panels (a)-(f) show the $T$-dependence plots,
respectively, of $\psi,~{}\phi,~{}\mu,~{}\tilde{F},~{}\alpha$ and $S$. Each
panel has plots at five different values of $\rho_{F}$, namely
$0.35,1/2,0.70,0.81,0.95$. Figure 4: The phase diagram of model (2) in
$(\rho_{F},T)$ plane showing the metallic superfluid ($\phi,~{}\psi>0$) and
insulating non-superfluid phases ($\phi=\psi=0$) phases (the insulating normal
liquid at $T=0$ and the insulating normal gas for $T>0$). These phases are
separated by lines of discontinuous transitions: at the phase boundary, the
minimum free energy $\tilde{F}$ is continuous with a derivative discontinuity
for $\rho\leq 1/2$ (first-order transition, dashed line), while $\tilde{F}$
has a jump for $\rho>1/2$ (zeroth-order transition, dotted line). Figure 5:
The temperatures $T_{1}$ and $T_{2}$ are plotted as functions of $\rho_{F}$.
As explained in the text, $T_{1}$ is the critical temperature for first-order
transition, where ${\tilde{F}}_{O}={\tilde{F}}_{D}$. It is not defined for
$\rho_{F}>1/2$, where ${\tilde{F}}_{O}<{\tilde{F}}_{D}$ for $T\leq T_{2}$. The
ordered minimum disappears at $T_{2}$. The critical temperature $T_{c}$ is
$T_{1}$ for $\rho_{F}\leq 1/2$ (first-order transition) and $T_{2}$ for
$\rho_{F}>1/2$ (zeroth-order transition). Note that $T_{2}>T_{1}$. This is
responsible for a segment of the phase boundary in Fig. 4 being vertical at
$\rho_{F}=1/2$.  Figure 6:  Plot of $\psi_{c}^{2}$, where $\psi_{c}$ is the
jump in the superfluid order parameter at the discontinuous transition
temperature $T_{c}$, as a function of $\rho_{F}$.  Figure 7:  Plot of
$T_{c}/B_{0}$ as a function of $\rho_{F}$. Here, $B_{0}=2\psi_{0}^{2}$ is the
Fermi band width at $T=0$zeroTarxiv ; zeroTPRO . Figure 8:  Plot of
$T_{c}/B_{c}$ as a function of $\rho_{F}$. Here, $B_{c}=2\psi_{c}^{2}$ is the
$T=T_{c}$ analogue of the $T=0$ Fermi band widths $B_{0}$.

Our numerical results are presented in figures 1-8. In Fig.1 we show the
behavior of disordered-phase entropy as a function of $\rho_{F}$. The entropy
in this case is $S_{D}=-\tilde{F}_{D}/T$, where $\tilde{F}_{D}$ is the
disordered-phase free energy minimum. As we discussed above, this is
$-\tilde{F}_{I}/T$ for $\rho_{F}\leq 1/2$ and $-\tilde{F}_{II}/T$ for
$\rho_{F}>1/2$. We can see that the entropy is $2\ln 2$ for $\rho_{F}\leq 1/2$
as one might expect in the disordered phase. However, the entropy is
anomalously large for $\rho_{F}>1/2$, which is also the regime where the
temperature-driven transition is zeroth order. We believe this is a
consequence of the filling constraint (1), that leads to the solution in
equation (25) for $\alpha$ so that $\alpha=T[\ln(1-\rho_{F})-\ln\rho_{F}]$ in
this case (see equation (26)).

To understand how temperature suppresses the metallic-superfluid order, we
solve the self-consistency equations (24) and (25) for $\psi,~{}\alpha$ using
an approach graphically described in Fig.2. As shown in the figure, at a
certain temperature $T<T_{2}$, the line $y=T$ intersects the curve $y=J(\psi)$
at two points $\psi_{1},\psi_{2}~{}(\psi_{0}\geq\psi_{1}>\psi_{2})$. The
solution $\psi=\psi_{1}$ corresponds to the free energy minimum at this
temperature. As the temperature increases, it is obvious from the figure that
this solution moves to the left, i.e. decreases, leading to a thermal
suppression of superfluidity. At a certain high temperature $T>T_{2}$, the
line $y=T$ has no intersection with the curve $y=J(\psi)$. Therefore there is
no solution to the self-consistency equation $T=J(\psi)$ at temperatures
higher than a certain $T_{2}$, at which point the line $y=T$ becomes a tangent
to the curve $y=J(\psi)$ at its maximum.

By numerically implementing this graphical method of solution, we can obtain
$\psi,~{}\phi,~{}\mu,~{}\tilde{F},~{}\alpha$ and $S$ (the entropy) as the
temperature $T$ is varied; these results are plotted in panels (a)-(f) of
Fig.3. Each panel shows temperature dependence of one of these quantities for
five different values of $\rho_{F}=0.35,~{}1/2,~{}0.70,~{}0.81,~{}0.95$. This
choice of $\rho_{F}$ values is the same as for the $T=0$ case, which span the
superfluid metallic phase as reported earlier zeroTarxiv ; zeroTPRO . The
figures show that at each of these fillings, there is a certain critical
temperature $T_{c}$ where the model has a discrete phase transition: the
quantities $\psi,\phi,\mu$ and $\alpha$ all show discontinuous changes at
$T_{c}$ (figures 3(a, b, c, e)). We can observe that at $\rho_{F}=0.35$ and
$1/2$, the free energy minimum $~{}\tilde{F}$ is continuous, but with a
derivative discontinuity, whereas it is discontinuous at
$\rho_{F}=0.70,~{}0.81$ and $0.95$ (Fig. 3(d)).

In figure 3(f) we show plots of entropy $S(T)=-\partial\tilde{F}/\partial T$,
computed by numerical differentiation of $\tilde{F}(T)$. The unusual feature
that can be readily seen is that the entropy becomes negative for certain
temperatures below $T_{c}$ when $\rho_{F}>1/2$. While the concept of negative
entropy has been applied to quantum information systems earlier cerf ;
delrio2011 , it’s relevance for physical systems has been discussed only
recently Chatzi2020 . Cerf and Adami showed that unlike in classical
information theory, quantum conditional entropies can be negative for quantum
entangled systemscerf . Subsequently, del Rio et al. explained its
thermodynamic meaning: negative entropy is related to a possible cooling of an
environment connected to a quantum information system, when quantum
information contained in the system is eraseddelrio2011 . In a very recent
studyChatzi2020 , it has been proposed that the results of two independent
inelastic neutron scattering experimentsOlsen ; Callear , which showed an
anomalous scattering from H2 molecules in nanoscale confined geometries, can
be explained on the basis of negative conditional entropy and quantum
thermodynamics.

Our numerical results for $\rho_{F}\leq 1/2$ show that at $T=T_{c}=T_{1}$, the
ordered and disordered phases coexist and we have
$\tilde{F}_{O}=\tilde{F}_{D}$. The ordered minimum survives well into the
disordered phase, and disappears at a temperature $T_{2}>T_{1}$. We have
therefore a standard first-order transition in this case. On the other hand,
for $\rho_{F}>1/2$, the ordered minimum disappears abruptly at a certain
temperature $T_{2}(\rho_{F})$. The system then assumes the only minimum
available to it, namely the disordered minimum. The free energy obviously
shows a discontinuous jump in this case. The transition is therefore zeroth
order. These numerical results are consistent with the qualitative remarks we
made above in Section III.

As a result, we can define $T_{1}$ only for $\rho_{F}\leq 1/2$, where clearly
$T_{c}=T_{1}<T_{2}$. These results are summarized in figures 4 and 5. Figure 4
shows the phase diagram of our model in $\rho_{F}-T$ plane, showing the
metallic superfluid under the dome, and two insulating non-superfluid phases
outside it. We refer to the insulating phase at $T=0$ for $0<\rho_{F}<0.3$ as
an insulating normal liquid and the insulating phase for $T>0$ as an
insulating normal gas. While the latter is dominated by thermal effects, the
former is an unusual phase of bosons at $T=0$ that is a non-superfluid because
of correlation and quantum effects.

Figure 5 shows the plots of $T_{1}(\rho_{F})$ and $T_{2}(\rho_{F})$. The
function $T_{2}(\rho_{F})$ can be defined for $0\leq\rho_{F}\leq 1$, and has a
maximum at a certain $\rho_{F}<1/2$. The temperature $T_{1}$ on the other hand
remains lower than $T_{2}$; it vanishes for $0\leq\rho_{F}\leq 0.3$, in
agreement with the $T=0$ resultszeroTarxiv ; zeroTPRO .

The line separating the two phases in Fig. 4 has a vertical segment at the van
Hove point $\rho_{F}=1/2$. This is as a consequence of two facts: (a) $T_{c}$
is the lower of $T_{1}$ and $T_{2}$, and (b) $T_{1}<T_{2}$, for $\rho_{F}\leq
1/2$.

As we saw in Fig.3, the superfluid order parameter $\psi$ is discontinuous
with a jump $\psi_{c}$ at $T_{c}$. In Fig.6 we show the plot of
$\psi_{c}^{2}(\rho_{F})$. The jump in the order parameter seems to be maximum
around the same $\rho_{F}$ where $T_{2}(\rho_{F})$ is maximum. It shows an
abrupt drop at $\rho_{F}=1/2$, and after a broad maximum, decreases slowly
towards zero at $\rho_{F}=1$.

Based on the zero-temperature Fermi band width of $B_{0}=2\psi_{0}^{2}$, the
ratio $T_{c}/B_{0}$ agreed very well with measured values for several types of
unconventional superfluids and superconductors in our earlier workzeroTarxiv ;
zeroTPRO . This value of $T_{c}\simeq 0.12$ in our earlier work was not based
on a calculation, but only an estimate based on the zero-temperature free
energy minimum for $\rho_{F}=1/2$. However, in our present work we have
performed explicit computation of $T_{c}$ based on $T_{1},~{}T_{2}$
calculations for the whole range of $\rho_{F}$ from $0$ to $1$, as shown in
figures (4) and (5). Figure 7 presents a plot of the ratio $T_{c}/B_{0}$ as a
function of $\rho_{F}$. We note that the calculated value of the ratio for
$\rho_{F}=1/2$ is in almost exact agreement with our earlier
estimatezeroTarxiv ; zeroTPRO . The ratio is between $0.01$ and $0.10$ for
most $\rho_{F}$ values in the range 0.3 to 1.0.

Figure 8 plots the ratio $T_{c}/B_{c}$ as a function of $\rho_{F}$, using
$B_{c}=2\psi_{c}^{2}$, the $T=T_{c}$ analogue of the $T=0$ Fermi band width
$B_{0}$. It is interesting to compare this with experimentally determined
values of the ratio $T_{c}$/$T_{F}$, of the superfluid or superconducting
transition temperature $T_{c}$ to the Fermi temperature $T_{F}$. This ratio
also scales approximately with the ratio $\Delta$/$E_{F}$, the pairing
strength of single-band superconductors, where $\Delta$ is the superconducting
gap and $E_{F}$ is the Fermi energy. In particular, it has been
recognizedUemura  that the unconventional superconductors show a relatively
large value of $T_{c}$/$T_{F}\sim 0.02-0.2$, while the conventional
superconductors show a small value of $T_{c}$/$T_{F}$ $\sim$ $10^{-4}$ to
$10^{-5}$. For example, elemental metal tin (Sn)Ashcroft  has a
$T_{c}$/$T_{F}$ $\sim$ 3$\times$$10^{-5}$, while the
high-$T_{c}$-cupratesYamamoto ; Brookes ; Xie  exhibit a $T_{c}$/$T_{F}$
$\sim$ 0.02-0.03. On the other hand, the iron based superconductors (estimated
from experimental dataLubashevsky ; Okazaki ; Kasahara ; Rinott ; MonoFeSe )
and the newly discovered hydride superconductors (as estimated from
experimental $T_{c}$ valuesDrozdov ; Drozdov2 ; Somayazulu  and band structure
calculationsBianconi ; Jarlborg ; Liu ) show $T_{c}$/$T_{F}$ $\sim$ 0.1 - 0.2.
Surprisingly, the corresponding value for a Fermi-Bose mixtureFerrier
estimated from experimental data is $T_{c}$/$T_{F}$ $\simeq 0.19$, although
the $T_{c}$$\sim$ 200 nK. Even for a purely ultracold Fermi atomic systemChin
, $T_{c}$/$T_{F}$ $\simeq 0.2$. Thus, the high-$T_{c}$ cuprates, iron-based
superconductors, hydrides and ultracold atomic systems are clearly classified
as unconventional superconductors/superfluids Millev90 .

A very recent study on an iron-based superconductor has reported evidence for
the first solid-state BEC superconductor, although the observed value of the
ratio $\Delta$/$E_{F}$ ($\simeq 0.12$, corresponding to a $T_{c}$/$T_{F}$
$\simeq 0.025$) is reduced across the BCS-BEC crossover, and the authors
interpret it as most probably arising from interband coupling effectsHashimoto
.

The $T_{c}$ values of these unconventional materials span nine orders of
magnitude: while the ultracold FBMs have $T_{c}\simeq 200$nK, the hydrides
have a $T_{c}\simeq 250$K. But the important point is that the ratio
calculated in our model is remarkably close to these values for several
classes of unconventional superfluids and superconductors. In particular, we
obtain $T_{c}/B_{0}\sim 0.01-0.1~{}(T_{c}/B_{c}\sim 0.01-0.25)$; the value of
this ratio is a very robust feature of our model, and compares well with the
range $0.03-0.22$ for most unconventional materials (see the Table 1 of our
previous paper for detailszeroTPRO ).

An important question is that of a potential experimental realization of the
composite hopping model. Consider a lattice with $N$ sites and $M$ electrons
($2N>M>N$) with nearest-neighbor hopping and on-site repulsion. In this case,
by the Pigeonhole principle, at least one site must have more than one
electron. Taking into account the repulsion between electrons, in the minimum
energy configuration we have $M-N$ local pairs (that can be considered as
hardcore bosons) and $2N-M$ electrons each occupiing a site. When one of the
electrons of a pair hops to a neighboring site with one electron, we have a
realization of composite hopping (see our earlier workzeroTarxiv ; zeroTPRO
for a more detailed explanation). This situation is perhaps also well
described by a one-band fermion Hubbard model above half filling, and the
composite hopping model might offer a good effective description. Our model
approximates the electrons by spinless fermions, but in subsequent work we
plan to treat the spin half case. In this case, the composite hopping strength
$t$ is just the single-electron hopping strength. For a narrow band Hubbard
model, we can take $t\sim 0.1-0.2$eV, and if we further use $T_{c}\simeq
0.06zt$ (corresponding to its peak value in Fig.4 at the van Hove point
$\rho_{F}=1/2$), we obtain a value of $T_{c}\simeq 250-500$K.

If room-temperature superconductivity is theoretically possible within a
composite-hopping framework, then it provides a strong reason to look for
practical examples of it. In the pigeonhole context discussed above, external
pressure might help overcome the repulsion between electrons and keep them
paired. It is interesting to explore if high-pressure room-temperature
superconductors like sulphidesdias2020  are indeed solid-state realizations of
the composite-hopping model, and are superconducting because of such a
‘pigeonhole’ pairing mechanism.

A second possibility might be realization of composite hopping in the context
of ultracold atoms in an optical lattice. Quantum simulation is an exciting
area of researchlewenstein2012 , and given the many unusual physical
properties displayed by our model such as negative compressibility (discussed
in our earlier workzeroTarxiv ; zeroTPRO ), negative entropycerf ; delrio2011
; Chatzi2020 , zeroth-order transitionMaslov2004 ; Gunasekaran ; Altamirano01
; Zou ; Hennigar ; Altamirano02 ; Amin ; kundu2020  and the remarkable
agreement of the calculated ratio of $T_{c}$/$T_{F}$ with a wide range of
unconventional superfluids and superconductors zeroTPRO , it would be
interesting to explore the physics of composite hopping in this perspective.

In our model, there is a clear distinction between the boson-dominated
($\rho_{F}\leq 1/2$) and the fermion-dominated ($\rho_{F}>1/2$) regimes: in
the former, the temperature-driven transition is first-order and the entropy
remains positive throughout, while in the latter regime the transition is
zeroth order and the entropy is negative in the ordered phase close to
$T_{c}$. In the latter regime, the zero-temperature bulk modulus becomes
negative over a range of $\rho_{F}$zeroTPRO .

An important question whether the zeroth order temperature-driven transition
for $\rho_{F}>1/2$ is a consequence of mean-field approximation (3), or an
inherent property of the composite-hopping model. This question assumes
importance given that the model offers an irreducible nontrivial description
of an FBM perhaps not explored before, and the circumstance that zeroth-order
transitions are relatively rare in the solid stateMaslov2004 ; kundu2020 . It
would therefore be of interest to study the model using other numerical
techniques of quantum many-body theory like Quantum Monte-Carlo calculations
(QMC)bhmqmc .

##  V Conclusion

In this work, we have extended the zero-temperature study of the composite
hopping model of FBMs to explore finite-temperature thermodynamics. We
computed the $\rho_{F}-T$ phase diagram and found that the temperature-driven
metallic superfluid phase to insulating normal phase transition is
discontinuous: for $\rho\leq 1/2$, it is first order, while for $\rho>1/2$ it
is zeroth order. We calculated the the temperature dependent superfluid
amplitude $\psi$, the fermion hopping amplitude $\phi$, the Fermi chemical
potential, and free energy within a mean-field approximation. We also computed
the entropy and found that it becomes negative for a certain range of
temperatures below $T_{c}$ when $\rho_{F}>1/2$. The ratios $T_{c}/B_{0}$ and
$T_{c}/B_{c}$ where $B_{0}$ and $B_{c}$ are Fermi band widths at $T=0$ and
$T=T_{c}$, respectively, could also be calculated. The calculated ratios of
$T_{c}$ to Fermi band widths match well with experimental values of
$T_{c}$/$T_{F}$ (where $T_{F}$ is the Fermi temperature) of unconventional
superfluids and superconductors, including Fermi-Bose mixtures, the
high-$T_{c}$ cuprates and iron-based superconductors, that span a $T_{c}$
range of nine orders. The results indicate the important role of composite
hopping in describing the superfluid properties of Fermi-Bose mixtures.

## Acknowledgments

A.C. thanks the Ministry of Science and Technology of the Republic of China,
Taiwan, for financially supporting this research under Contract No. MOST
108-2112-M-213-001- MY3.

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