Is the classical relation between energy and momentum valid in QM?

[shankk]

TL;DR Summary

When can we use kinetic energy $$T = E - V = \frac{1}{2}mv^2$$ in QM? If it is valid then that would imply definite momentum in stationary states which doesn’t sound right.

Here we are talking about non-relativistic quantum physics. So we all know kinetic energy $$T = E - V = \frac{1}{2}mv^2$$ in classical physics. Here V is the potential energy of the particle and E is the total energy. Now what I am seeing is that this exact same relation is being used in quantum physics with impunity. For example when it comes to a free particle, this relation was used to get relation between its wave number and angular frequency.

From what I understand, this classical relation still holds, but the only thing new here is that E and p are not known precisely but instead we have a statistical distribution for them and so therefore even though the classical relation holds we cannot use it (because it requires exact values to be fed into it and not expectation values). Am I getting it right?

But then again we know that for a stationary state energy E is fixed and so by above reasoning p must always be fixed for a stationary state as well. This doesn’t sound right. So what am I missing?

 

[PeterDonis]

From what I understand, this classical relation still holds, but the only thing new here is that E and p are not known precisely but instead we have a statistical distribution for them

No, the new thing in QM is that $E$ and $p$ are not numbers any more, they're operators. So what you are calling the "kinetic energy" is now a relationship between operators, specifically, that $\hat{T} = \hat{p}^2 / 2 m$. The total energy operator, the Hamiltonian, is then $\hat{H} = \hat{p}^2 / 2m + V$.

we know that for a stationary state energy E is fixed and so by above reasoning p must always be fixed for a stationary state as well.

No. First, when you say that "E is fixed" in a stationary state, what you actually mean is that the wave function is an eigenstate of the Hamiltonian $\hat{H}$. But an eigenstate of the Hamiltonian does not have to be an eigenstate of $\hat{p}$. A stationary state must be an eigenstate of $\hat{p}^2$ since that is what appears in $\hat{H}$. But there are many different eigenstates of $\hat{p}$ that are eigenstates of $\hat{p}^2$ with the same eigenvalue (states with the same magnitude of momentum but in different directions), and a superposition of multiple such states will not be an eigenstate of $\hat{p}$, so it won't have a "fixed" momentum, but it will be an eigenstate of $\hat{p}^2$, so it will have a fixed energy.

 

[shankk]

No, the new thing in QM is that $E$ and $p$ are not numbers any more, they're operators. So what you are calling the "kinetic energy" is now a relationship between operators, specifically, that $\hat{T} = \hat{p}^2 / 2 m$. The total energy operator, the Hamiltonian, is then $\hat{H} = \hat{p}^2 / 2m + V$.
No. First, when you say that "E is fixed" in a stationary state, what you actually mean is that the wave function is an eigenstate of the Hamiltonian $\hat{H}$. But an eigenstate of the Hamiltonian does not have to be an eigenstate of $\hat{p}$. A stationary state must be an eigenstate of $\hat{p}^2$ since that is what appears in $\hat{H}$. But there are many different eigenstates of $\hat{p}$ that are eigenstates of $\hat{p}^2$ with the same eigenvalue (states with the same magnitude of momentum but in different directions), and a superposition of multiple such states will not be an eigenstate of $\hat{p}$, so it won't have a "fixed" momentum, but it will be an eigenstate of $\hat{p}^2$, so it will have a fixed energy.

Okay. So that cleared up some of my doubts. Now I think that the relation of $T = E - V = \frac{p^2}{2m}$ still holds but here E is now the eigenvalue of the hamiltonian, p is the positive square root of the eigenvalue of $\hat{p}^2$ and V is the potential energy. But now here, p would depend on x coordinate in case V is a function of x. For the free particle we were able to write p as a constant because V was constant (0). Am I getting it now?

 

[PeroK]

Okay. So that cleared up some of my doubts. Now I think that the relation of $T = E - V = \frac{p^2}{2m}$ still holds but here E is now the eigenvalue of the hamiltonian, p is the positive square root of the eigenvalue of $\hat{p}^2$ and V is the potential energy. But now here, p would depend on x coordinate in case V is a function of x. For the free particle we were able to write p as a constant because V was constant (0). Am I getting it now?

This doesn't look very quantum mechanical to me. You have an equation of expectation values:
$$\langle T \rangle = \langle E \rangle - \langle V \rangle$$ But, in general, the quantities $T, E$ and $V$ are not well-defined in QM (*). As above, you have operators representing these observables.

(*) Well-defined in terms of the state, You can, of course, get single-measurement values for these.

In terms of the free particle, the energy and momentum eigenstates are not physically realisable, so you will always have a superposition of eigenstates (a so-called wave packet).

 

[PeterDonis]

Now I think that the relation of $T = E - V = \frac{p^2}{2m}$ still holds but here E is now the eigenvalue of the hamiltonian

No. The relationship is a relationship between operators, not eigenvalues. There is also a relationship between expectation values, as @PeroK points out, which is implied by the relationship between operators. But expectation values are not the same as eigenvalues.

now here, p would depend on x coordinate in case V is a function of x.

Again, p is an operator, not a number or a function. It makes no sense to say it "depends" on a coordinate.