Solving Commutator [E,x]: Stuck on One Tiny Portion

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In summary: I think E is usually commuted with the energy as well, but I'm not positive).In summary, E and x are supposed to commute, but I'm stuck on one tiny portion when I work through this commutator...E and x do not commute according to the given equation. Commutation relations between the Hamiltonian and the energy are not always straightforward.
  • #1
Geezer
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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):

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

...which becomes...


[tex][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})[/tex]

So, here's my question: must [tex]\frac{\partial x}{\partial t}\Psi[/tex]

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...
 
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  • #2
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
Geezer said:
So, here's my question: must [tex]\frac{\partial x}{\partial t}\Psi[/tex]

necessarily equal zero?
Yes. Because [itex]\partial x / \partial t = 0[/itex].
 
  • #4
Pengwuino said:
the time derivative is not a valid operator so commutator in this form makes no sense.
However, the time derivative is a valid partial operator on the set of Hilbert-space-valued functions of t.
 
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  • #5
I posted this last night before going to bed. While brushing my teeth, it occurred to me that [tex]i\hbar\frac{\partial}{\partial t}[/tex] is the energy operator associated with the time independent Schrodinger equation; if [tex]\Psi[/tex] is "time independent," then its derivative WRT time must be zero.
 
  • #6
Pengwuino said:
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.

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"
 
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  • #7
Please set me straight. I always thought that E is a constant not an operator. Starting from the Schrodinger equation

[tex]i \hbar\frac{\partial \Psi }{\partial t} = H \Psi[/tex]

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

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

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

[tex]H\psi = E \psi[/tex]

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

[tex]i \hbar\frac{\partial }{\partial t} = H [/tex]

So [E, x] = 0, but [H, x] may or may not be zero.
 
  • #8
Pengwuino said:
First of all, they do not commute (think simple harmonic oscillator).

But [tex]i\hbar\frac{\partial}{\partial t}[/tex] 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...)
 
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  • #9
kuruman said:
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

[tex]i \hbar\frac{\partial }{\partial t} = H [/tex]

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

That's what I'm thinking. The Hamiltonian incorporates a potential energy, too...
 
  • #10
I do not believe defining
[tex]\hat{E} = i \hbar \frac{\partial}{\partial t}[/tex]​
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
[tex]\hat{H} = i \hbar \frac{\partial}{\partial t}[/tex]​
is definitely wrong. The equation
[tex]\hat{H} \Psi = i \hbar \frac{\partial \Psi}{\partial t}[/tex]​
is an equation we are seeking to solve -- it is not an identity of operators and vectors.
 
  • #11
Hurkyl said:
I do not believe defining
[tex]\hat{E} = i \hbar \frac{\partial}{\partial t}[/tex]​
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
[tex]\hat{H} = i \hbar \frac{\partial}{\partial t}[/tex]​
is definitely wrong. The equation
[tex]\hat{H} \Psi = i \hbar \frac{\partial \Psi}{\partial t}[/tex]​
is an equation we are seeking to solve -- it is not an identity of operators and vectors.

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)...
 
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  • #12
Geezer said:
Yes, I'll agree that [tex]\hat{H}[/tex] is not an operator
Was that a typo?
 
  • #13
Hurkyl said:
Was that a typo?

Whoops...yes, that was a typo. Sorry. I'll correct it.
 
  • #14
Hurkyl said:
I do not believe defining
[tex]\hat{E} = i \hbar \frac{\partial}{\partial t}[/tex]​
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
[tex]\hat{H} = i \hbar \frac{\partial}{\partial t}[/tex]​
is definitely wrong. The equation
[tex]\hat{H} \Psi = i \hbar \frac{\partial \Psi}{\partial t}[/tex]​
is an equation we are seeking to solve -- it is not an identity of operators and vectors.

You have set me straight. Thanks.
 
  • #15
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  • #16
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
turin said:
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.

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
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
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 [itex]\psi[/itex] denotes a time-varying ket... that is a function [itex]\mathbb{R} \to \mathcal{H}[/itex], then:

For each t, [itex]\psi(t)[/itex] is a ket to which we can1 apply X. So [itex]X \psi[/itex] makes perfect sense as a time-varying ket: specifically
[itex](X \psi)(t) = X (\psi(t))[/itex]​
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 [itex]X \psi[/itex] is a time-varying ket, [itex]E X \psi[/itex] makes sense. Similarly, [itex]X E \psi[/itex] makes sense, as does [itex][E,X]\psi[/itex].

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.
[itex]\partial / \partial t[/itex] 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 [itex]\partial / \partial t[/itex] into that connection (multiplied by the appropriate constant).

1: I'm neglecting issues about the domain of X here.
 
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  • #20
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).

Hurkyl said:
I think you might be overcomplicating things.
Most likely.

Hurkyl said:
... [itex]X E \psi[/itex] makes sense, ...
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 weird things like wave-function collapse). I believe that's just something that I need to think about (although your insight is thouroughly appreciated)!.

Hurkyl said:
But you need a new thing -- there is some sort of connection on your vector bundle.
I suspected as much.
 
  • #21
turin said:
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.
Right(ish) -- the ordinary definition of the derivative of a vector-valued function on R works no matter what the target vector space is. E is given by
[tex](E \psi)(t) = i \hbar \lim_{h \to 0} \frac{\psi(t+\epsilon) - \psi(t)}{\epsilon}[/tex]

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).
And like ordinary derivatives, some functions won't be differentiable. And some of those times, the derivative will exist in a larger space -- e.g. a time-varying wavefunction might be time differentiable, but the derivative might not be square-integrable.

The operators P and X have the same issues, even without involving time-varying stuff. [itex]X\psi[/itex] and [itex]P\psi[/itex] are undefined for many kets. And sometimes, they will be defined for wavefunctions, but the result won't be square-integrable.


Anyways, it is true that since this derivative appears in the Schrödinger equation, any non-differentiable time-varying ket cannot represent the time evolution of a quantum system.




By the way, I would like to retract an earlier statement -- it turns out that for a time-varying bra [itex]\langle \psi |[/itex], [itex]\langle \psi | E[/itex] is almost never a time-varying bra. Fortunately, we don't actually have to worry about what kind of object it is, as long as we focus on how E applies to time-varying kets.
 

1. What is a commutator in relation to solving equations?

A commutator is a mathematical operation used to measure the difference between two operations. In the context of solving equations, a commutator [E,x] is used to measure the difference between two equations, with x being the variable in question.

2. How do you solve a commutator [E,x]?

To solve a commutator [E,x], you must first understand the properties of commutators. Then, you can use algebraic techniques such as expanding and simplifying to manipulate the equations and isolate the variable x.

3. What does it mean to be stuck on one tiny portion of a commutator [E,x]?

Being stuck on one tiny portion of a commutator [E,x] means that you are unable to manipulate the equations further to solve for the variable x. This could be due to a lack of understanding of commutators or a more complex equation that requires advanced techniques to solve.

4. Can you provide an example of solving a commutator [E,x]?

Sure, let's say we have the commutator [2x,3x]. We can expand this to (2x)(3x) and simplify to 6x^2. This shows that the difference between 2x and 3x is 6x^2. From here, we can use algebraic techniques to solve for x.

5. Why are commutators important in solving equations?

Commutators are important in solving equations because they allow us to measure the difference between two operations, which can help us manipulate the equations and solve for the variable. They also have applications in other areas of mathematics and physics, such as in quantum mechanics.

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