Amplitude to go from one state to another

[Opus_723]

I have a question about probability amplitudes to go from one state to another.

I think it'll be clearest if I start with the case that I understand. Suppose I start in some initial state |ψ>, and I let it evolve over time t to some state e^iHt/ħ|ψ>. Now, if I want to know the probability that I measure some particular value of, say, momentum, then I project it onto the momentum eigenbasis and the probability amplitude of measuring a momentum of p is

<p|e^iHt/ħ|ψ>

where p is, and I stress this because it is key to the confusion that follows, a momentum eigenstate. If I want the probability that I measure a value a for some quantity other than momentum, I do the same thing but I use <a|, which is an eigenstate with well-defined a.

Now my confusion is that I often see discussions where people talk about "the probability amplitude for going from state |ψ> to state |φ>" where |φ> is quite often NOT an eigenstate of any kind. And they'll just write this as

<φ|e^iHt/ħ|ψ>

But I don't really see any sensible way to interpret the above statement if |φ> is not an eigenstate of some sort. For example, I have seen the above expression used where |φ> is a wave packet. And I just don't understand why, mathematically, the above gives you the probability that state |ψ> evolves to state |φ>. Even when |φ> is an eigenstate, I wouldn't call that "the probability amplitude that |ψ> evolves to |φ>", because the final state could have all sorts of momentum components if it is, for example, a wave packet. I would just get the probability amplitude of measuring a momentum of p. But the above expressions seems to be used FAR more generally than my narrow interpretation, and I'm at a loss as to why.

Thanks.

 

[PeterDonis]

I don't really see any sensible way to interpret the above statement if |φ> is not an eigenstate of some sort.

First, every state is an eigenstate of some operator. So requiring $\vert \psi \rangle$ to be an eigenstate doesn't actually impose any restriction at all.

Second, in your original statement, the bra $\langle p \vert$ was an eigenstate of momentum, but you weren't thinking of $\vert \psi \rangle$ as an eigenstate of anything--it certainly wasn't an eigenstate of momentum, if I'm interpreting you correctly. So why wouldn't your argument, if it were valid, apply equally well to your original statement, and say that an expression like $\langle p \vert e^{i H t / \hbar} \vert \psi \rangle$ makes no sense unless $\vert \psi \rangle$ is an eigenstate?

Third, even if $\vert \psi \rangle$ is an eigenstate, that does not guarantee that $e^{i H t / \hbar} \vert \psi \rangle$ will be an eigenstate, so your argument, if it were valid for $\vert \psi \rangle$, would say that you can never include time evolution in your analysis at all.

In short, I think you need to take a step back and rethink why you think expressions like the one you wrote down only make sense with eigenstates, because the above considerations seem to show that such a viewpoint can't be maintained.

 

[PeterDonis]

Oh, and fourth: $\langle p \vert$ is an eigenstate of momentum, but it's not an eigenstate of, say, position. So your argument, if it were valid, would seem to say we can never compute the probability of, for example, measuring a position of $x$ at time $t_1$ and then a momentum of $p$ at some later time $t_2$.

 

[Opus_723]

Okay, for definiteness, say that |ψ> is a wave packet with some spread of momentum. Then, for example, <x|e^iHt/ħ|ψ> makes sense to me as the probability amplitude of finding the particle at position x after a time t, during which the wave packet propagated, dispersed, etc. This makes sense to me because the expression is simply one of the components of the evolved wave packet in the |x> basis, which form a complete set of states because they are eigenstates of a Hermitian operator.

If, on the other hand, I take the inner product of e^iHt/ħ|ψ>with |φ>, which is perhaps another wave packet, so it's not an eigenstate of any observable that I know of, then I don't understand how to interpret this number. Apparently you can interpret it as the "probability amplitude that |ψ> evolves to |φ>", according to a few books I've read, but it's not obvious to me why. I'm not arguing that this is untrue, I just don't understand why it works, unless |φ> is the eigenstate of some measurable quantity, in which case I think I get it.

 

[PeterDonis]

Apparently you can interpret it as the "probability amplitude that |ψ> evolves to |φ>", according to a few books I've read, but it's not obvious to me why.

It's because "the probability amplitude that $\vert \psi \rangle$ evolves to $\vert \phi \rangle$" is the definition of the expression you wrote down. Your example where you put $\langle x \vert$ is just setting $\vert \phi \rangle = \vert x \rangle$. It doesn't change the definition at all.

I just don't understand why it works, unless |φ> is the eigenstate of some measurable quantity

Go back and read the "first" statement I made in post #2.

 

[Opus_723]

If it's just defined that way for any state, I haven't seen a text explicitly state that. In Shankar, for example, the closest I can find in the postulates is this:

"If the particle is in a state |ψ>, measurement of the variable (corresponding to) Ω will yield one of the eigenvalues ω with probability P(ω) proportional to |<ω|ψ>|² . The state of the system will change from |ψ> to |ω> as a result of the measurement."

Earlier, Shankar specifically refers to Ω as a function of the X and P operators.

Now, if it's true that EVERY state is an eigenstate of some Hermitian operator that is a function of X and P, then I see what you're getting at, but it's far from obvious to me that this is true. Could you elaborate? For example, what such operator has a wave packet as an eigenstate?

 

[PeterDonis]

If it's just defined that way for any state, I haven't seen a text explicitly state that.

Most texts don't explicitly state everything. But from the formulation where the bra on the left is an eigenstate, plus the fact that any state can be expressed as a linear combination of eigenstates, plus the fact that all operators in QM are linear, including the time evolution operator, you should be able to prove that if the definition applies when the bra on the left is an eigenstate, it must apply equally well when the bra on the left is any state whatever.

it's far from obvious to me that this is true

It's a mathematical theorem about vectors in a Hilbert space and Hermitian operators. The theorem basically says that every vector in a Hilbert space is an eigenvector of some Hermitian operator. I can't find a specific reference at the moment.

For example, what such operator has a wave packet as an eigenstate?

If the wave packet is a coherent state (which is a common way to model a wave packet), then it's an eigenstate of the annihilation operator.

 

[Orodruin]

Now, if it's true that EVERY state is an eigenstate of some Hermitian operator that is a function of X and P, then I see what you're getting at, but it's far from obvious to me that this is true. Could you elaborate?

Just as an example. For any state $|\psi\rangle$, the projection operator $|\psi\rangle\langle\psi|$ is a Hermitian operator with $|\psi\rangle$ as an eigenstate.

If the wave packet is a coherent state (which is a common way to model a wave packet), then it's an eigenstate of the annihilation operator.

The annihilation operator is not Hermitian, which is what the OP was looking for.

 

[kith]

To expand a bit on what Orodruin said: If you have a Hermitean operator $A$ with eigenvalues $a_i$ and corresponding eigenstates $|a_i\rangle$ you can write this operator as
$$A = \sum_i a_i |a_i\rangle \langle a_i|$$
I remember that realizing this improved my understanding of operators and eigenstates quite a bit.

Using this, you can easily construct observables with arbitrary eigenstates. It doesn't tell you how to actually measure these observables in the lab, however.

 

[PeterDonis]

The annihilation operator is not Hermitian

Ah, yes, good point.