Quantum Operators Indices Confusion

In summary, the conversation discusses the properties of the operators A and B, where A is a hermitian operator and B is a unitary projector operator. The conversation also touches upon the use of indices in summation and the use of the summation convention. It is also mentioned that the equation <\alpha_n | B | \alpha_m> = \delta_{nm} <\alpha_m | B | \alpha_m> is only true if m = n. Finally, the conversation discusses the equation | \langle \Delta A \Delta B \rangle |^2 = \frac{1}{4}| \langle [ A,B] \rangle |^2 + \frac{
  • #1
barnflakes
156
4
My lecturer has written [itex]A | \alpha_n> = a_n |\alpha_n> => A = \sum_n a_n | \alpha_n>< \alpha_n |[/itex]

and

[itex]B | \alpha_k> = b_k |\alpha_k> => B = \sum_k b_k | \alpha_k>< \alpha_k |[/itex]

Where A is a hermitian operator. I understand he's used the properties of the unitary projector operator here, but is ok to include the subscript n three times in a sum? It goes on to say:

[itex] AB = \sum_{n,k} a_n b_k |\alpha_n>< \alpha_n | \alpha_k >< \alpha_k| = \sum_n a_n b_n | \alpha_n>< \alpha_n| [/itex]

Now to me I can't see how the indices are correct there, since the middle [itex]< \alpha_n | \alpha_k > = \delta_{nk}[/itex] can only get rid of the b_k index and not the one in the vector as well.

Can anyone shed any light please?
 
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  • #2
All of that is correct. When you perform the sum over k, only the terms will k=n will contribute since all the others are =0. You can certainly have same index appear three times in the sum. The only time when we don't want to see the same index three times is when we want to use the summation convention that says that we never write any summation sigmas and always sum over those indices that appear exactly twice. The result you got only implies that this would be a very bad time to try to use that convention.
 
Last edited:
  • #3
Thank you Fredrik, that's cleared it up nicely. Another line further on has me somewhat confused:

[itex] <\alpha_n | B | \alpha_m> = \delta_{nm} <\alpha_m | B | \alpha_m>[/itex]

How is that possible? Unless you are able to change the index in the state vector?

The closest I got was :
[itex] <\alpha_n | B | \alpha_m> = \sum_m \delta_{nm} <\alpha_m | B | \alpha_m>[/itex]

Has he just been lazy and missed out the sum?
 
  • #4
That equation is obviously true if the left-hand side is =0 when [itex]m\neq n[/itex], and you can easily verify that it is:

[tex]\langle\alpha_n|B|\alpha_m\rangle=b_m \langle\alpha_n|\alpha_m\rangle= b_m\delta_{nm}[/tex]

The result you got is wrong. It would imply that [itex]\langle\alpha_n|B|\alpha_m\rangle[/itex] is independent of m (since your right-hand side is just [itex]\langle\alpha_n|B|\alpha_n\rangle=b_n[/itex]).
 
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  • #5
Thanks Fredrik, I think I'm slowly starting to understand. Can I just confirm:

[itex]\sum_{n,k} |\alpha_n>< \alpha_k| [/itex] is equal to the identity matrix? If [itex] <\alpha_n | \alpha_k > = \delta_{nk}[/itex] of course.
 
  • #7
barnflakes said:
Also, for equation 6 on this document: https://www.physicsforums.com/library.php?do=view_item&itemid=208

Is it just using the fact that [itex]|z|^2 = |Imz|^2 + |Rez|^2 \geq |Imz|^2[/itex] or is something more subtle going on?

no what you have is:

a + b >= a

there is a curly bracket missing though:

[tex]
| \langle \Delta A \Delta B \rangle |^2 = \frac{1}{4}| \langle [ A,B] \rangle |^2 + \frac{1}{4} | \langle \{ \Delta A,~ \Delta B} \rangle |^2 \geq \frac{1}{4}| \langle [A,B] \rangle |^2~~~-~(6)
[/tex]

should read

[tex]
| \langle \Delta A \Delta B \rangle |^2 = \frac{1}{4}| \langle [ A,B] \rangle |^2 + \frac{1}{4} | \langle \{ \Delta A,~ \Delta B \} \rangle |^2 \geq \frac{1}{4}| \langle [A,B] \rangle |^2~~~-~(6)
[/tex]

also these posts are good:
https://www.physicsforums.com/library.php?do=view_item&itemid=207

http://mathworld.wolfram.com/SchwarzsInequality.html
 
  • #8
barnflakes said:
[itex]\sum_{n,k} |\alpha_n>< \alpha_k| [/itex] is equal to the identity matrix? If [itex] <\alpha_n | \alpha_k > = \delta_{nk}[/itex] of course.
Try having that operator act on [itex]|\alpha_m\rangle[/itex] and see what you get.

(Also, your bras and kets would look better if you use \langle and \rangle instead of < and >).
 
  • #9
Actually, you have pointed it correctly. or you see the effect, you will get it fully~
 
  • #10
Can someone explain why [tex]\langle \bold{r} | \bold {p} \rangle = \frac{e^{i \bold{p}.\bold{r}/ \hbar}}{(2 \pi \hbar)^{3/2}}[/tex] I simply don't know how to get that result
 
  • #11
barnflakes said:
Can someone explain why [tex]\langle \bold{r} | \bold {p} \rangle = \frac{e^{i \bold{p}.\bold{r}/ \hbar}}{(2 \pi \hbar)^{3/2}}[/tex] I simply don't know how to get that result

what has this to do anything with your first question and the title of the thread?

p-hat (in r-space) = -i d/dr

and in 3 dimensions

p-hat = -i nabla

so it is just a simple differential equation, with the normalization condition, edit: I dropped the hbar since I am a particle physicst ;)
 

What are quantum operators?

Quantum operators are mathematical tools used to describe the behavior of quantum systems. They represent physical observables, such as position, momentum, or energy, and are used to calculate the probabilities of obtaining a certain outcome in a quantum measurement.

How do quantum operators work?

Quantum operators work by acting on a quantum state to produce a new quantum state. This new state represents the result of a measurement of the corresponding observable. The operator itself is a mathematical representation of the physical observable and is typically represented by a matrix.

What is the role of indices in quantum operators?

Indices in quantum operators represent the different possible outcomes of a measurement. For example, the index 0 might represent the spin of an electron being up, while the index 1 might represent the spin being down. These indices are used to calculate the probabilities of obtaining a particular outcome in a measurement.

Why is there confusion surrounding quantum operator indices?

There may be confusion surrounding quantum operator indices because they can be abstract and require a solid understanding of quantum mechanics to fully grasp. Additionally, different textbooks or sources may use different conventions for representing indices, leading to confusion for those learning about quantum operators.

How can one avoid confusion with quantum operator indices?

To avoid confusion with quantum operator indices, it is important to have a strong understanding of the underlying principles of quantum mechanics. Additionally, it may be helpful to consult multiple sources and compare the conventions used for representing indices. Practice and familiarity with quantum operators can also help in understanding and avoiding confusion with indices.

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