# Sakurai 1.7

1. Oct 29, 2009

### Bill Foster

1. The problem statement, all variables and given/known data

Show this is the null operator:

$$\prod_{a'}\left(A-a')$$

2. Relevant equations

The null operator X is one that when it operates on an arbitrary ket $$|\alpha\rangle$$ results in 0:

$$X|\alpha\rangle = 0$$

3. The attempt at a solution

Multiply the expression by $$|a'\rangle$$:

$$\prod_{a'}\left(A-a')|a'\rangle$$
$$=\prod_{a'}\left(A|a'\rangle-a'|a'\rangle)$$
$$=\prod_{a'}\left(a'|a'\rangle-a'|a'\rangle)=0$$

That works out. But that is assuming that $$A|a'\rangle = a'|a'\rangle$$

What if that assumption is false?

2. Oct 29, 2009

### gabbagabbahey

Assuming $a'$ represent the eigenvalues of $A$, then that equation will definitely be true...you then also need to appeal to the fact that any arbitrary state $|\alpha\rangle$ (in the Hilbert space spanned by $A$) can be decomposed as a superposition of the eigenkets of $A$ (i.e. $|\alpha\rangle=\sum_{a'}\langle a'|\alpha\rangle|a'\rangle$).

3. Oct 29, 2009

### gabbagabbahey

You can't do this...$a'$ is essentially a dummy variable in the operator's definition...

4. Oct 29, 2009

### Redbelly98

Staff Emeritus
I have Sakurai's book in front of me. According to the problem statement, |a'> are the eigenkets of A, a Hermitian operator. (Also, there is no degeneracy.)

Last edited: Oct 29, 2009
5. Oct 30, 2009

### Bill Foster

Can I do this?

$$\prod_{a'}\left(A-a'\right)\sum_{a'}\langle a'|\alpha\rangle|a'\rangle$$

6. Oct 30, 2009

### gabbagabbahey

It's not very good notation to use the same dummy variable in both the product and the sum....try using

$$\prod_{a'}\left(A-a'\right)|\alpha\rangle=\prod_{a'}\left(A-a'\right)\sum_{a''}\langle a''|\alpha\rangle|a''\rangle$$

7. Oct 30, 2009

### Bill Foster

$$\prod_{a'}\left(A-a'\right)\sum_{a''}\langle a''|\alpha\rangle|a''\rangle$$
$$=\prod_{a'}\left(A\sum_{a''}\langle a''|\alpha\rangle|a''\rangle-a'\sum_{a''}\langle a''|\alpha\rangle|a''\rangle\right)$$
$$=\prod_{a'}\left(\sum_{a''}\langle a''|\alpha\rangle A|a''\rangle-a'\sum_{a''}\langle a''|\alpha\rangle a'|a''\rangle\right)$$
$$=\prod_{a'}\left(\sum_{a''}\langle a''|\alpha\rangle A|a''\rangle-\sum_{a''}\langle a''|\alpha\rangle \delta\left(a' - a''\right)\right)$$
$$=\prod_{a'}\left(\sum_{a''}\langle a''|\alpha\rangle A|a''\rangle-\langle a'|\alpha\rangle \right)$$
$$=\prod_{a'}\left(\sum_{a'}\langle a'|\alpha\rangle A|a'\rangle-\langle a'|\alpha\rangle \right)$$
$$=\prod_{a'}\left(\langle a'|\alpha\rangle \left(\sum_{a'} A|a'\rangle-1\right) \right)$$

If this is true, then this works out:

$$\sum_{a'} A|a'\rangle =1$$

8. Oct 30, 2009

### Bill Foster

No, I can't take that inner product out of the sum. Damn it.

9. Oct 30, 2009

### gabbagabbahey

Just start by taking the product inside the sum (which you can you since they are over different indices), and then realize that

$$\prod_{a'}\left(A-a'\right)=(A-a_1)(A-a_2)\ldots(A-a'')\ldots$$

10. Oct 30, 2009

### Bill Foster

Which will give me a term in each element of the sum like this:

$$\langle a''|\alpha\rangle A|a''\rangle - \langle a''|\alpha\rangle a''|a''\rangle$$

Which is zero if

$$A|a''\rangle = a''|a''\rangle$$

11. Oct 30, 2009

### gabbagabbahey

You need to be careful with this....is $(A-a'')$ always going to be the rightmost operator in the product? If not, you will also need to show that it commutes with all the other terms in the product, so that

$$(A-a_1)(A-a_2)\ldots(A-a'')\ldots(A-a_i)(A-a_{i+1})\ldots|a''\rangle=(A-a_1)(A-a_2)\ldots(A-a_i)(A-a_{i+1})\ldots(A-a'')|a''\rangle=0$$

12. Oct 30, 2009

### Bill Foster

I see. If it commutes, then my most recent "solution" is correct?

13. Oct 30, 2009

### gabbagabbahey

Yes, so just show that $A-a'$ commutes with $A-a''$ (for arbitrary scalars $a'$ and $a''$), and you're done.

14. Oct 30, 2009

### Bill Foster

Not quite. I've got parts (b) & (c) to do, which states:

What is the significance of $$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}$$?

I already know it's the projection operator. I started by multiplying it by the arbitrary ket $$|\alpha\rangle$$:

$$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}|\alpha\rangle$$
$$=\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\sum_{n}\langle a_n|\alpha\rangle | a_n\rangle$$
$$=\sum_{n}\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_n|\alpha\rangle | a_n\rangle$$
$$=\sum_{n}\prod_{m\ne n}\frac{\langle a_n|\alpha\rangle A| a_n\rangle-\langle a_n|\alpha\rangle a_m| a_n\rangle}{a_n-a_m}$$
$$=\sum_{n}\prod_{m\ne n}\frac{\langle a_n|\alpha\rangle a_n| a_n\rangle-\langle a_n|\alpha\rangle a_m| a_n\rangle}{a_n-a_m}$$
$$=\sum_{n}\prod_{m\ne n}\frac{ a_n| a_n\rangle- a_m| a_n\rangle}{a_n-a_m}\langle a_n|\alpha\rangle$$
$$=\sum_{n}\prod_{m\ne n}\frac{ a_n - a_m}{a_n-a_m}|a_n\rangle \langle a_n|\alpha\rangle$$
$$=\sum_{n}\prod_{m\ne n}|a_n\rangle \langle a_n|\alpha\rangle$$
$$=\sum_{n}|a_n\rangle \langle a_n|\alpha\rangle$$

There should not be a sum there. That is not the projection operator; that is

$$\sum_{n}|a_n\rangle \langle a_n|=1$$

15. Oct 31, 2009

### gabbagabbahey

Again, this is bad notation to have the same index appear more than twice in a single expression (when it appears once, it's a free index. When it appears twice, it is a dummy index. When it appears more than twice it is just confusing)....try summing over $j$ instead...

16. Oct 31, 2009

### Bill Foster

$$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}|\alpha\rangle$$
$$=\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\sum_{j}\langle a_j|\alpha\rangle | a_j\rangle$$
$$=\sum_{j}\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_j|\alpha\rangle | a_j\rangle$$
$$=\sum_{j}\prod_{m\ne n}\frac{\langle a_j|\alpha\rangle A| a_j\rangle-\langle a_j|\alpha\rangle a_m| a_j\rangle}{a_n-a_m}$$
$$=\sum_{j}\prod_{m\ne n}\frac{\langle a_j|\alpha\rangle a_j| a_j\rangle-\langle a_j|\alpha\rangle a_m| a_j\rangle}{a_n-a_m}$$
$$=\sum_{j}\prod_{m\ne n}\frac{ a_j| a_j\rangle- a_m| a_j\rangle}{a_n-a_m}\langle a_j|\alpha\rangle$$
$$=\sum_{j}\prod_{m\ne n}\frac{ a_j - a_m}{a_n-a_m}|a_j\rangle \langle a_j|\alpha\rangle$$

It has to be summed over n so this term will cancel out:

$$\frac{ a_j - a_m}{a_n-a_m}$$

...unless you know of another way to get rid of it, the product, and the sum.

17. Oct 31, 2009

### gabbagabbahey

You can't do this step; for the same reason that $(A-a_1)(A-a_2)|\alpha\rangle\neq (A|\alpha\rangle-a_1|\alpha\rangle)(A|\alpha\rangle-a_2|\alpha\rangle)$

You can, however use the commutativity relation you used in part (a) to say

$$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}=\left{\begin{array}{lr}\left(\prod_{m\ne n,j}\frac{A-a_m}{a_n-a_m}\right)\frac{A-a_j}{a_n-a_j} & ,j\neq m \\ \prod_{m\ne n,j}\frac{A-a_m}{a_n-a_m} &,j = m\end{array}$$

Do you see why?

Last edited: Oct 31, 2009
18. Oct 31, 2009

### Redbelly98

Staff Emeritus
In the product over m≠n, one of the factors will have m=j (except when j is n). That may help simplify things here.

19. Oct 31, 2009

### Bill Foster

Yes. All terms will be zero except for the one where $$j=n$$.

That will leave me with the following:

$$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_n|\alpha\rangle | a_n\rangle$$

which I can write as:

$$= \prod_{m\ne n}\frac{A-a_m}{a_n-a_m}| a_n\rangle\langle a_n|\alpha\rangle$$

Which is a problem because I still have this factor that I somehow need to get rid of:

$$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}$$

20. Oct 31, 2009

### gabbagabbahey

Well, what is $(A-a_m)|a_n\rangle$ for arbitrary $m$?