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The thermodynamic expectation value for an observable $A$ is defined as $$\langle A \rangle = \frac{1}{Z} \sum_n \langle\psi_n| e^{-\beta H} A|\psi_n \rangle, \qquad (1)$$ where $\beta=1/k_bT$, the $\psi_n$ are a basis for the Hilbert space and $$Z= \sum_n \langle\psi_n| e^{-\beta H} |\psi_n\rangle. $$ Now, in the limit $T\rightarrow 0$ (or $\beta \rightarrow \infty$), only the ground state should contribute, hence I would expect that $$\langle A \rangle = \langle A \rangle_0 = \langle \psi_0 | A |\psi_0 \rangle,$$ where $\psi_0$ is the ground state.

What I want to know is, whether this is correct and if it is correct, how can I proof this, starting from eq. (1). I started with assuming that the $\psi_n$ are the energy eigenstates of the Hamiltonian $H$ (with $\psi_0$ being the eigenstate corresponding to the lowest energy $E_0$ [with $E_0<E_1<E_2\ldots$ ]) and rewrote the expectation value as $$\langle A \rangle = \frac{ \langle A \rangle_0 + \sum_{n=1}^N e^{-\beta (E_n-E_0)} \cdot \langle \psi_n| A|\psi_n \rangle}{\sum_{n=0}^N e^{-\beta (E_n-E_0)}}.$$

But from here I cannot see what happens, if I take $\beta\rightarrow \infty$.

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2 Answers 2

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I suspect that you have made a mistake in the derviation of your expression. The $\langle A \rangle_0$ in front of the sum term should not be there. Look at the exponential factors in your expression:

$$ e^{-\beta (E_n - E_0)} . $$

Let's try to evaluate the terms in the sum first and then take the limit $\beta \rightarrow \infty$. If $n=0$, we have $$ e^{-\beta (E_0 - E_0)} = 1 . $$ However, if $n \neq 0$, we have $$ e^{-\beta (E_n - E_0)} = e^{-\beta \Delta E} \rightarrow 0 , $$ where $\Delta E = E_n - E_0$ is positive and in the right hand side I have taken the limit $\beta \rightarrow 0$. Now the sum in your formula becomes $$ \langle A \rangle = \langle A \rangle_0 \langle \psi_0 \vert A \vert \psi_0 = \langle A \rangle_0^2 $$ since all other terms in the sum expect the one with $n=0$ vanish. If you get rid of the extra $\langle A \rangle_0$ in your expression, this yields the correct result.

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  • $\begingroup$ Thanks, I fixed it in my question. This answers everything. Thank you. $\endgroup$
    – Merlin1896
    Commented Oct 19, 2015 at 11:32
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If $\beta\rightarrow\infty$ you can omit all terms in the Partition function sum except for the one with the lowest energy:

$$ \lim\limits_{\beta\rightarrow\infty} \frac{\sum\limits_{n=1}^N e^{-\beta(E_n-E_0)} \langle \psi_n|A|\psi_n \rangle }{\sum\limits_{n=1}^N e^{-\beta(E_n-E0)}} = \frac{e^{-\beta(E_\text{min}-E_0)} \langle \psi_\text{min}|A|\psi_\text{min} \rangle }{e^{-\beta(E_\text{min}-E_0)}} = \langle \psi_\text{min}|A|\psi_\text{min} \rangle = \langle A \rangle_0 $$ This is true because $$\lim\limits_{\beta\rightarrow\infty} \frac{e^{-\beta(E_n-E_0)}}{e^{-\beta(E_\text{min}-E_0)}} = \delta_{n,\text{min}}$$

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  • $\begingroup$ I dont understand your notation. I labeled the state with the lowest energy with $E_0$ and assumed that the states are ordered, so that $E_0<E_1<E_2\ldots$. What is your $E_{min}?$ in contrast to $E_0$? $\endgroup$
    – Merlin1896
    Commented Oct 19, 2015 at 10:01
  • $\begingroup$ then $E_0=E_\text{min}$ $\endgroup$
    – Jannick
    Commented Oct 19, 2015 at 10:59
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    $\begingroup$ Please read again your post. With $E_0=E_{min}$, some statements don't make much sense. Also you are missing some $\beta$. $\endgroup$
    – Merlin1896
    Commented Oct 19, 2015 at 11:01

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