# Commutator: [E,x]

1. Aug 15, 2009

### Geezer

Okay, I *know* that E and x are supposed to commute, but I'm stuck on one tiny portion when I work through this commutator...

So, here's my work. Feel free to point out my error(s):

$$[E,x]\Psi=(i\hbar\frac{\partial}{\partial t}x-xi\hbar\frac{\partial}{\partial t})\Psi$$

...which becomes...

$$[E,x]\Psi=(i\hbar\frac{\partial x}{\partial t}\Psi+i\hbar x\frac{\partial\Psi}{\partial t}-i\hbar x\frac{\partial\Psi}{\partial t})$$

So, here's my question: must $$\frac{\partial x}{\partial t}\Psi$$

necessarily equal zero? Clearly it should if the commutator is going to be equal to zero...

Also, I recognize that something like dx/dt in quantum is fairly meaningless, but dx/dt is a lot lot <p>/m, which isn't meaningless in quantum...

Feel free to set me straight...

2. Aug 15, 2009

### Pengwuino

First of all, they do not commute (think simple harmonic oscillator). Second, X is not a function. Psi is the function you're acting upon. Third, as one can find in a previous thread on this topic, https://www.physicsforums.com/showthread.php?t=326154 , the time derivative is not a valid operator so commutator in this form makes no sense.

3. Aug 15, 2009

### Hurkyl

Staff Emeritus
Yes. Because $\partial x / \partial t = 0$.

4. Aug 15, 2009

### Hurkyl

Staff Emeritus
However, the time derivative is a valid partial operator on the set of Hilbert-space-valued functions of t.

Last edited: Aug 15, 2009
5. Aug 15, 2009

### Geezer

I posted this last night before going to bed. While brushing my teeth, it occurred to me that $$i\hbar\frac{\partial}{\partial t}$$ is the energy operator associated with the time independent Schrodinger equation; if $$\Psi$$ is "time independent," then its derivative WRT time must be zero.

6. Aug 15, 2009

### Geezer

What's your take on this (link below)? They seem to say that they do commute...

http://quantummechanics.ucsd.edu/ph130a/130_notes/node116.html#example:comEx"

Last edited by a moderator: Apr 24, 2017
7. Aug 15, 2009

### kuruman

Please set me straight. I always thought that E is a constant not an operator. Starting from the Schrodinger equation

$$i \hbar\frac{\partial \Psi }{\partial t} = H \Psi$$

one observes that the partial diff. eq. separates by assuming a solution

$$\Psi(x,t) = \psi (x)e^{- i E t/\hbar}$$

where E is the separation constant. In that case, we get

$$H\psi = E \psi$$

but this does not mean that H = E. Looking at the Schrodinger Equation, it seems to me that the correct way to write the operator is

$$i \hbar\frac{\partial }{\partial t} = H$$

So [E, x] = 0, but [H, x] may or may not be zero.

8. Aug 15, 2009

### Geezer

But $$i\hbar\frac{\partial}{\partial t}$$ is not the full energy for the harmonic oscillator, because the harmonic oscillator is subject to a potential (i.e. V=1/2 kx^2), so this commutator, applied to the harmonic oscillator, doesn't make sense...

(Folks, feel free to let me know if I'm on the right track or not...)

Last edited: Aug 15, 2009
9. Aug 15, 2009

### Geezer

That's what I'm thinking. The Hamiltonian incorporates a potential energy, too....

10. Aug 15, 2009

### Hurkyl

Staff Emeritus
I do not believe defining
$$\hat{E} = i \hbar \frac{\partial}{\partial t}$$​
to be that uncommon. If nothing else, you would need that definition for relativistic QM, because energy is the time-component of the momentum 4-vector.

The equation
$$\hat{H} = i \hbar \frac{\partial}{\partial t}$$​
is definitely wrong. The equation
$$\hat{H} \Psi = i \hbar \frac{\partial \Psi}{\partial t}$$​
is an equation we are seeking to solve -- it is not an identity of operators and vectors.

11. Aug 15, 2009

### Geezer

Yes, I'll agree that H is not an operator, but we often do commutation relations with the Hamiltonian (or, at least with the kinetic energy operator)...

Last edited: Aug 15, 2009
12. Aug 15, 2009

### Hurkyl

Staff Emeritus
Was that a typo?

13. Aug 15, 2009

### Geezer

Whoops...yes, that was a typo. Sorry. I'll correct it.

14. Aug 15, 2009

### kuruman

You have set me straight. Thanks.

15. Aug 15, 2009

### Geezer

Last edited: Aug 15, 2009
16. Aug 18, 2009

### turin

Suppose that |x> and |y> are generic position vectors. I assume that the position vectors span the extended Hilbert space. So, what is <x|E|y>? Do all of these matrix elements vanish? If so, then I don't think that E is a valid operator (in ordinary nonrelativistic QM). What is the context of the problem? Is it relativistic QM? If so, then I don't understand the meaning of x as an operator.

17. Aug 18, 2009

### Hurkyl

Staff Emeritus
E is not an operator on kets. E is a (partial) operator on ket-valued functions on R.

If we use |y> to denote the constant (generalized-)ket-valued function, then <x|E|y> is, indeed, zero-distribution-valued constant function. But the |y> do not span the space of all (generalized-)ket-valued functions.

18. Aug 19, 2009

### turin

I don't think that I understood what you wrote. Some ideas just popped into my head, though. If you don't mind, I would appreciate your comments.

I'm guessing that R is the space of the time parameter. Then, it seems like there exists a different Hilbert space for each value of time. When a Hilbert space operator, such as position or momentum, acts on a Hilbert space, the matrix elements are taken between two elements of that same space. However, E characterizes the difference between two Hilbert spaces, and the matrix elements of E are taken between elements of two different Hilbert spaces. With that said, I am even more confused about the meaning of [E,X], because it can only act in one direction consistently. For instance, E and X can both act to the right on the Hilbert space at t. But then E must act to the left on the Hilbert space at t+dt, while X acts to the left on the Hilbert space at t. So, it seems that the matrix element of this commutator, <ψf|[E,X]|ψi>, does not make sense, because one of |ψi> or |ψf> must be a member of two different Hilbert spaces at the same time. I guess I'm wrong.

Another thought that I had was that I can make an arbitrary change of basis: |x>→eif(x,t)|x>, where I will choose f(x,t) to be real-valued for simplicity of html tags. Then, <y|E|x>→-(∂f/∂t)ei(f(x,t)-f(y,t))<y|x>→-(∂f/∂t)δ(x-y). This can be nonzero, as far as I can tell. However, E is probably represented differently in this new basis, in just such a way that <y|E|x> still vanishes, and what I just wrote is probably incorrect.

19. Aug 22, 2009

### Hurkyl

Staff Emeritus
I think you might be overcomplicating things.

What you're describing is a sort of differential-geometric viewpoint on time-varying kets. While I'm sure that is a reasonable thing to do, it's going to be a lot more complicated, and I'm not sure what benefit you'd get.

If $\psi$ denotes a time-varying ket... that is a function $\mathbb{R} \to \mathcal{H}$, then:

For each t, $\psi(t)$ is a ket to which we can1 apply X. So $X \psi$ makes perfect sense as a time-varying ket: specifically
$(X \psi)(t) = X (\psi(t))$​
and so we can use X as an operator not only on kets, but also on time-varying kets.

It makes no sense to try and apply E to a ket, because it is an operator on time-varying kets:
(E \psi)(t) = i \hbar \psi'(t)​

X and E both make sense as operators acting on time-varying kets. Since $X \psi$ is a time-varying ket, $E X \psi$ makes sense. Similarly, $X E \psi$ makes sense, as does $[E,X]\psi$.

By considering conjugation, one can make similar statements for them as acting on time-varying bras.

In short, applying X makes sense, because they are ket-valued, and applying E makes sense because they are functons of time.

If you're going to stick with the differential-geometric picture, you're going to have to put in place some extra scaffolding.

You've replaced time-varying kets with "ket fields" over R.
$\partial / \partial t$ is going to be an ordinary tangent vector field on R.
X is an operator field -- essentially the same thing as a rank (1,1) tensor. (But as applied to kets, not as to tangent vectors)

But you need a new thing -- there is some sort of connection on your vector bundle. This is an object that takes a ket field and a tangent vector field, and returns something that captures the idea of a "directional derivative" of the ket field along the tangent vector field. E would be the differential operator formed by feeding $\partial / \partial t$ into that connection (multiplied by the appropriate constant).

1: I'm neglecting issues about the domain of X here.

Last edited: Aug 22, 2009
20. Aug 22, 2009

### turin

OK, I understand now to my level of satisfaction. Thank you very much, Hurkyl! However, now the commutator seems trivial; basically like a commutator of a c-number (i.e. that it is simply defined to vanish, and the only reason I was confused was that I was trying to prove a definition).

Most likely.

That one takes a little bit of convincing on my part, but I think I can accept that. I am thinking of Eψ(t) as proportional to the sum of two Hilbert space vectors: the vector ψ(t+δt) and the vector inverse of ψ(t), with part of the proportionality factor being 1/δt. I argue to myself that this sum must also be a Hilbert space element (because the Hilbert space is a vector space), so X is allowed to act on Eψ(t). The only thing that I am a little bit uneasy about in your whole argument, then, is convincing myself that the limiting procedure δt→0 does not disturb the proportionality. I guess I could argue physically that this limit must be well-defined (but then I worry about wierd things like wave-function collapse). I believe that's just something that I need to think about (although your insight is thouroughly appreciated)!.

I suspected as much.