# Quantum Operators Indices Confusion

1. Nov 24, 2009

### barnflakes

My lecturer has written $A | \alpha_n> = a_n |\alpha_n> => A = \sum_n a_n | \alpha_n>< \alpha_n |$

and

$B | \alpha_k> = b_k |\alpha_k> => B = \sum_k b_k | \alpha_k>< \alpha_k |$

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:

$AB = \sum_{n,k} a_n b_k |\alpha_n>< \alpha_n | \alpha_k >< \alpha_k| = \sum_n a_n b_n | \alpha_n>< \alpha_n|$

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

Can anyone shed any light please?

2. Nov 24, 2009

### Fredrik

Staff Emeritus
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: Nov 24, 2009
3. Nov 24, 2009

### barnflakes

Thank you Fredrik, that's cleared it up nicely. Another line further on has me somewhat confused:

$<\alpha_n | B | \alpha_m> = \delta_{nm} <\alpha_m | B | \alpha_m>$

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

The closest I got was :
$<\alpha_n | B | \alpha_m> = \sum_m \delta_{nm} <\alpha_m | B | \alpha_m>$

Has he just been lazy and missed out the sum?

4. Nov 25, 2009

### Fredrik

Staff Emeritus
That equation is obviously true if the left-hand side is =0 when $m\neq n$, and you can easily verify that it is:

$$\langle\alpha_n|B|\alpha_m\rangle=b_m \langle\alpha_n|\alpha_m\rangle= b_m\delta_{nm}$$

The result you got is wrong. It would imply that $\langle\alpha_n|B|\alpha_m\rangle$ is independent of m (since your right-hand side is just $\langle\alpha_n|B|\alpha_n\rangle=b_n$).

Last edited: Nov 25, 2009
5. Nov 25, 2009

### barnflakes

Thanks Fredrik, I think I'm slowly starting to understand. Can I just confirm:

$\sum_{n,k} |\alpha_n>< \alpha_k|$ is equal to the identity matrix? If $<\alpha_n | \alpha_k > = \delta_{nk}$ of course.

6. Nov 25, 2009

### barnflakes

7. Nov 25, 2009

### ansgar

no what you have is:

a + b >= a

there is a curly bracket missing though:

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

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

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

http://mathworld.wolfram.com/SchwarzsInequality.html

8. Nov 26, 2009

### Fredrik

Staff Emeritus
Try having that operator act on $|\alpha_m\rangle$ and see what you get.

(Also, your bras and kets would look better if you use \langle and \rangle instead of < and >).

9. Nov 27, 2009

### Sillyboy

Actually, you have pointed it correctly. or you see the effect, you will get it fully~

10. Dec 13, 2009

### barnflakes

Can someone explain why $$\langle \bold{r} | \bold {p} \rangle = \frac{e^{i \bold{p}.\bold{r}/ \hbar}}{(2 \pi \hbar)^{3/2}}$$ I simply don't know how to get that result

11. Dec 13, 2009

### ansgar

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 ;)