Quantum commutator problems

1. Jun 6, 2012

Qiward

I now show some derivations regarding quantum commutators,leading to some inconsistencies. Can someone tell what went wrong? What causes the inconsistencies? and what is the correct way of understanding/handling the concepts?

Issue # 1 - Hamiltonian and commutation with time

(1) The Ehrenfest Theorem: $\frac{d<A>}{dt}$ = $\frac{1}{i\hbar}$<[A, H] + <$\frac{∂A}{∂t}$>

(2) The Heisenberg Equation: $\frac{d<A(t)>}{dt}$ = $\frac{1}{i\hbar}$<[A(t), H] + <$\frac{∂A(0)}{∂t}$>

Now, when A or A(0) does not depend on time t explicitly, we have the usual form [A, H] = i$\hbar$$\frac{dA}{dt}$.

Question #1: let's let A = t, therefore, $\frac{dA}{dt}$=$\frac{∂A}{∂t}$ = 1. Plugging them in the two formulas above, we have [t, H] = 0, instead of the correct form [t, H] = i$\hbar$. What's wrong?

Issue # 2 - Momemtum and position commutators

$\frac{d<A>}{dx}$ = ($\frac{∂<|}{∂x}$)A|> + <$\frac{∂A}{∂x}$> + <|A($\frac{∂|>}{∂x}$) = <|-$\frac{i}{\hbar}p$A|> + <$\frac{∂A}{∂x}$> + <|A$\frac{i}{\hbar}p$|> = $\frac{1}{i\hbar}[p, A]$ + <$\frac{∂A}{∂x}$>

The correct (or usual) form is $[p, A]$ = $-i\hbar$$\frac{∂A}{∂x}$. Getting this answer leads to my questions below.

Question #2: Does this mean that we have to force $\frac{d<A>}{dx}$ = 0?!

Question # 3: Let A = x, similar to issue #1, we would have $[p, x]$ = 0?! because $\frac{dx}{dx}$ = $\frac{∂x}{∂x}$ = 0.

Again, what is wrong here? Thanks for any hint or discussion.

Question #4 - a minor question: Why do we normally have H = $i\hbar\frac{d}{dx}$, while $p = i\hbar \frac{∂}{∂x}$? In other words, why one is full derivative while the other partial? I think the answer to this question is closely related to the derivations I gave above.

2. Jun 6, 2012

Matterwave

Issue #1: There is no such "time operator" in quantum mechanics. The time is simply a parameter. It's not an observable. Therefore, it makes no sense to take A=t since A must be an operator.

Issue #2: I'm not following your derivation. Also, where's the second half of the commutator? I only see the pA term and not the -Ap term.

Question #2: Dunno, since I can't follow what you're doing.

Question #3: dx/dx=1 and not 0.

Question #4: The Hamiltonian is usually not what you write, but the derivative is taken with respect to t, not x, and it's usually a partial derivative as well.

3. Jun 6, 2012

Qiward

Question #1: time t is not an observable? why? Can you shed some light? If p and x are observables, then t should be observable as well since p = mdx/dt. In addition, t = x/c if you use speed of light as a unit. In other words, from four-vector's perspective, it seems to me there is no foundamental difference between t and x. Am I wrong?

Question #3: My mistake. I meant $\frac{dx}{dx}$ = $\frac{∂x}{∂x}$ = 1, and therefore [p, x]=0. This is the crux. (See re-derivation at the bottom)

Question #4: My mistake again. I really meant, as used in many textbooks such as Sakurai's, why is Hamiltonian normally full derivative w.r.t. time, while p is partial derivative w.r.t. to x. (I am ignoring the unit i*hbar.)

In Issue #2's derivation, < or <| means wave function ψ*, and > or |> means ψ.

$\frac{d<A>}{dx}$ = ($\frac{∂<ψ|}{∂x}$)A|ψ> + <ψ|$\frac{∂A}{∂x}$|ψ> + <ψ|A($\frac{∂|ψ>}{∂x}$) = <ψ|$\frac{-i}{\hbar}$pA|ψ> + <ψ|$\frac{∂A}{∂x}$|ψ> + <ψ|A$\frac{i}{\hbar}p$|ψ>
=$\frac{i}{\hbar}$<ψ|[A, p]|ψ> + <ψ|$\frac{∂A}{∂x}$|ψ>

The common form is [A, p] = $i\hbar\frac{∂A}{∂x}$. Getting this answer leads to my questions below.

Question #2: Does this mean that we have to force d<A>dx = 0?!

Question # 3: Let A = x, similar to issue #1, we would have [p,x] = 0?! because $\frac{dx}{dx}$ = $\frac{∂x}{∂x}$ = 1 on both sides of the above "long" derivation, they cancel out to give [x, p] = 0?! Again, any problem? I just feel something is wrong.

4. Jun 6, 2012

Matterwave

1) The "time" is treated differently than position in non-relativistic quantum mechanics just as it is treated differently in non-relativistic classical mechanics. If you want to go to the relativisic limit, you adopt quantum field theory rather than standard quantum mechanics. In quantum field theory, x and t are both just parameters (i.e. x is no longer an operator). In standard quantum mechanics, x is an operator while t is not.

2) Can you give a source for your "common form" of [A,p]? I'm still only seeing the pA term on the right hand side. I would have thought that:

$$[A,p]=i\hbar\left(A\frac{\partial}{\partial x}-\frac{\partial}{\partial x} A\right)$$

In fact, I don't believe your quoted formula works out, for if A=p itself then your formula suggests:

$$[p,p]=(i\hbar)^2\frac{d^2}{dx^2}$$
Which makes no sense.

Your "derivation" looks fine to me - it looks just like Ehrenfest's theorem except you replaced the translations in time with translations in space.

3) Perhaps your error is that you are looking at $\frac{dx}{dx}$ rather than $\frac{d\langle x\rangle}{dx}$. The expectation value of x is not just x, it is just some number (possibly dependent on t). My feeling is that you are erroring by switching back and forth from having expectation values and not having expectation values.

So, if you want to set A=x, you would have:

$$\frac{d\langle x\rangle}{dx}=\frac{i}{\hbar}\langle [x,p]\rangle+\langle 1\rangle=-1+1=0$$

I plugged in the usual commutator [x,p]=ihbar.

Last edited: Jun 6, 2012
5. Jun 7, 2012

Qiward

Thanks a lot for your explanation. If t and x are what you said, then I guess non-relativistic QM is a bit too ad hoc, and hence not a satisfactory theory to me. I will study quantum field theory. Thanks for the info you give.

Regarding my derivation, there are indeed both Ap and pA terms. That's why I had [A, p] in the end. They show on my computer and I don't know why one is missing on yours.

Two more questions:

(1) Can one operator operate on another? For example, p and x are both operators. <ψ|px|ψ> = <ψ|(-$i\hbar$$\frac{∂x}{∂x}$|)ψ> + <ψ|x(-$i\hbar$$\frac{∂}{∂x}$|ψ>) = -$i\hbar$ + <ψ|xp|ψ>. Do you agree with my derivation? Notice, p operating on x and I have assumed both to be operators.

(2) In your response, at the end you gave $\frac{d<x>}{dx}$ = 0. How do you explain this result physically? The picture I have is ----- (a) the particle moves dx in distance; or (b) the reference system shifts by -dx. Ignoring the negative sign. In either case, the average position <x> should change accordingly. Thus, d<x>/dx is non-zero.

Now, the only thing that may undermine my logic above is ----- wave functions ψ and *ψ will also change as a result. I just don't know whether the change in ψ will be so such that d<x>/dx = 0 is always held.

Thanks very much!

6. Jun 8, 2012

Matterwave

You don't really "operate on the operator", you operate one operator and then you operate the second one. So, something like px|psi> should be interpreted as acting x on the ket and then acting p on the resulting ket, you don't "act p on x" like you seem to be trying to do.

Of course, if you are working with wave functions in position space, then the x operator is simply "multiply by x (the variable)", and so you act p on that.

For example, for a wave function f(x), we have:
$$\hat{p}\hat{x}f(x)=i\hbar\frac{\partial}{\partial x}[xf(x)]$$

When I say "you are missing the Ap term", I'm not talking about in your derivation, but in your statement "The usual form of [A,p] is...". My previous post has my objection to that equation.

<x> is just a number. It is the average value of x for the given wave function. It's a number like 4 or 5. It's not a variable like x. Therefore d<x>/dx=0 just like d(4)/dx=0.

7. Jun 14, 2012

Qiward

MatterWave, Great. You have made this clear now. I appreciate your convincing explanation.