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## Main Question or Discussion Point

Hi everyone, I'm working through an example in my textbook, and it's making very little sense to me. The problem is:

Let [itex]Z_1(m)[/itex] be the partition function for a single quantum particle of mass m in a volume V. First, calculate the partition function for two of these particles if they are bosons (and then also for fermions).

So, it tells me earlier in the chapter that, for bosons for example, the wave state has to be symmetric. So, for two particles (a and b) in "plane wave states" [itex]k_1[/itex] and [itex]k_2[/itex] where the energy for a plane wave state k is [itex]E = \frac{\hbar^2 k^2}{2m}[/itex], the total state is [itex]\left|k_1,k_2\right\rangle = \frac{\left|k_1\right\rangle \left|k_2\right\rangle - \left|k_2\right\rangle \left|k_1\right\rangle}{\sqrt{2}}[/itex] (for k1 not equal to k2. For k1=k2, it's just [itex]\left|k_1\right\rangle \left|k_2\right\rangle[/itex])

I guess I understand that. A plane wave (let's just say 1D, along the x axis) is of the form [itex]e^{ikx}[/itex], so in this case, the total state would be (in the position basis)

[tex]\left|k_1,k_2\right\rangle = \frac{e^{ik_1x_a}e^{ik_2x_b} - e^{ik_1x_b}e^{ik_2x_a}}{\sqrt{2}}[/tex]

Is that right?

Anyway, my real trouble is next. I think I'm trying to do it more rigorously than I have to, but I want to understand this explicitly. So, for bosons, they say

[tex]Z_2 = tr(e^{-βH})[/tex], where tr() is the trace of a matrix. Now, they never really tell us why this is, earlier in the book. They just seem to introduce it randomly. Could anyone tell me why this is a definition of the partition function?

Anyway, I'll trust this for now. Going on, they say:

[tex]tr(e^{-βH}) = \sum_{k_1,k_2} \left\langle k_1,k_2\right|e^{-βH}\left|k_1,k_2\right\rangle = \sum_{k_1>k_2}\frac{\left\langle k_1 \right|\left\langle k_2\right| + \left\langle k_2 \right|\left\langle k_1\right|}{\sqrt{2}}e^{-βH}\frac{\left|k_1 \right\rangle \left| k_2\right\rangle + \left| k_2\right\rangle\left|k_1\right\rangle}{\sqrt{2}} + \sum_{k}\left\langle k \right|\left\langle k\right|e^{-βH}\left|k \right\rangle \left| k\right\rangle[/tex]

So this is just expanding from before. Now, the Hamiltonian acts on the product kets to the right, and because they both have the same pair of k values, produces the same thing, [itex]e^{-\frac{\beta\hbar^2}{2m}(k_1^2 + k_2^2)}[/itex] for the first sum and [itex]e^{-\frac{2\beta\hbar^2}{2m}k^2}[/itex] for the 2nd one.

So it seems like we now get

[tex]\sum_{k_1>k_2}e^{-\frac{\beta\hbar^2}{2m}(k_1^2 + k_2^2)}(\frac{\left\langle k_1 \right|\left\langle k_2\right| + \left\langle k_2 \right|\left\langle k_1\right|}{\sqrt{2}})(\frac{\left|k_1 \right\rangle \left| k_2\right\rangle + \left| k_2\right\rangle\left|k_1\right\rangle}{\sqrt{2}}) + \sum_{k}e^{-\frac{2\beta\hbar^2}{2m}k^2}(\left\langle k \right|\left\langle k\right|)(\left|k \right\rangle \left| k \right\rangle)[/tex]

So it seems to me that in the first sum, only two of those products (when you foil out the terms inside the parentheses) are nonzero. So that cancels the sqrt(2)'s on the bottom. In the right, it seems like it's just 1. Is this right?

Then, in the book, they change the indices of the first sum from k1>k2 to k1,k2 and add a factor of 1/2 in front of the sum, as is standard. However, they do the same for the second sum, but I don't see why. The second one is just going over all k, so there is no double counting, right?

Anyway, both of these sums are now basically some variation of the individual partition function (I see this part easily). So they finally get

[tex]Z = \frac{1}{2}[Z_1^2(m) + Z_1(m/2)][/tex]

The fermi one confuses me even more, but I'll be brief with it. Basically, the first step of setting up the PF as the trace of the exponent is the same, and they plug in the antisymmetric state. But here the ones where k1=k2 should be zero, unlike the bosons. Still, they somehow end up with a sum over k, where they mysteriously skip a bunch of steps. Could anyone explain this to me?

Thanks!

Let [itex]Z_1(m)[/itex] be the partition function for a single quantum particle of mass m in a volume V. First, calculate the partition function for two of these particles if they are bosons (and then also for fermions).

So, it tells me earlier in the chapter that, for bosons for example, the wave state has to be symmetric. So, for two particles (a and b) in "plane wave states" [itex]k_1[/itex] and [itex]k_2[/itex] where the energy for a plane wave state k is [itex]E = \frac{\hbar^2 k^2}{2m}[/itex], the total state is [itex]\left|k_1,k_2\right\rangle = \frac{\left|k_1\right\rangle \left|k_2\right\rangle - \left|k_2\right\rangle \left|k_1\right\rangle}{\sqrt{2}}[/itex] (for k1 not equal to k2. For k1=k2, it's just [itex]\left|k_1\right\rangle \left|k_2\right\rangle[/itex])

I guess I understand that. A plane wave (let's just say 1D, along the x axis) is of the form [itex]e^{ikx}[/itex], so in this case, the total state would be (in the position basis)

[tex]\left|k_1,k_2\right\rangle = \frac{e^{ik_1x_a}e^{ik_2x_b} - e^{ik_1x_b}e^{ik_2x_a}}{\sqrt{2}}[/tex]

Is that right?

Anyway, my real trouble is next. I think I'm trying to do it more rigorously than I have to, but I want to understand this explicitly. So, for bosons, they say

[tex]Z_2 = tr(e^{-βH})[/tex], where tr() is the trace of a matrix. Now, they never really tell us why this is, earlier in the book. They just seem to introduce it randomly. Could anyone tell me why this is a definition of the partition function?

Anyway, I'll trust this for now. Going on, they say:

[tex]tr(e^{-βH}) = \sum_{k_1,k_2} \left\langle k_1,k_2\right|e^{-βH}\left|k_1,k_2\right\rangle = \sum_{k_1>k_2}\frac{\left\langle k_1 \right|\left\langle k_2\right| + \left\langle k_2 \right|\left\langle k_1\right|}{\sqrt{2}}e^{-βH}\frac{\left|k_1 \right\rangle \left| k_2\right\rangle + \left| k_2\right\rangle\left|k_1\right\rangle}{\sqrt{2}} + \sum_{k}\left\langle k \right|\left\langle k\right|e^{-βH}\left|k \right\rangle \left| k\right\rangle[/tex]

So this is just expanding from before. Now, the Hamiltonian acts on the product kets to the right, and because they both have the same pair of k values, produces the same thing, [itex]e^{-\frac{\beta\hbar^2}{2m}(k_1^2 + k_2^2)}[/itex] for the first sum and [itex]e^{-\frac{2\beta\hbar^2}{2m}k^2}[/itex] for the 2nd one.

So it seems like we now get

[tex]\sum_{k_1>k_2}e^{-\frac{\beta\hbar^2}{2m}(k_1^2 + k_2^2)}(\frac{\left\langle k_1 \right|\left\langle k_2\right| + \left\langle k_2 \right|\left\langle k_1\right|}{\sqrt{2}})(\frac{\left|k_1 \right\rangle \left| k_2\right\rangle + \left| k_2\right\rangle\left|k_1\right\rangle}{\sqrt{2}}) + \sum_{k}e^{-\frac{2\beta\hbar^2}{2m}k^2}(\left\langle k \right|\left\langle k\right|)(\left|k \right\rangle \left| k \right\rangle)[/tex]

So it seems to me that in the first sum, only two of those products (when you foil out the terms inside the parentheses) are nonzero. So that cancels the sqrt(2)'s on the bottom. In the right, it seems like it's just 1. Is this right?

Then, in the book, they change the indices of the first sum from k1>k2 to k1,k2 and add a factor of 1/2 in front of the sum, as is standard. However, they do the same for the second sum, but I don't see why. The second one is just going over all k, so there is no double counting, right?

Anyway, both of these sums are now basically some variation of the individual partition function (I see this part easily). So they finally get

[tex]Z = \frac{1}{2}[Z_1^2(m) + Z_1(m/2)][/tex]

The fermi one confuses me even more, but I'll be brief with it. Basically, the first step of setting up the PF as the trace of the exponent is the same, and they plug in the antisymmetric state. But here the ones where k1=k2 should be zero, unlike the bosons. Still, they somehow end up with a sum over k, where they mysteriously skip a bunch of steps. Could anyone explain this to me?

Thanks!