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Calculating the transition amplitude in Euclidean spacetime is useful because from it we can extract the ground state energy and ground state wave-functions values.

For example, let's assume we are calculating: $\left<x=b\left.\right|e^{-\hat{H}T}\left.\right|x=a\right>$. Assume we can calculate this quantity via the path-integral formalism and we have found it to be equal to $f(T)$.

  • We insert a complete set of energy eigenstates into the transition amplitude: $$ \left<x=b\left.\right|e^{-\hat{H}T}\left.\right|x=a\right> = \left<x=b\left.\right|e^{-\hat{H}T}\sum_n\left|n\right>\left<n\right|\left.\right|x=a\right> = \sum_n e^{-E_nT}\left<x=b\right|\left.n\right>\left<n\right|\left.x=a\right> = f\left(T\right) $$

The usual trick now is to take the limit $T\to\infty$ of the above equation and say that the most important term is the lowest energy eigenvalue, and so: $$ \lim_{T\to\infty} \sum_n e^{-E_nT}\left<x=b\right|\left.n\right>\left<n\right|\left.x=a\right> \stackrel{?}{=} \lim_{T\to\infty} e^{-E_0T}\left<x=b\right|\left.0\right>\left<0\right|\left.x=a\right> = \lim_{T\to\infty} f\left(T\right)$$

$f\left(T\right)$ usually is of the form $A\times e^{- \varepsilon T} $ and so we say $E_0 = \varepsilon$.

I feel funny about this procedure because taking this limit, if $E_0$ is positive then this term should go to zero as well. In this sense, is the equality with the question mark merely an estimate, and not a true equality? If so $E_0$, which we obtain in this way, is just an approximation? Why is it then that for the SHO we get the exact result in this way $\frac{\omega}{2}$?

Even worse, when $f\left(T\right)$ is of the form $f\left(T\right)=A\times e^{- \varepsilon_0 T}+B\times e^{- \varepsilon_1 T}$, then I have seen (if I understand correctly) that the conclusion is then that $E_0=\varepsilon_0$ and $E_1=\varepsilon_1$ (i.e. the 2nd lowest energy eigenvalue is extracted as well) (for example, see Coleman "Aspects of Symmetry" page 274 equation 2.31).

Can anyone shed some light at the mathematical background of this "trick"? Is it not the same limit as the epsilon delta limit? Can I extract the whole spectrum in this way, just from knowledge of $f\left(T\right)$? If I can calculate the path-integral for various different values (not just $x=a$ and $x=b$) then could I, in principle, extract in this way also the whole set of energy eigenstates at all points? Is the information obtained in this way always an approximation? How can I find out the margin of error on this approximation then?

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  • $\begingroup$ At least one relevant aspect to this is the fact that we may choose the energy scale arbitrarily, so that we may WLOG assume that $E_0 = 0$. Then we get an actual equality $<x=b|0><0|x=a> = \lim_{T\to\infty} f(T)$ and the question of whether this is an approximation or not is off the table. $\endgroup$ – PPR Feb 15 '15 at 17:48
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This question is a bit old but, since Community bumped it up, the following answer might still be useful.

To extract the ground state energy $E_0$, computing $\lim_{T\rightarrow\infty} f(T)$ is the wrong thing to do, precisely for the reason mentioned by the OP.

One needs to use https://en.wikipedia.org/wiki/Asymptotic_analysis and the fact $f(T)\sim Ae^{-TE_0}$ when $T\rightarrow \infty$.

So the correct ansatz instead is $$ E_0=\lim_{T\rightarrow\infty}\left\{ -\frac{1}{T}\log f(T)\right\}\ . $$

BTW, in the Euclidean path integral formulation (i.e., mapping to the realm of statistical mechanics) the above computation usually amounts to that of the free energy density of a statistical mechanical system.

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It should not be a problem whether the groundstate energy is positive or negative (you can add a finite constant energy to the Hamiltonian without changing the physics (if we forget about GR, which is clearly the case here)). Same thing if the first excited state has negative energy (but is larger than the ground state).

Using the notation $\epsilon_n$ for the eigenvalues of the Hamiltonian with $\epsilon_0<\epsilon_1<\cdots$, we have $$f(T)=\sum_{n=0}^\infty A_n e^{-\epsilon_n T},\\ =A_0 e^{-\epsilon_0 T}\left(1+\sum_{n=1}^\infty \frac{A_n}{A_0}e^{-(\epsilon_n-\epsilon_0)T}\right). $$ In the limit $T\to\infty$, the sum in the second line vanishes and we can extract both the amplitude $A_0$ and the groundstate energy.

By subtracting $A_0 e^{-\epsilon_0 T}$, and using the same procedure, one can extract $\epsilon_1$ and $A_1$, etc.

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  • $\begingroup$ Hi, thanks for your response! I still have a few questions: (1) Why do you assume $f\left(T\right)$ can be expanded as you did? I never saw such an expansion. (2) In the $T\to\infty$ limit, not only the sum vanishes, but also the exponent outside the parenthesis should go to zero. I understand it is the leading term, but that was exactly what my question was about, I guess. How do you define, in a mathematically precise way, the procedure of "comparing leading terms"? (3) In this process, are we obtaining merely an estimate for the energy, or is the computation exact? $\endgroup$ – PPR May 1 '14 at 9:11
  • $\begingroup$ @Psycho_pr: (1) just use the eigenbasis of the Hamiltonian and insert the identity. $A_n$ depends on the quantity you are looking at. For example, $A_n=1$ for the partition function Z. (2) You can definitely compare the terms in the brackets, which are exponentially smaller than $1$. (3) This gives exactly the groundstate energy (with exponentially small error as $T\to0$. Starting from the partition function, we have $\epsilon_0=-T^{-1}\ln Z+O\left(e^{-(\epsilon_n-\epsilon_0)/T}/T\right)$. $\endgroup$ – Adam May 1 '14 at 13:27
  • $\begingroup$ Hi. I feel somewhat uneasy with your explanation. I'm sorry to be thick and you should know I do appreciate your effort but I must insist: (1) I don't understand where I'm supposed to insert the identity. I have a function $f(T)$, the identity is $\sum_n \psi_n(x) \int dx \psi_n(x)^*$ where $\psi_n(x)$ are the energy eigenfunctions of $\hat{H}$. Where should I insert it and how would it help? $\endgroup$ – PPR May 5 '14 at 16:44
  • $\begingroup$ (2) When you say compare terms in the brackets, you mean compare between 1 and $\sum_{n=1}^{\infty}\frac{A_n}{A_0}e^{-(\epsilon_n-\epsilon_0)T}? Because that's not the comparison I was talking about. I was talking about comparing leading terms from the LHS with comparing terms from RHS. How do you make such a comparison rigorous? (as if you were comparing linearly independent terms) $\endgroup$ – PPR May 5 '14 at 16:44
  • $\begingroup$ (3) The sentence "..gives exactly..with exponentially small error.." doesn't make any sense to me. Either it is an approximation, in which case it is not exact, or it is exact, in which case there should be no error. In a way, this was the point of my whole question, to make mathematically precise this whole procedure. $\endgroup$ – PPR May 5 '14 at 16:45

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