What Is the Null Operator in Quantum Mechanics?

[Bill Foster]

Homework Statement

Show this is the null operator:

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

Homework Equations

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

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

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?

 

[gabbagabbahey]

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

What if that assumption is false?

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. $|\alpha\rangle=\sum_{a'}\langle a'|\alpha\rangle|a'\rangle$).

 

[gabbagabbahey]

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

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

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

 

[Redbelly98]

... that is assuming that $$A|a'\rangle = a'|a'\rangle$$

What if that assumption is false?

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.)

 

[Bill Foster]

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

Can I do this?$$\prod_{a'}\left(A-a'\right)\sum_{a'}\langle a'|\alpha\rangle|a'\rangle$$

 

[gabbagabbahey]

Can I do this?


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

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

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

 

[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:

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

 

[Bill Foster]

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

 

[gabbagabbahey]

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

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

 

[Bill Foster]

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

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

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

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

 

[gabbagabbahey]

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

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

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

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

 

[Bill Foster]

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

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

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

 

[gabbagabbahey]

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

 

[Bill Foster]

Yyou're done.

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

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

 

[gabbagabbahey]

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 $$|\alpha\rangle$$:

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

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...

 

[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.

 

[gabbagabbahey]

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

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?

 

[Redbelly98]

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

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.

 

[Bill Foster]

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.

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:

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

 

[gabbagabbahey]

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

 

[Bill Foster]

If I have this:
$$\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}| a_n\rangle\langle a_n|\alpha\rangle$$

I can write it like this:
$$\frac{A-a_1}{a_n-a_1}\frac{A-a_2}{a_n-a_2}\dots\frac{A-a_k}{a_n-a_k}| a_n\rangle\langle a_n|\alpha\rangle$$

Since those terms commute, I can multiply any of them by $$| a_n\rangle\langle a_n|\alpha\rangle$$

Say, for example, the first one:

$$\frac{\left(A-a_1\right)| a_n\rangle\langle a_n|\alpha\rangle}{a_n-a_1}\frac{A-a_2}{a_n-a_2}\dots\frac{A-a_k}{a_n-a_k}$$
$$=\frac{\left(A| a_n\rangle\langle a_n|\alpha\rangle-a_1| a_n\rangle\langle a_n|\alpha\rangle\right)}{a_n-a_1}\frac{A-a_2}{a_n-a_2}\dots\frac{A-a_k}{a_n-a_k}$$
$$=\frac{\left(a_n|a_n\rangle\langle a_n|\alpha\rangle-a_1| a_n\rangle\langle a_n|\alpha\rangle\right)}{a_n-a_1}\frac{A-a_2}{a_n-a_2}\dots\frac{A-a_k}{a_n-a_k}$$
$$=\frac{\left(a_n-a_1\right)}{a_n-a_1}| a_n\rangle\langle a_n|\alpha\rangle\frac{A-a_2}{a_n-a_2}\dots\frac{A-a_k}{a_n-a_k}$$
$$=\frac{A-a_2}{a_n-a_2}|a_n\rangle\langle a_n|\alpha\rangle\dots\frac{A-a_k}{a_n-a_k}$$

Can I do that? Can I keep applying that $$|a_n\rangle\langle a_n|\alpha\rangle$$ to each term, converting it to "1"?

 

[gabbagabbahey]

Yep...

 

[Bill Foster]

Then I'm done.

Spasibo

 

[bjnartowt]

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

Before reading this thread, I thought you could simply consider:
(A - I*a')

...where a' is the corresponding eigenvalue, and I is the identity matrix, |a'><a'| ... and thus (A - I*a') acting upon an individual eigenket would give 0, and surely, then, the product would give zero, no matter what eigenket it acted upon.

Could you describe more clearly what I am missing? This whole problem seemed innocuous until I visited this thread...

 

[gabbagabbahey]

Before reading this thread, I thought you could simply consider:
(A - I*a')

...where a' is the corresponding eigenvalue, and I is the identity matrix, |a'><a'| ... and thus (A - I*a') acting upon an individual eigenket would give 0, and surely, then, the product would give zero, no matter what eigenket it acted upon.

Could you describe more clearly what I am missing? This whole problem seemed innocuous until I visited this thread...

The point is that $\prod_{a'}(A-Ia')=(A-Ia_1)(A-Ia_2)\ldots(A-Ia_n)$, that is $a'$ is a dummy variable (the prime is a dummy index) being multiplied over. So, $$\prod_{a'}(A-Ia')|a'\rangle=(A-Ia_1)(A-Ia_2)\ldots(A-Ia_n)|a_1\rangle|a_2\rangle\ldots|a_n\rangle$$.