Given a wave function $\psi(x) = \langle \psi | x \rangle$. It can be expanded in terms of orthogonal states:

$$ \langle \psi | x \rangle = \sum_n \langle \psi | n \rangle \langle n |x \rangle $$


  1. Can this always be done for an arbitary action?

  2. Given a general action $S(x,\dot{x})$, is there a general formula to find $\langle \psi |n \rangle$ and $\langle x |n \rangle$ in terms of the action, S, and the wavefunction $\psi(x)$?

All I know is that the ground state can be given formally by the path integral:

$$\langle x | 0 \rangle = \int\limits^{y_0=x}_{y_{-\infty}=0} e^{iS(y,\dot{y})} Dy$$

and so (I think)

$$\langle \psi | 0 \rangle = \int \psi(y_0)e^{iS(y,\dot{y})} Dy$$

  1. Are there any general formulae for $|n\rangle$ for $n>0$ ?

  2. Also, can this be done in the case of full QFT or just for free fields?

  • 1
    $\begingroup$ The expansion is just a matter of using the fact that the states |n> form a complete set, you can express that as the so-called closure relation: $$\sum_n |n\rangle\langle n| = 1$$ The path integral gives the time evolution, the matrix element you wrote down is to be interpreted in the Heisenberg picture, so it gives the amplitude for a particle at position 0 to evolve to position x. You can manipulate that via various closure relations and express it in terms of wavefunctions. $\endgroup$ – Count Iblis Oct 18 '15 at 19:42
  • $\begingroup$ To get the ground state from the path integral, you can use the Gell-Mann and Low theorem $\endgroup$ – Count Iblis Oct 18 '15 at 19:42
  • $\begingroup$ Hi, yes, I got the ground state using the path integral. I just wondered if there is a way to get the other states using the path integral? Maybe there is no general way of doing it? The states are eigenstates of the energy if this helps? Maybe there is no general formula? $\endgroup$ – zooby Oct 18 '15 at 19:52

(1) Can this always be done for an arbitrary action?

The answer to this was given by CountIblis in his comment. The expansion you are asking about follows from the closure formula for basis $\{|n\rangle\}$, $I = \sum_n{|n\rangle\langle n|}$: $$ \langle \psi|x\rangle = \langle \psi|I|x\rangle = \sum_n{\langle\psi|n\rangle\langle n|x\rangle} $$ It is independent of the action that governs the evolution of $|\psi\rangle$.

(2) Given a general action $S(x,\dot{x})$, is there a general formula to find $\langle \psi|n\rangle$ and $\langle x|n\rangle$ in terms of the action, $S$, and the wavefunction $\psi(x)$?

It we know the wavefunctions $\langle x| \psi \rangle$ and $\langle x| n \rangle$, the overlap $\langle n |\psi\rangle = \langle \psi|n\rangle^*$ can be found (trivially) as $$ \langle n |\psi\rangle = \int{dx \;\langle n |x\rangle \langle x|\psi\rangle} = \int{dx\; \langle x |n\rangle^* \langle x|\psi\rangle} $$
You can look at this as another application of the closure relation, this time for the $\{|x\rangle\}$ basis, $I = \int{dx\; |x\rangle \langle x|}$. Again, it is independent of the action for the evolution of $|\psi\rangle$.

The problem of finding the wavefunctions $\langle x| n \rangle$ for excited states in the path integral representation is another issue, and not a simple one. You may want to take a look at these papers to get an idea of what's involved:

  • T. E. Sorensen & W. B. England, Molecular Physics 89, 1577 (1996)
  • A.G. Ushveridze, Physics Letters A 110(4), 217–220 (1985)
  • $\begingroup$ Unfortunately this did not answer the question. I did not say the $\langle x | n \rangle$ was known. This is one of the unknowns that we are trying to find! All that is known is S and the wavefunction. All you've done is express one unknown in terms of the other unknown! $\endgroup$ – zooby Oct 20 '15 at 13:25
  • 1
    $\begingroup$ In answering (2) I said "If we know ...$\langle x|n\rangle$ ...", followed by (last paragraph) "The problem of finding the wavefunctions $\langle x|n\rangle$ ... is another issue". This means "we can calculate $\langle n| \psi\rangle$ provided we know $\langle x|n\rangle$ in addition to $\langle x|\psi\rangle$, but $\langle x|n\rangle$ itself is not so easy to calculate in the first place in path integral representation even if we know S exactly". However, I did provide 2 refs. that consider exactly this path integral calculation for excited states, closing the problem you asked about. $\endgroup$ – udrv Oct 20 '15 at 18:48

I don't think there is a general way. Also, probably the state basis is only applicable probably to free field theories. Just the quadratic part of the action.

In general it is a very difficult if not impossible task to diagonalise the Hamiltonian. That is why path integrals are used for interactions.


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