Statistical Mechanics: Four non-interacting particles are confined in a box

[tanaygupta2000]

Homework Statement

Four non-interacting particles are confined in a box of volume V with only the following energies, where nx, ny, nz are non-zero positive integers defining a quantum state (nx, ny, nz). Find the energy of the system at absolute zero of temperature if
(1.) Any number of particles can be in a quantum state
(2.) Not more than one particle can be in a quantum state

Relevant Equations

E = (π^2)(ℏ^2)/2mV^(2/3) * (nx^2 + ny^2 + nz^2)
= C*(nx^2 + ny^2 + nz^2)

Regarding the first part, I proceeded as:
nx ny nz
4 0 0 => E1 = 16C
0 4 0 => E2 = 16C
0 0 4 => E3 = 16C
3 1 0 => E4 = 10C
3 0 1 => E5 = 10C
0 3 1 => E6 = 10C
1 3 0 => E7 = 10C
0 1 3 => E8 = 10C
1 0 3 => E9 = 10C
2 1 1 => E10 = 6C
1 2 1 => E11 = 6C
1 1 2 => E12 = 6C
Hence total energy is coming out to be 3(16C) + 6(10C) + 3(6C) = 126C

For the second part, I noticed that there is no case having exactly 0 or 1 particles in all of the three quantum states.
So total energy = 0
Am I right in attempting the solution?
Also I don't understand the role of temperature in the question.

 

[TSny]

Homework Statement:: Four non-interacting particles are confined in a box of volume V with only the following energies, where nx, ny, nz are non-zero positive integers defining a quantum state (nx, ny, nz). Find the energy of the system at absolute zero of temperature if
(1.) Any number of particles can be in a quantum state
(2.) Not more than one particle can be in a quantum state
Relevant Equations:: E = (π^2)(ℏ^2)/2mV^(2/3) * (nx^2 + ny^2 + nz^2)
= C*(nx^2 + ny^2 + nz^2)

Regarding the first part, I proceeded as:
nx ny nz
4 0 0 => E1 = 16C
0 4 0 => E2 = 16C
0 0 4 => E3 = 16C
3 1 0 => E4 = 10C
3 0 1 => E5 = 10C
0 3 1 => E6 = 10C
1 3 0 => E7 = 10C
0 1 3 => E8 = 10C
1 0 3 => E9 = 10C
2 1 1 => E10 = 6C
1 2 1 => E11 = 6C
1 1 2 => E12 = 6C
Hence total energy is coming out to be 3(16C) + 6(10C) + 3(6C) = 126C

For the second part, I noticed that there is no case having exactly 0 or 1 particles in all of the three quantum states.
So total energy = 0
Am I right in attempting the solution?
Also I don't understand the role of temperature in the question.

$n_x$, $n_y$, and $n_z$ can be any positive integer. So, $n_x = 1, 2, 3, 4, 5, ... \infty$ and the same for $n_y$ and $n_z$. Note that $0$ is not an allowed value for $n_x$, $n_y$, or $n_z$. Also, all three of $n_x$, $n_y$, and $n_z$ can have the same value.

So, there are an infinite number of quantum states available in the box.

Since the temperature is at absolute zero, any particles in the box will settle into the quantum states that have the least energy subject to the conditions mentioned in cases (1.) and (2.) of the problem statement.

As a warm-up question, suppose there is only one particle in the box. What would be its energy if the temperature is absolute zero?

 

[tanaygupta2000]

So I guess I have to make the table like this:
nx ny nz
2 1 1
1 2 1
1 1 2

If there is only one particle in this box, then we have the following possibilities:
nx ny nz
1 0 0
0 1 0
0 0 1
irrespective of temperature being absolute zero.
But how's this possible since nx, ny, nz cannot have the value '0'?

 

[TSny]

It appears to me that you might be mistaking the values of nx, ny, and nz for the number of particles. But nx, ny, and nz are just quantum numbers that label different quantum states. They do not represent numbers of particles. Suppose you had just one particle in the box. Then the particle could occupy any of the possible quantum states. For example, the particle might be in the quantum state where nx = 7, ny = 2, and nz = 9. The values of nx, ny, and nz can be used to calculate the energy of the particle in that state.

If the temperature is known to be absolute zero, then the particle would occupy the quantum state with the least energy. Which quantum state would this be?

 

[tanaygupta2000]

At absolute zero, single particle would occupy nx = ny = nz = 1.

 

[TSny]

At absolute zero, single particle would occupy nx = ny = nz = 1.

Yes.

 

[tanaygupta2000]

So for 4 particles,
(1.) For the first part,
nx ny nz
4 4 4
(2.) But for the second part, the question requires that
nx ny nz
1 1 1
But how's this possible, considering that the particles cannot divide ?

 

[TSny]

So for 4 particles,
(1.) For the first part,
nx ny nz
4 4 4

What is your reasoning for taking nx = ny = nz = 4?

 

[tanaygupta2000]

Because for each particle, nx = ny = nz = 1
and there are 4 particles
and given that any number of particles can be in a quantum state.

 

[TSny]

Because for each particle, nx = ny = nz = 1

Yes. The 4 particles would be in the same quantum state represented by nx = ny = nz = 1. [This is for part (1) of the question.]

What is the value of the energy associated with the state nx = ny = nz = 1?

 

[tanaygupta2000]

For nx = ny = nz = 1, E = 3C
So for 4 particles, Total energy = 12C
Is that right?

 

[TSny]

For nx = ny = nz = 1, E = 3C
So for 4 particles, Total energy = 12C
Is that right?

Yes. Good. Note how you never had to deal with the state nx = ny = nz = 4.

 

[tanaygupta2000]

So for the second part of the question, should I do 4×[(1)² + (2)² + (3)²]C
which equals 56C ?

 

[TSny]

So for the second part of the question, should I do 4×[(1)² + (2)² + (3)²]C
which equals 56C ?

No. You want to get the lowest possible total energy with no more than 1 particle in a state.

 

[tanaygupta2000]

So I think this time n_x = n_y = n_z = 1
giving E = 3C
I am confused in this part

 

[tanaygupta2000]

or

nx	ny	nz
2	1	1
1	2	1
1	1	2

giving E = 3×6C = 18C
(The numbers in the table indicate number of states occupied by single particle, each in n_x, n_y and n_z)

 

[TSny]

or

nx	ny	nz
2	1	1
1	2	1
1	1	2

giving E = 3×6C = 18C

Close. That takes care of 3 particles (one particle in each state that you listed). But what about the 4^th particle?

 

[tanaygupta2000]

or

nx	ny	nz
2	1	1
1	2	1
1	1	2

giving E = 3×6C = 18C
(The numbers in the table indicate number of states occupied by single particle, each in n_x, n_y and n_z)

In I row, 2 particles are in first and second states of n_x respectively, while 1 particle each in first state of n_y and n_z.
Similarly,
In II row, 2 particles are in first and second states of n_y, while 1 particle each in first state of n_x and n_z.
In III row, 2 particles are in first and second states of n_z, while 1 particle each in first state of n_x and n_y.

 

[tanaygupta2000]

I'm also wondering if this time I have to take n_x = n_y = n_z = 4

 

[TSny]

In I row, 2 particles are in first and second levels of state n_x respectively, while 1 particle each in first level of n_y and n_z.
Similarly,
In II row, 2 particles are in first and second levels of state n_y, while 1 particle each in first level of n_x and n_z.
In III row, 2 particles are in first and second levels of state n_z, while 1 particle each in first level of n_x and n_y.

No. You seem to be thinking that the values that you pick for nx, ny, and nz must add up to 4; i.e., they must add up to equal the number of particles. This is not the case. The numbers nx, ny, and nz just label the different quantum states of the system.

I could put one particle in the state labeled by (nx, ny, nz) = (3, 1, 5). I could put another particle in the state
(8, 7, 22). The third particle could go in the state (10, 9, 1). The fourth particle might occupy the state (1, 9, 10). The third and fourth particles would have the same energy even though they are in different states. For each particle, you can calculate the energy of the particle by using the formula for E given in the problem. Thus, you can determine the total energy.

You want to put the particles in states so that the total energy of the 4 particles is the least possible. So, you have to think about which states (nx, ny, nz) the particles should occupy in order to minimize the total energy.

Here is a rough analogy. Suppose you have a stairway with an infinite number of steps. You have four blocks of the same mass. You want to place the 4 blocks on the steps in such a way that you minimize the total gravitational potential energy of the system. But you can't put more than one block on anyone step. How would you place the 4 blocks?

 

[TSny]

I'm also wondering if this time I have to take n_x = n_y = n_z = 4

You could put one particle in the state (nx, ny, nz) = (4, 4, 4). The energy of this particle would be 48C, a rather large energy. This is not a good choice if you want to minimize the total energy. There are 63 states that have less energy than the state (4, 4, 4).

 

[tanaygupta2000]

No. You seem to be thinking that the values that you pick for nx, ny, and nz must add up to 4; i.e., they must add up to equal the number of particles. This is not the case. The numbers nx, ny, and nz just label the different quantum states of the system.

I could put one particle in the state labeled by (nx, ny, nz) = (3, 1, 5). I could put another particle in the state
(8, 7, 22). The third particle could go in the state (10, 9, 1). The fourth particle might occupy the state (1, 9, 10). The third and fourth particles would have the same energy even though they are in different states. For each particle, you can calculate the energy of the particle by using the formula for E given in the problem. Thus, you can determine the total energy.

You want to put the particles in states so that the total energy of the 4 particles is the least possible. So, you have to think about which states (nx, ny, nz) the particles should occupy in order to minimize the total energy.

Here is a rough analogy. Suppose you have a stairway with an infinite number of steps. You have four blocks of the same mass. You want to place the 4 blocks on the steps in such a way that you minimize the total gravitational energy of the system. But you can't put more than one block on anyone step. How would you place the 4 blocks?

I will put I block on I step, II on II, III on III and IV block on IV step.
So should I take

	nx	ny	nz
I Particle	1	1	1
II Particle	2	2	2
III Particle	3	3	3
IV Particle	4	4	4

or

	nx	ny	nz
I Particle	1	2	3
II Particle	1	2	3
III Particle	1	2	3
IV Particle	1	2	3

 

[TSny]

Your second table won't work since it puts all 4 particles in the same state (nx, ny, nz) = (1, 2, 3). But you can't put more than one particle in a state.

The first table puts only one particle in a state, which is good. But, your choice of the four states doesn't produce the minimum energy. Your first state (nx, ny, nz) = (1, 1, 1) is a good choice since that state has the lowest energy of all possible states. You need to think about which state (or states) would have the next lowest energy. Note that there is no rule which says that the three values of nx, ny, and nz must be the same. (You chose the three values nx, ny, and nz to be the same in each row of your first table above.)

 

[tanaygupta2000]

Your second table won't work since it puts all 4 particles in the same state (nx, ny, nz) = (1, 2, 3). But you can't put more than one particle in a state.

The first table puts only one particle in a state, which is good. But, your choice of the four states doesn't produce the minimum energy. Your first state (nx, ny, nz) = (1, 1, 1) is a good choice since that state has the lowest energy of all possible states. You need to think about which state (or states) would have the next lowest energy. Note that there is no rule which says that the three values of nx, ny, and nz must be the same. (You chose the three values nx, ny, and nz to be the same in each row of your first table above.)

Okay so I understand
For Particle I => (1,1,1)
For Particle II => (1,1,2)
For Particle III => (1,2,1)
For Particle IV => (2,1,1)

 

[tanaygupta2000]

Okay so I understand
For Particle I => (1,1,1)
For Particle II => (1,1,2)
For Particle III => (1,2,1)
For Particle IV => (2,1,1)

Is it that the specific quantum state of each particle is the set of specific coordinates (n_x, n_y, n_z) corresponding to it ?

 

[TSny]

Okay so I understand
For Particle I => (1,1,1)
For Particle II => (1,1,2)
For Particle III => (1,2,1)
For Particle IV => (2,1,1)

Yes. Good.

 

[TSny]

Is it that the specific quantum state of each particle is the set of specific coordinates (n_x, n_y, n_z) corresponding to it ?

I would say that the specific coordinates (nx, ny, nz) provide a label for the specific state of the particle.