Is there such a thing as the spectrum of an operator in the path integral formalism?

Question

It is often claimed that the path-integral formalism and the Schrodinger one are equivalent, and the proof usually involves calculating amplitudes in both ways and checking that they are the same, or, alternatively, starting from the Schrodinger formulation and deriving the path integral from there in terms of the evolution operator. However, there is more to canonical quantization than amplitudes and expected values. In particular, we have the spectrum of an operator, which tells us which values an observable may take. Is there a way of arriving to a similar result ("the energy of the harmonic oscillator is discretized") strictly from the path integral?

Answer 1

The distinction you are referring to between the path integral formalism and Schrodinger one should really be between the functional formalism and the operator formalism. The path integral allows you to calculate the probability amplitude from an initial to a final state, so it encodes exactly the same information$^\dagger$ as the wavefunction $\psi(x,t)$ whose time evolution is governed by the Schrodinger equation

$$ i\hbar\frac{\partial}{\partial t} \psi(x,t) = \left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\right ] \psi(x,t) $$

which indeed can be derived from the path integral. Note that we are not explicitly dealing with operators or vector spaces here - this is just a second order partial differential equation.

We can then find for example the stationary states by finding $\psi(x,t)$ that satisfy the above equation as well as

$$ i\hbar\frac{\partial}{\partial t} \psi(x,t) = E \psi(x,t) $$

and the possible values of $E$ gives us the energy spectrum (which will be discrete in the harmonic oscillator case).

More generally, we can ask what is the probability amplitude of finding the system in some state $\phi_j$ given the wavefunction $\psi$. This is given by (omitting the time dependance)

$$ \langle \phi_j | \psi \rangle = \int dx \phi_j^*(x)\psi(x)$$

where in the LHS we are expressing this as an inner product between vectors in a Hilbert space, and in the RHS still just using wavefunctions.

When we switch to the language of operators acting on vectors in a Hilbert space, we can always describe $| \phi_j \rangle$ as an eigenvector of some Hermitian operator $\mathcal{\hat O}$. We can therefore say that this corresponds to a measurement of this operator (with an associated eigenvalue $\lambda_j$).

Note that Since the eigenvectors of Hermitian operators form an orthogonal basis, we can write $| \psi \rangle$ as a linear combinations of them, $| \psi \rangle = \sum_i \alpha_i | \phi_i \rangle$ with $\sum_i |\alpha_i|^2=1$, so indeed $ |\langle \phi_j | \psi \rangle|^2 $ forms a probability measure over the eigenstates of $\mathcal{\hat O}$.

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$^\dagger$ The path integral gives the amplitude $K(x,x';t)$ for a particle to propagate from $x$ to $x'$ in a time $t$, which is equivalent to

$$\psi(x',t) = \int dx \psi(x,0) K(x,x';t) $$

Answer 2

As already hinted in the comments, path integral can be used to calculate the Green's function. In particular, in one-particle case, the Feynman propagator is but the retarded Green's function of the time-dependent Schrodinger equation:

If we have $$ H\phi_n(x)=\epsilon_n(x),\psi(x,0)=\sum_nc_n\phi_n(x),$$ then $$\psi(x,t)=\sum_ne^{-\frac{i\epsilon_n t}{\hbar}}c_n\phi_n(x)= \sum_ne^{-\frac{i\epsilon_n t}{\hbar}}\int dx' \phi_n^*(x')\psi(x',0)\phi_n(x)= \int dx' K(x,x',t)\psi(x',0) $$ where $$ K(x,x',t)=\theta(t)\sum_ne^{-\frac{i\epsilon_n t}{\hbar}}\phi_n^*(x')\phi_n(x),$$ or in Fourier space: $$ K(x,x',\omega)=\sum_n\frac{\phi_n^*(x')\phi_n(x)}{\omega - \epsilon_n/\hbar+i0^+} $$ (a factor might be missing here.)

Thus, we can find the full one -particle spectrum from the propagator. Analysis of multi-particle Green's functions allows also to obtain spectrum of multi-particle states.

Remark
In presence of interactions, single-particle states are only approximate, and the poles of the propagator acquire finite complex component "lifetime":$\gamma_n=\frac{1}{\tau_n}$. For more discussion see my answers to:
If a non-interacting particle behaves like an undamped wave, can an interacting particle behave like a damped wave?
Particle/hole excitations have finite lifetime