Bra-ket of uncertainty commutator (Sakurai 1.18)

In summary: 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
  • #1
Silicon-Based
51
1
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.
 
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  • #2
Silicon-Based said:
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.
 
  • #3
tnich said:
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.
 
  • #4
Silicon-Based said:
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.
 
  • #5
tnich said:
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.)
 
  • #6
Silicon-Based said:
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##?
 
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  • #7
tnich said:
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?
 
  • #8
Silicon-Based said:
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.
 
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  • #9
tnich said:
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.
 

Related to Bra-ket of uncertainty commutator (Sakurai 1.18)

1. What is the "Bra-ket of uncertainty commutator" in Sakurai 1.18?

The "Bra-ket of uncertainty commutator" in Sakurai 1.18 is an equation that describes the uncertainty principle in quantum mechanics. It is used to calculate the commutator between the position and momentum operators, which represents the uncertainty in measuring both of these quantities simultaneously.

2. Why is the "Bra-ket of uncertainty commutator" important?

The "Bra-ket of uncertainty commutator" is important because it helps us understand the fundamental principle of uncertainty in quantum mechanics. It also allows us to make predictions about the behavior of quantum systems and make calculations for various physical quantities.

3. How is the "Bra-ket of uncertainty commutator" derived?

The "Bra-ket of uncertainty commutator" is derived using the commutation relation between the position and momentum operators. It involves using the properties of the inner product and the Hermitian adjoint of operators.

4. What does the "Bra-ket of uncertainty commutator" tell us about a system?

The "Bra-ket of uncertainty commutator" tells us that there is a limit to how precisely we can measure both the position and momentum of a particle. It also indicates that the more accurately we know one quantity, the less accurately we know the other.

5. Can the "Bra-ket of uncertainty commutator" be applied to any quantum system?

Yes, the "Bra-ket of uncertainty commutator" can be applied to any quantum system that involves the measurement of both position and momentum. It is a fundamental principle in quantum mechanics and is applicable to various physical phenomena.

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