How to deal with index in the problem of thermal field?

In summary, the problem of harmonic oscillator for single mode can be solved by considering the density operator and using the partition function to find the average energy. For multimode, all modes must be considered using the partition function and the average energy can be found by taking the derivative of the partition function. The correct expression for the partition function is a product of sums over the occupation numbers for each mode.
  • #1
KFC
488
4
In the problem of harmonic oscillator, for single mode, that is, the energy

[tex]H = \hbar\omega(n + 1/2)[/tex]

It is easy to find the average of energy by considering the density operator

[tex]<H> = \frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}[/tex]

But for multimode (assume no polarization), we have to consider all modes specified with [tex]k[/tex]

[tex]H = \sum_k \hbar\omega_k(n_k + 1/2)[/tex]

So the average energy would be

[tex]<H> = \prod_k\frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}
= \prod_k\frac{\sum_{k_1} \hbar\omega_{k_1}(n_{k_1} + 1/2) \exp[\beta\sum_{k_2}\hbar\omega_{k_2}(n_{k_2}+1/2)] }{\exp[\beta\sum_{k_3}\hbar\omega_{k_3}(n_{k_3}+1/2)]}
[/tex]

Since each sum and prod should have independent index so I use k, k1, k2, k3. I am very confusing how to deal with the index to get the simplifed expression.

In some textbook, it first let the partition function be
[tex]Z=Tr(e^{\beta H})[/tex]

so the energy average be

[tex]
<H> = -\frac{\partial \ln Z}{\partial \beta} = \sum_{k}\dfrac{\hbar\omega_{k}}{\exp\{\beta\hbar\omega_{k}\}-1}
[/tex]

My question is how to deal with index if I don't use the partion function.
 
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  • #2
KFC said:
In the problem of harmonic oscillator, for single mode, that is, the energy

[tex]H = \hbar\omega(n + 1/2)[/tex]

It is easy to find the average of energy by considering the density operator

[tex]<H> = \frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}[/tex]

But for multimode (assume no polarization), we have to consider all modes specified with [tex]k[/tex]

[tex]H = \sum_k \hbar\omega_k(n_k + 1/2)[/tex]

So the average energy would be

[tex]<H> = \prod_k\frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}
= \prod_k\frac{\sum_{k_1} \hbar\omega_{k_1}(n_{k_1} + 1/2) \exp[\beta\sum_{k_2}\hbar\omega_{k_2}(n_{k_2}+1/2)] }{\exp[\beta\sum_{k_3}\hbar\omega_{k_3}(n_{k_3}+1/2)]}
[/tex]

Since each sum and prod should have independent index so I use k, k1, k2, k3. I am very confusing how to deal with the index to get the simplifed expression.

In some textbook, it first let the partition function be
[tex]Z=Tr(e^{\beta H})[/tex]

so the energy average be

[tex]
<H> = -\frac{\partial \ln Z}{\partial \beta} = \sum_{k}\dfrac{\hbar\omega_{k}}{\exp\{\beta\hbar\omega_{k}\}-1}
[/tex]

My question is how to deal with index if I don't use the partion function.

same way as with the partition function. it's just easier to calculate the partition function and take a derivative.
 
  • #3
olgranpappy said:
same way as with the partition function. it's just easier to calculate the partition function and take a derivative.

But there is three different index, how can I make it into one index?
 
  • #4
Well, I found one problem. For single mode, which way is the correct one to write the partition function?

[tex]Z= \prod_k\exp[\beta\sum_{k}\hbar\omega_{k}(n_{k}+1/2)][/tex]

or

[tex]Z= \prod_k\exp[\beta\hbar\omega_{k}(n_{k}+1/2)][/tex]

?

And what about the hamiltonian in the numerator?
 
  • #5
KFC said:
Well, I found one problem. For single mode, which way is the correct one to write the partition function?

[tex]Z= \prod_k\exp[\beta\sum_{k}\hbar\omega_{k}(n_{k}+1/2)][/tex]

or

[tex]Z= \prod_k\exp[\beta\hbar\omega_{k}(n_{k}+1/2)][/tex]

?

And what about the hamiltonian in the numerator?

Neither is the correct expression for the partition function. This is apparent since the occupation numbers still appear on the LHS...

But, before going on, note that
[tex]
\prod_j e^{f_j}
=e^{f_1}e^{f_2}\ldots e^{f_N}\ldots=e^{f_1+f_2+\ldots}
=e^{\sum_j f_j}\;.
[/tex]

The correct partition function is
[tex]
\sum_{n_1}\sum_{n_2}\ldots\sum_{n_N}\ldots
e^{-\beta(\sum_k(n_k+1/2)\hbar\omega_k)}
=\sum_{n_1}\sum_{n_2}\ldots\sum_{n_N}\ldots
e^{-\beta((n_1+1/2)\hbar\omega_k)}
e^{-\beta((n_2+1/2)\hbar\omega_k)}\ldots
=
\sum_{n_1}e^{-\beta((n_1+1/2)\hbar\omega_k)}\sum_{n_2}e^{-\beta((n_2+1/2)\hbar\omega_k)}\ldots
[/tex]
and this equals
[tex]
\prod_k \sum_{n_k}e^{-\beta(n_k+1/2)\hbar\omega_k}\;,
[/tex]
where the sum, for bosons, runs from 0 to infinity.
 

Related to How to deal with index in the problem of thermal field?

1. What is an index in the problem of thermal field?

An index in the problem of thermal field refers to the numerical value used to represent the thermal properties of a material, such as its thermal conductivity or specific heat capacity. It is an important factor in understanding how heat is transferred and distributed in a given system.

2. How do I determine the index of a material?

The index of a material can be determined by conducting experiments and measurements on its thermal properties. These can include measuring the material's temperature change over time, its response to heat transfer, and its ability to store and release thermal energy. Alternatively, the index may also be provided by the material manufacturer or found in reference tables.

3. Why is the index important in dealing with thermal fields?

The index is important in dealing with thermal fields because it helps us understand how heat is transferred and distributed in a given system. By knowing the index of a material, we can predict its behavior in response to changes in temperature and make informed decisions about how to manage and control thermal fields in that system.

4. How does the index affect the design of thermal systems?

The index of a material directly affects the design of thermal systems as it dictates how heat will be transferred and distributed within that system. For example, a material with a high thermal conductivity index would be ideal for conducting heat away from a heat source, while a material with a low index may be better suited for insulating against heat transfer.

5. Can the index of a material change?

Yes, the index of a material can change depending on factors such as temperature, pressure, and composition. Some materials may also exhibit different thermal properties in different directions, known as anisotropy, leading to different indices depending on the direction of heat transfer. It is important to consider these potential changes when designing and managing thermal systems.

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