Quantum statistics of the two-state system

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This article derives important quantum statistical results for a two-level system . Since there are no real two-level systems, the model is purely theoretical , a so-called toy model . However, the results are approximately valid for two well-isolated energy levels.

Model description

Representation of the toy model

The model is based on particles with two possible energy levels each . To simplify the description, we put our energy zero point on the lower energy level. Then the upper energy level is on an energy . That means that each of our particles can either have the energy 0 or . ${\displaystyle n}$ ${\displaystyle E_{0}}$ ${\displaystyle E_{0}}$

Energy of the system

The Hamilton operator of such a system is easy to set up.

${\displaystyle {\hat {H}}=\sum _{k=1}^{n}{E_{0}{\hat {n}}_{k}}}$

where is an operator which indicates whether the -th particle is in the excited state or not. ${\displaystyle {\hat {n}}_{k}}$ ${\displaystyle k}$

Calculation of the sums of state

The partition function is obtained by inserting the Hamilton operator into the canonical partition function  :

${\displaystyle Z_{k}(N,V,T)=\operatorname {Sp} \left(\mathrm {e} ^{-{\frac {\hat {H}}{k_{\mathrm {B} }T}}}\right).}$

${\displaystyle Z_{k}(N,V,T)=\operatorname {Sp} \left(\mathrm {e} ^{-{\frac {\sum _{k=1}^{n}{E_{0}{\hat {n}}_{k}}}{k_{\mathrm {B} }T}}}\right).}$

The sum can be taken from the exponential function and becomes the product.

${\displaystyle Z_{k}(N,V,T)=\prod _{k=1}^{n}\operatorname {Sp} \left(\mathrm {e} ^{-{\frac {E_{0}{\hat {n}}_{k}}{k_{\mathrm {B} }T}}}\right).}$

Now we replace the trace with a sum over and the operator with its eigenvalue n _k . ${\displaystyle n_{k}}$ ${\displaystyle {\hat {n}}_{k}}$

${\displaystyle Z_{k}(N,V,T)=\prod _{k=1}^{n}\sum _{n_{k}}\left(\mathrm {e} ^{-{\frac {E_{0}n_{k}}{k_{\mathrm {B} }T}}}\right).}$

Analogously we get for the grand canonical partition function  :

${\displaystyle Z_{g}(\mu ,V,T)=\prod _{k=1}^{n}\sum _{n_{k}}\left(\mathrm {e} ^{-{\frac {(E_{0}-\mu )n_{k}}{k_{\mathrm {B} }T}}}\right).}$

The particle number operator is required for this , which is used in the general formula for the grand canonical partition function: ${\displaystyle {\hat {n}}=\sum _{k=0}^{n}{\hat {n}}_{k}}$ ${\displaystyle n}$

${\displaystyle Z_{g}(\mu ,V,T)=\operatorname {Sp} \left(\mathrm {e} ^{-{\frac {{\hat {H}}-\mu {\hat {n}}}{k_{\mathrm {B} }T}}}\right)}$

used.

Bosons

The grand canonical partition sum obtained in this way is still being reformulated. The system described contains particles on two energy levels, so there are usually several particles in one level, so it is a bosonic system. The sum over can therefore run from zero to infinity . For each summand, however, you can see that the same energy terms are always multiplied. Therefore the exponential function can also be rewritten: ${\displaystyle n}$ ${\displaystyle n_{k}}$

${\displaystyle Z_{g}(\mu ,V,T)=\prod _{k=1}^{n}\sum _{n_{k}}\left(\mathrm {e} ^{-{\frac {(E_{0}-\mu )}{k_{\mathrm {B} }T}}}\right)^{n_{k}}.}$

We recognize this form as a geometric series . It follows :

${\displaystyle Z_{g}(\mu ,V,T)=\prod _{k=1}^{n}{\frac {1}{1-\left(\exp \left(-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}\right)\right)}}.}$ (Bosonic partition function)

Fermions

To get the fermionic partition function we need a trick. The system will only fermionisch when in each state a maximum of a particle is located. To do this, we consider our system for , i.e. for a single particle. This means that our product is no longer available for the time being, and the sum above only contains two options: the particle is excited or not. Therefore it can only be 0 or 1. ${\displaystyle n=1}$ ${\displaystyle n_{k}}$ ${\displaystyle n_{k}}$

The partition function then becomes:

${\displaystyle Z_{g}(\mu ,V,T)=\sum _{n_{k}=0}^{1}\left(\mathrm {e} ^{-{\frac {(E_{0}-\mu )n_{k}}{k_{\mathrm {B} }T}}}\right).}$

Inserting n _k yields:

${\displaystyle Z_{g}(\mu ,V,T)=1+\left(\exp \left(-{\frac {(E_{0}-\mu )}{k_{\mathrm {B} }T}}\right)\right).}$

It was used that results. ${\displaystyle \mathrm {e} ^{0}=1}$

This gives us the partition function for a particle. At this point we leave our model and say that we originally wanted to describe a -particle system . By considering a system of models with one particle each, we obtain a fermionic system. The sum of states of the overall system thus consists of factors that each represent our previous result: ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle n}$

${\displaystyle Z_{g}(\mu ,V,T)=\prod _{k=1}^{n}\left(1+\left(\mathrm {e} ^{-{\frac {(E_{0}-\mu )}{k_{\mathrm {B} }T}}}\right)\right).}$ (Fermionic partition function)

Thermodynamic potential

We calculate the thermodynamic potential via:

${\displaystyle \Omega (T,V,\mu )=-k_{\mathrm {B} }T\log({Z_{g}}).}$

Bosons

Inserting the previously calculated grand-canonical partition function gives:

${\displaystyle \Omega (T,V,\mu )=-k_{\mathrm {B} }T\log \left({\prod _{k=1}^{n}{\frac {1}{1-(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}})}}}\right).}$

The product can be taken as the sum of the logarithm , inside a change of sign reverses the fraction. It follows:

${\displaystyle \Omega (T,V,\mu )=k_{\mathrm {B} }T\sum _{k=1}^{n}\log(1-(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}})).}$

Fermions

Inserting and bringing forward the sum results in:

${\displaystyle \Omega (T,V,\mu )=-k_{\mathrm {B} }T\sum _{k=1}^{n}\log(1+(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}})).}$

We see that the potential is the same except for the signs .

${\displaystyle \Omega (T,V,\mu )=_{-}^{+}k_{\mathrm {B} }T\sum _{k=1}^{n}\log(1_{+}^{-}(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}))\,\,\,\,_{\mathrm {Fermionen} }^{\mathrm {Bosonen} }.}$

Distribution functions

The expected value for the number of particles is obtained by deriving the thermodynamic potential from the negative chemical potential.

${\displaystyle \langle {\hat {N}}\rangle =-\left({\frac {\partial \Omega }{\partial \mu }}\right)_{\beta ,V}}$ With ${\displaystyle \beta ={\frac {1}{k_{\mathrm {B} }T}}.}$

Bosons

The derivation of the logarithm gives:

${\displaystyle {\frac {1}{1-\left(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}\right)}}.}$

The inner derivation gives:

${\displaystyle -{\frac {1}{k_{\mathrm {B} }T}}\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}.}$

In summary, we then get:

${\displaystyle \langle {\hat {N}}\rangle ={\frac {k_{\mathrm {B} }T}{k_{\mathrm {B} }T}}\sum _{k=1}^{n}{\frac {1}{1-\left(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}\right)}}\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}.}$

After shortening k _B T and combining the fraction, we get the Bose distribution :

${\displaystyle \langle {\hat {N}}\rangle =\sum _{k=1}^{n}{\frac {1}{\left(\mathrm {e} ^{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}\right)-1}}.}$ (Bose distribution)

Fermions

Similarly, by inserting the potential for fermions after deriving:

${\displaystyle \langle {\hat {N}}\rangle ={\frac {k_{\mathrm {B} }T}{k_{\mathrm {B} }T}}\sum _{k=1}^{n}{\frac {1}{1+\left(\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}\right)}}\mathrm {e} ^{-{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}}.}$

Summarizing now gives the Fermi distribution :

${\displaystyle \langle {\hat {N}}\rangle =\sum _{k=1}^{n}{\frac {1}{\left(\mathrm {e} ^{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}\right)+1}}.}$ (Fermi distribution)

Here, too, we see that the distributions only differ by one sign:

${\displaystyle \langle {\hat {N}}\rangle =\sum _{k=1}^{n}{\frac {1}{\left(\mathrm {e} ^{\frac {E_{0}-\mu }{k_{\mathrm {B} }T}}\right)\pm 1}}.}$

The generally valid Fermi and Bose distributions result from the model.