Discrepancy in uncertainty among various operators

[hokhani]

TL;DR

Momentum uncertainty isn't identical for position and Hamiltonian.

Consider a particle in an eigenstate of the Hamiltonian when the particle is in a 1D box with very large length $L$. The operators $\hat{x}$, $\hat{p}$ and $\hat{H}$ don't commute together. If we consider the uncertainty for Hamiltonian and momentum, then we have $\Delta p→\infty$ but if we take uncertainty between momentum and position we have $\Delta p→0$. I would be grateful if help me what is wrong here?

 

[anuttarasammyak]

Momentum and coordinate are conjugate variables which satisfy the Kennard inequality. Hamiltonian and momentum, Hamiltonian and coordinate are not conjugate variable pairs.

 

[Lord Jestocost]

Momentum uncertainty isn't identical for position and Hamiltonian

Maybe, this can help. Check chapter “10.2 Compatibility” in the lecture “Commutators and Uncertainty Principle”.

Durham University: https://www.maths.dur.ac.uk/users/kasper.peeters/mathphys/uncertainty.html

 

[PeterDonis]

If we consider the uncertainty for Hamiltonian and momentum, then we have $\Delta p→\infty$ but if we take uncertainty between momentum and position we have $\Delta p→0$.

How are you obtaining these statements? Please show your work.

 

[hokhani]

How are you obtaining these statements? Please show your work.

The uncertainty is given by: $$\langle (\Delta A)^2\rangle\langle (\Delta B)^2\rangle \ge \frac{1}{4} |\langle [A,B]\rangle|^2$$ with $\Delta A=A-\langle A\rangle$.
If the particle is in the $n$th energy eigenstate, we have $\langle H \rangle =0$. Then taking $A=H$ and $B=p$ and knowing that $[H,p]\ne0$ we have, $\langle (\Delta H)^2 \rangle \to 0$ and so, $\langle (\Delta p)^2 \rangle\to \infty$.
If we take $A=x$ and $B=p$, since $[x,p]=i\hbar$ and $\langle (\Delta x)^2 \rangle =L^2( \frac{1}{12}-\frac{1}{8n^2 \pi^2})\approx L^2$, for large $L$ we have $\langle (\Delta p)^2 \rangle \to 0$.

P.S. Sorry, in the main post I used $\langle \Delta A\rangle$ instead of $\langle (\Delta A)^2 \rangle$.

 

[PeterDonis]

If the particle is in the $n$th energy eigenstate, we have $\langle H \rangle =0$.

And this invalidates the rest of your reasoning regarding $\Delta p$ being infinite. The correct conclusion from $\langle H \rangle = 0$ is not that $\Delta p \to \infty$. It is that the uncertainty relations are undefined for this case. (Note that, unfortunately, many QM textbooks gloss over this and blithely state, for example, that in a position eigenstate, "momentum uncertainty is infinite", and vice versa, instead of correctly telling you that the uncertainty relations are undefined for these cases.)

In other words, in an energy eigenstate, we simply can't use any commutators with $H$ to evaluate the uncertainty in another observable. So your reasoning that says $\Delta p \to \infty$ for this case is simply not valid. You can't conclude anything about $\Delta p$ from the uncertainty relation involving $H$ for this case.

Note that similar remarks apply to $\Delta x$; by your reasoning, that should also be infinite in an energy eigenstate since $[H, x] \neq 0$. But of course you recognize that it isn't, since you go on to use finite values for $\Delta x$ in order to take the limit as $L$ gets very large. If your claim about $\Delta p$ being infinite were correct, that claim would also apply to $\Delta x$, which would have to be infinite no matter what the value of $L$ was.

In fact, the only uncertainty relation that is useful in this case is the one between $\Delta x$ and $\Delta p$, which you correctly use to argue that $\Delta p$ gets very small as $L$ gets very large. But again, to use that reasoning at all, you have to first ignore the uncertainty relations of both $x$ and $p$ with $H$, since both of them are undefined in this case, as stated above.

 

[Matterwave]

In fact, the only uncertainty relation that is useful in this case is the one between $\Delta x$ and $\Delta p$, which you correctly use to argue that $\Delta p$ gets very small as $L$ gets very large.

You can't use uncertainty relations to argue a single uncertainty will become small though? The uncertainty relation only gives a lower bound, nothing about it says you need to hit that lower bound. Perhaps you meant "can get very small"?

 

[PeterDonis]

You can't use uncertainty relations to argue a single uncertainty will become small though?

You can't argue anything from an uncertainty relation where one of the factors is zero. That breaks the math.

 

[PeterDonis]

The uncertainty relation only gives a lower bound

Yes, it assumes that the only source of uncertainty is that inherent in the quantum wave function. That's the argument I understood the OP to be making in that case. In practice you're right, the actual uncertainties will often be much larger because of other factors that have nothing to do with QM.

 

[Matterwave]

You can't argue anything from an uncertainty relation where one of the factors is zero. That breaks the math.

I fully agree with the rest of your post. My only concern was about the quoted part.

In practice you're right, the actual uncertainties will often be much larger because of other factors that have nothing to do with QM.

But even purely in QM, the generalized uncertainty relation $\sigma_a\sigma_b \geq \frac{1}{2}\langle [A,B]\rangle$ is an inequality and most quantum states won't attain the minimum. My contention is that you can't use this inequality (by itself) to correctly argue $\sigma_a$ or $\sigma_b$ "will become small" as the OP tried to do in writing $\Delta p\rightarrow 0$.

 

[PeterDonis]

even purely in QM, the generalized uncertainty relation $\sigma_a\sigma_b \geq \frac{1}{2}\langle [A,B]\rangle$ is an inequality and most quantum states won't attain the minimum.

That's true. I haven't computed the actual deltas from the energy eigenstates the OP is using, but of course that could be done to see where they actually fall. I would expect the ground state $n = 0$ to have the minimum uncertainty, so the OP's reasoning would apply there, but not necessarily the excited states. But you're right that that should be checked by doing the actual computations.

 

[anuttarasammyak]

(10.18) in post #3 link suggests that in energy eigenstates $<V’>=0$. Force by PE is zero in average. Thus

$$0\cdot\delta \ge 0$$
has no information about $\delta$.

(10.19) suggests that in energy eigenstates $<p>=0$. In changing IFR, e.g., particle in a moving box, it does not stand. $\triangle x=\infty$ in motion could recover it.

 

[hokhani]

And this invalidates the rest of your reasoning regarding $\Delta p$ being infinite. The correct conclusion from $\langle H \rangle = 0$ is not that $\Delta p \to \infty$.

Thank you very much. It seems also that the right of the inequality is zero, i.e., $\langle [H,p]\rangle =0$ which strengthens more your reasoning. However, I am not sure if it is correct to take $H=\frac{p^2}{2m}$.

Note that similar remarks apply to $\Delta x$; by your reasoning, that should also be infinite in an energy eigenstate since $[H, x] \neq 0$.

With similar reasoning it seems that also $\langle [H,x]\rangle =0$.

 

[Matterwave]

However, I am not sure if it is correct to take $H=\frac{p^2}{2m}$.

It's right to be suspicious here since that's just the free Hamiltonian. Where did the "in a box" constraint go? Hint: boundary conditions are important.

 

[hokhani]

It's right to be suspicious here since that's just the free Hamiltonian. Where did the "in a box" constraint go? Hint: boundary conditions are important.

The boundary conditions enter through the wave function which is zero outside the box.

 

[PeterDonis]

I am not sure if it is correct to take $H=\frac{p^2}{2m}$.

That's only correct for a free particle, i.e., zero potential. That's not the case in your example; you have a potential confining the particle to a finite region of space.

The boundary conditions enter through the wave function which is zero outside the box.

You can't just wave your hands and arbitrarily say the wave function is zero outside the box. The wave function has to be a solution of Schrodinger's Equation. A solution which is zero outside a finite region is only a solution of Schrodinger's Equation with a nonzero potential.

 

[hokhani]

That's only correct for a free particle, i.e., zero potential. That's not the case in your example; you have a potential confining the particle to a finite region of space.

Right, I didn't consider it. But at least one can say that the right part of the inequality is also undefined.