Bra-ket of uncertainty commutator (Sakurai 1.18)

[Silicon-Based]

Homework Statement

Prove that the equality sign in the generalized uncertainty relation holds if the state in question satisfies $\Delta A|\alpha\rangle = \lambda \Delta B|\alpha\rangle$ with $\lambda$ purely imaginary.

Relevant Equations

$\langle(\Delta A)^2\rangle \langle(\Delta B)^2\rangle \geq \frac{1}{4}|\langle \alpha |[A,B]|\alpha\rangle|^2$

$\langle \alpha |[A,B]|\alpha\rangle = \langle \alpha |\Delta A \Delta B - \Delta B \Delta A|\alpha\rangle$

It's easy to show that $[\Delta A, \Delta B] = [A,B]$. I'm specifically having issues with evaluating the bra-ket on the RHS of the uncertainty relation:

$\langle \alpha |[A,B]|\alpha\rangle = \langle \alpha |\Delta A \Delta B - \Delta B \Delta A|\alpha\rangle$
​

The answer is supposed to be $-2\lambda \langle \alpha |(\Delta B)^2|\alpha\rangle$, but I don't know how to evaluate the bra-ket to reach it; do I apply $\Delta A \Delta B$ and $\Delta B \Delta A$ each to both the ket and the bra, or just the ones that are immediately associated with them, i.e. not the one separated by the minus sign? If it's the former what do I do with e.g. the term $\Delta A \Delta B|\alpha \rangle$? The trick was supposed to be to notice that $\langle\alpha| \Delta A = -\lambda \langle\alpha|\Delta B$, but I seem to lack a more fundamental understanding. Any help is appreciated.

 

[tnich]

Homework Statement : . . . $\lambda$ purely imaginary.

I think you could start with
$\langle \alpha |\Delta A \Delta B - \Delta B \Delta A|\alpha\rangle = \langle \alpha |\Delta A \Delta B|\alpha\rangle - \langle \alpha |\Delta B \Delta A|\alpha\rangle$
and substitute for $\Delta A|\alpha\rangle$ (or its conjugate) in each term.

 

[Silicon-Based]

I think you could start with
$\langle \alpha |\Delta A \Delta B - \Delta B \Delta A|\alpha\rangle = \langle \alpha |\Delta A \Delta B|\alpha\rangle - \langle \alpha |\Delta B \Delta A|\alpha\rangle$
and substitute for $\Delta A|\alpha\rangle$ (or its conjugate) in each term.

So if I do that I do indeed get the required result, provided I only find the eigenvalues of the kets/bras that associate with $\Delta A$, so I neglect the example I've given before. Now I only need an explanation why I'm allowed to do this.

 

[tnich]

so I neglect the example I've given before.

You seem to be applying unnecessary restrictions on bra-ket algebra. Maybe it would help you to convert the problem to matrix algebra in Hilbert space. There you can apply the normal rules of linear algebra and see how they convert to bra-ket notation.

 

[Silicon-Based]

You seem to be applying unnecessary restrictions on bra-ket algebra. Maybe it would help you to convert the problem to matrix algebra in Hilbert space. There you can apply the normal rules of linear algebra and see how they convert to bra-ket notation.

I think that would confuse me more. I'm only asking why I need not consider $\Delta A \Delta B|\alpha \rangle$. (Or maybe I do and there is something else I don't see.)

 

[tnich]

I think that would confuse me more. I'm only asking why I need not consider $\Delta A \Delta B|\alpha \rangle$. (Or maybe I do and there is something else I don't see.)

The associative property still works. There is no need to associate $\langle \alpha |\Delta A \Delta B|\alpha\rangle$ as $(\langle \alpha |)(\Delta A \Delta B|\alpha\rangle)$ when you can do it as $(\langle \alpha |\Delta A )(\Delta B|\alpha\rangle)$.
You seem to think that you would be ignoring some eigenvalues of $\Delta A \Delta B$. Have you considered how the eigenvalues of $\Delta A$ are related to those of $\Delta B$ and $\Delta A \Delta B$?

 

[Silicon-Based]

The associative property still works. There is no need to associate $\langle \alpha |\Delta A \Delta B|\alpha\rangle$ as $(\langle \alpha |)(\Delta A \Delta B|\alpha\rangle)$ when you can do it as $(\langle \alpha |\Delta A )(\Delta B|\alpha\rangle)$.
You seem to think that you would be ignoring some eigenvalues of $\Delta A \Delta B$. Have you considered how the eigenvalues of $\Delta A$ are related to those of $\Delta B$ and $\Delta A \Delta B$?

Thanks, that is very helpful. Does that mean that for a general operator $A$ acting on a bra-ket, I can choose it to act only on the ket or the bra and not on both?

 

[tnich]

Thanks, that is very helpful. Does that mean that for a general operator $A$ acting on a bra-ket, I can choose it to act only on the ket or the bra and not on both?

I would not interpret it that way. You can decide which one to apply it to first. If you have $\langle \alpha |A|\beta\rangle$, and $\langle \alpha |A = \langle \delta$ and $|\psi\rangle= A|\beta\rangle$. Then you can either say $\langle \alpha |A|\beta\rangle= \langle \delta|\beta\rangle$ or $\langle \alpha |A|\beta\rangle=\langle \alpha |\psi\rangle$. But $\langle \delta|\beta\rangle = \langle \alpha |\psi\rangle$.
I suggested that you look at it in Hilbert space because there the all of these operations become multiplications of vectors and matrices. $A$ is a matrix, the bras are row vectors, and the kets are column vectors. If you can think of it like that, then it might not be so confusing.

 

[Silicon-Based]

I would not interpret it that way. You can decide which one to apply it to first. If you have $\langle \alpha |A|\beta\rangle$, and $\langle \alpha |A = \langle \delta$ and $|\psi\rangle= A|\beta\rangle$. Then you can either say $\langle \alpha |A|\beta\rangle= \langle \delta|\beta\rangle$ or $\langle \alpha |A|\beta\rangle=\langle \alpha |\psi\rangle$. But $\langle \delta|\beta\rangle = \langle \alpha |\psi\rangle$.
I suggested that you look at it in Hilbert space because there the all of these operations become multiplications of vectors and matrices. $A$ is a matrix, the bras are row vectors, and the kets are column vectors. If you can think of it like that, then it might not be so confusing.

I'm aware of the row/column vector representation, but I didn't put much attention in thinking about how these change under those operations, so I'll try to look into it. Could you elaborate on what you mentioned regarding eigenvalues. I don't like the language of eigenvalues, bases etc. when describing those things, which is what Sakurai does all the time, and it doesn't make a lot of sense to me.