How do we make sense of exponentiating an operator in quantum mechanics?

[Chopin]

Most quantum textbooks will tell you that converting between the Schrödinger and Heisenberg pictures involves something like the following:

|\Psi(t)\rangle = e^{i\hat{H}(t-t_0)}|\Psi(t_0)\rangle

This does make sense to me conceptually: we define eigenstates of the Hamiltonian |E\rangle, where \hat{H}|E\rangle = E|E\rangle, and then we define the initial conditions in terms of them: |\Psi(t_0)\rangle = \int{dE\:\psi(E)|E\rangle}. Then we just run each eigenstate forward at the appropriate frequency, giving us |\Psi(t)\rangle = \int{dE\:e^{iE(t-t_0)}\psi(E)|E\rangle}.

What I don't understand is how we can make mathematically rigorous the concept of exponentiating an operator like that. Do we simply define that action to be a shortcut for the procedure I just illustrated, or is there some more mathematically consistent way to define it? It looks like QFT starts to rely on this concept more and more when we talk about things like path integrals, so I'm trying to make sure I have a firm grasp on how all of the math works out.

 

[arkajad]

If the operator A is Hermitian, then it has spectral decomposition in the form

A=\int_{-\infty}^{\infty}dE(x), where E is a projection-valued measure. Then \exp(iHt) is defined via

\exp(iAt)=\int_{-\infty}^{\infty}\exp(ixt)dE(t)

Note: I am using here x as integration variable, E as a spectral measure. You will probably put H instead of my A, E (energy) as your integration variable, and you will then use P for the spectral measure - to avoid confusion. So your formula will look like

H=\int H dP(H)
e^{iHt}=\int e^{iEt}dP(E)

Search Google for "spectral theorem". You are interested in unbounded Hermitian operators.

Note: De facto the integration goes effectively only over the spectrum of your operator. So, if you know that your H is a positive operator, you can safely integrate only over the positive part of the real line. The crucial term in all that is "spectral measure".

 

[Avodyne]

Do we simply define that action to be a shortcut for the procedure I just illustrated,

Yes. That's what arkajad's fancy notation amounts to.

 

[arkajad]

Yes. That's what arkajad's fancy notation amounts to.

Not exactly. When we write \psi(E) it somehow assumes that \psi(E) exists (in some generalized sense). But it exists only if there is no degeneracy in energy eigenstates. In my fancy notation I have P(E), and P(E) can be any projection, not necessarily onto a one-dimensional subspace. In more than one space dimensions energy eigenstates are usually degenerate, therefore we would have to write \psi(E,\alpha) even if \alpha does not do anything.

These are essentially cosmetical issues, but even with the cosmetics one needs to be careful.

 

[Chopin]

Yeah, I wrote \psi(E) as a generic way to refer to the fact that we were just expanding the wavefunction in a basis in which the Hamiltonian was diagonal, because then we can take the eigenvalues and use them directly in the exponential. I understand that you have to integrate across the entire basis, however many variables it takes to span the space.

Ok, so this sounds like it's sort of like \delta(x), where there's a rigorous definition, but in practice it often just ends up being code for a notational trick. I assume that it's possible to prove that this notation obeys most of the normal rules the way you'd expect, for instance \frac{d}{dt}e^{i\hat{H}t} = i\hat{H}e^{i\hat{H}t}?

 

[arkajad]

I assume that it's possible to prove that this notation obeys most of the normal rules the way you'd expect, for instance \frac{d}{dt}e^{i\hat{H}t} = i\hat{H}e^{i\hat{H}t}?

Sure. In fact you can safely use the standard power series expansion

e^{iHt}=I+iHt+...

and do not worry about its convergence. It will converge when it needs to converge. Usually between some <a| and some |b> - if they are not too nasty.

 

[Chopin]

That's effectively what a path integral is doing, right? Linearizing the effect of the Hamiltonian over a short distance and then summing up all the possible paths from start to finish?

 

[arkajad]

I would not mix the path integral into this particular question. Path integral can be thought of as a particular and clever way of solving the Schrodinger equation, even when the Hamiltonian has an explicit dependence on time, so that you have H(t) and your time evolution is not a simple action of an exponential exp(iHt). But essentially, yes, on each sufficiently small time interval you may consider H as a constant and assume the linear expansion.