Consider the amplitude $$Z=\big\langle 0\big|\exp\big(-\frac{iHT}{\hbar}\big)\big|0\big\rangle.$$ This is same as $$Z=\int\limits_{-\infty}^{+\infty} dq\int\limits_{-\infty}^{+\infty} dq^\prime\langle 0|q^\prime\rangle\langle q^\prime|e^{-iHT/\hbar}|q\rangle\langle q|0\rangle=\int\limits_{-\infty}^{+\infty} dq\int\limits_{-\infty}^{+\infty}dq^\prime~ \phi^*_0(q^\prime)\phi_0(q)K(q,t;q^\prime,t^\prime)$$ where $\phi_0(q)$ is the position space wavefunction for the ground state $|0\rangle$ evaluated at $q$ and $$K(q,t;q^\prime,t^\prime)\equiv \langle q^\prime|e^{-iHT/\hbar}|q\rangle=\int_{q(t)=q}^{q(t^\prime)=q^\prime} Dq(t) \exp\Big[i\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big].$$ This integral has informations about the ground state i.e., we need the ground state wavefunction to evaluate this.

But $Z$ has an alternative expression in terms of a path-integral as $$Z=\int Dq(t) \exp\Big[\frac{i}{\hbar}\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big].$$ Unlike the first expression of $Z$, from the second expression has anything to do with the ground state or knows about the ground state. I find it amusing that the second relation apparently oblivious to the ground state?

  • Is there any boundary condition for $q(0)$ and $q(T)$ in $\int Dq(t)$? This is not clear. Now, we are not evaluating $\big\langle q_i\big|\exp\big(-\frac{iHT}{\hbar}\big)\big|q_f\big\rangle$. So I see no reason of fixing anything.

  • It is possible to derive the second relation from the first or the first relation from the second?

  • Is there a similar path integral expression for $\big\langle f\big|\exp\big(-\frac{iHT}{\hbar}\big)\big|i\big\rangle$ where $i$ and $f$ are arbitrary initial or final state?

  • 1
    $\begingroup$ Second expression is under specified. It begs to ask what paths you are integrating over. First expression is a special case of the last expression. $\endgroup$
    – Sunyam
    Commented Jan 5, 2020 at 18:25
  • $\begingroup$ Yes. I am asking how to relate these two expressions of $Z$. $\endgroup$ Commented Jan 5, 2020 at 18:26
  • $\begingroup$ If you see answer of Qmechanic for your other recent question. For second expression to make sense (even at Physicists level of rigour), you have to specify what boundary conditions paths satisfy. If you are integrating over all paths without constraints, it just means you are evaluating $\int_{-\infty}^{+\infty}dq_f\int_{-\infty}^{+\infty}dq_i\langle q_f| e_{}^{- i H t}|q_i\rangle$ for whatever purpose you are after. $\endgroup$
    – Sunyam
    Commented Jan 5, 2020 at 18:37
  • $\begingroup$ Related: physics.stackexchange.com/q/409907/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jan 5, 2020 at 18:37
  • $\begingroup$ @Sunyam What boundary condition do you suggest here? In the previous answer I wasn't evaluating vacuum to vacuum amplitude but amplitude of going from $|q_i\rangle$ to $|q_f\rangle$. There boundary conditions would be $q(0)=q_i$ and $q_f=q(T)$. But I agree that as a path integral second expression is not meaningful unless we have a boundary condition. But what boundary condition would appear here and how? $\endgroup$ Commented Jan 5, 2020 at 18:42

1 Answer 1


Let's be more careful: $$ K(q,t|q^\prime, t^\prime)=\int_{q(t)=q}^{q(t^\prime)=q^\prime} Dq(t) \exp\Big[i\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big]. $$ In this integral you need integrate over all continuous trajectories with boundary conditions at $t$ and $t^\prime$: $q(t)=q$ and $q^\prime(t^\prime)=q^\prime$.

Using $K$ you easily can calculate evaluation of wave function:

$$ \psi(q,t) =\int dq^\prime K(q,t|q^\prime, t^\prime)\psi(q^\prime, t^\prime) $$

So for amplitude you obtain:

$$ \big\langle \psi^\prime(t)\big|\psi(t)\big\rangle = \big\langle \psi^\prime(t)\big|\exp\big(-\frac{iH(t-t^\prime)}{\hbar}\big)\big|\psi(t^\prime)\big\rangle = \int dq \underbrace{\int dq^\prime K(q,t|q^\prime, t^\prime)\psi(q^\prime, t^\prime)}_{\psi(q,t)}(\psi^\prime(q,t))^* $$

For vacum-vacum amplitude you need put ground state wave function.

For partition function (which usually denoted as Z):

$$ Z =Tr_{\mathcal{H}}e^{-iHT/\hbar} = \int dq \big\langle q\big|\exp\big(-\frac{iHT}{\hbar}\big)\big|q\big\rangle =\int dq\int_{q(0)=q}^{q(T)=q} Dq(t) \exp\Big[i\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big] $$

Useful reference Path Integral

Also I recommend you read appendix in book by Joseph Polchinski

  • $\begingroup$ I made made a small edit. Please see if you are okay with it. If not, revert it. $\endgroup$ Commented Jan 5, 2020 at 18:56
  • $\begingroup$ It's okay. You more careful wrote boundary conditions. $\endgroup$
    – Nikita
    Commented Jan 5, 2020 at 18:57
  • $\begingroup$ I updated my answer. $\endgroup$
    – Nikita
    Commented Jan 5, 2020 at 19:11
  • $\begingroup$ So basically $Z=\int dq \int dq^\prime \psi^{\prime}_0(q^\prime)\psi^*_0(q)K(q,t;q^\prime,t^\prime)$. How does that reduce to my expression for $Z$? This was my second question. $\endgroup$ Commented Jan 5, 2020 at 19:13
  • $\begingroup$ Could you be more concrete?, I can't understand question:( $\endgroup$
    – Nikita
    Commented Jan 5, 2020 at 19:22

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