Quantum particle in a magnetic field

[Oxvillian]

Perhaps a summary is in order?

(1) $A(x)$ is a c-number;

(2) $A(\hat{x})$ is an operator;

(3) In the x-representation, $\langle x \lvert A(\hat{x}) \lvert x'\rangle = A(x)\delta(x-x')$;

(4) Sometimes it is said that "in the x-representation, $A(\hat{x})$ is represented by $A(x)$", when what is really meant is (3);

(5) [irrelevant to this thread] In field theory, $\hat{A}(x)$ is an operator, and this time $x$ is simply a label.

Anyone agree/disagree?

 

[Fredrik]

I agree with that, in particular that what you're saying in (1) and (2) is appropriate, but we haven't been sticking to that in this thread. We have been using notations like A(x) for an operator, and we haven't always been concerned about what x really is.

The idea behind the notations in (1) and (2) is that we can take any nice enough function $A:\mathbb R\to\mathbb C$, and then assign a meaning to $A(\hat x)$ by some method. I think that in the Sakurai-style argument, that method would be $A(\hat x)=\int\mathrm dx\, A(x)|x\rangle\langle x|$. This definition ensures that
$$A(\hat x)|x\rangle =\int\mathrm dx' A(x')|x'\rangle\langle x'|x\rangle =\int dx'\, A(x')|x'\rangle \delta(x'-x) =A(x)|x\rangle.$$ Does that make sense? I feel weird about using the delta function like this when we're dealing with a "Hilbert space valued" function like $x\mapsto A(x)|x\rangle$...and I'm using the term "Hilbert space" very loosely here, since the $|x\rangle$ aren't elements of a Hilbert space. I'm a bit concerned about the fact that with this definition of $A(\hat x)$, we have
$$\langle x'|A(\hat x)|x\rangle =\int\mathrm dx''A(x'')\delta(x'-x'')\delta(x''-x),$$ and products of distributions are problematic. But this isn't supposed to be rigorous anyway, so maybe I should just stop worrying and treat the first delta as a function. Then we can continue $=A(x)\delta(x'-x)$.

 

[Oxvillian]

We have been using notations like A(x) for an operator

I think that's the central problem in this thread. You can't possibly characterize an operator using $A(x)$ because $A(x)$ is the continuous version of a row vector, not of a matrix. You need instead to specify the matrix element $\langle x \lvert \hat{A} \lvert x' \rangle$, which involves two parameters, $x$ and $x'$.

The idea behind the notations in (1) and (2) is that we can take any nice enough function $A:\mathbb R\to\mathbb C$, and then assign a meaning to $A(\hat x)$ by some method. I think that in the Sakurai-style argument, that method would be $A(\hat x)=\int\mathrm dx\, A(x)|x\rangle\langle x|$. This definition ensures that
$$A(\hat x)|x\rangle =\int\mathrm dx' A(x')|x'\rangle\langle x'|x\rangle =\int dx'\, A(x')|x'\rangle \delta(x'-x) =A(x)|x\rangle.$$ Does that make sense? I feel weird about using the delta function like this when we're dealing with a "Hilbert space valued" function like $x\mapsto A(x)|x\rangle$...and I'm using the term "Hilbert space" very loosely here, since the $|x\rangle$ aren't elements of a Hilbert space. I'm a bit concerned about the fact that with this definition of $A(\hat x)$, we have
$$\langle x'|A(\hat x)|x\rangle =\int\mathrm dx''A(x'')\delta(x'-x'')\delta(x''-x),$$ and products of distributions are problematic. But this isn't supposed to be rigorous anyway, so maybe I should just stop worrying and treat the first delta as a function.

I think the apparent lack of rigor in this thread is notational only. That is to say, we could if we wanted to replace all the delta "functions" and integrals with appropriate well-defined statements about distributions, and even mathematicians would be happy.

This is to be contrasted with the situation in quantum field theory, where it's still not clear whether certain statements physicists write down have a well-defined mathematical meaning

Then we can continue $=A(x)\delta(x'-x)$.

I vote for that

 

[Fredrik]

I think the apparent lack of rigor in this thread is notational only. That is to say, we could if we wanted to replace all the delta "functions" and integrals with appropriate well-defined statements about distributions, and even mathematicians would be happy.

Yes, I wouldn't be really surprised if it's possible to make sense of every step of the Sakurai-style calculation, and if it's not, then there's a similar calculation based on a fancy spectral theorem that yields the same result. A Sakurai-style calculation is almost certainly better (in the sense of being more similar to what a mathematician would do) than what I did in the other thread. Basically I just pretended that $A(\hat x)$ can be expressed as a power series and then used the commutation relations.

So one option for the OP (who I assume is not able to cover 800 pages of topology and functional analysis this week) is to take a look at Sakurai, and learn the way of doing calculations that's taught there. Perhaps one of the other QM books is better on this, but I just looked in Ballentine, and it looks like he doesn't explain this result at all.

 

[ChrisVer]

then I think the problem/misconception is that you see $\psi(x)$ as a number, while I see it as a wavefunction or let's say generally a function of $x$.

Operators by definition act on some functions and give some other function as a result...So when the $A(\hat{x})$ will act on the function $\psi(x)$ it's supposed to give some function... this function is supposed to be $A(x) \psi(x)$, because exactly as you also pointed out $\hat{x}\psi(x) = x \psi(x)$ and this can go on for any power of $x$.. The last function is only dependent on $x$...
As for changing everything to 3D nothing much changes, you just need to write all the time $\vec{x}$ instead... Although it doesn't play a role, because even your wavefunction is $\psi(\vec{x})$ and thus you will have $\nabla \vec{A}(\vec{x}) \psi(\vec{x})$ ... I don't think that you have any anisotropy so I guess the result is the triple of what you found in 1D...
That's also why I don't understand the terminology of "operator acts on operator", although someone can use that thing (for example you can work the commutation relations like that) one should always keep in mind that something will be acted on later...
However I'll have a look at Sakurai...