Sakurai 1.7

  • #1
338
0

Homework Statement



Show this is the null operator:

[tex]\prod_{a'}\left(A-a')[/tex]

Homework Equations



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

[tex]X|\alpha\rangle = 0[/tex]

The Attempt at a Solution



Multiply the expression by [tex]|a'\rangle[/tex]:

[tex]\prod_{a'}\left(A-a')|a'\rangle[/tex]
[tex]=\prod_{a'}\left(A|a'\rangle-a'|a'\rangle)[/tex]
[tex]=\prod_{a'}\left(a'|a'\rangle-a'|a'\rangle)=0[/tex]

That works out. But that is assuming that [tex]A|a'\rangle = a'|a'\rangle[/tex]

What if that assumption is false?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
That works out. But that is assuming that [tex]A|a'\rangle = a'|a'\rangle[/tex]

What if that assumption is false?

Assuming [itex]a'[/itex] represent the eigenvalues of [itex]A[/itex], then that equation will definitely be true...you then also need to appeal to the fact that any arbitrary state [itex]|\alpha\rangle[/itex] (in the Hilbert space spanned by [itex]A[/itex]) can be decomposed as a superposition of the eigenkets of [itex]A[/itex] (i.e. [itex]|\alpha\rangle=\sum_{a'}\langle a'|\alpha\rangle|a'\rangle[/itex]).
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Multiply the expression by [tex]|a'\rangle[/tex]:

[tex]\prod_{a'}\left(A-a')|a'\rangle[/tex]

You can't do this...[itex]a'[/itex] is essentially a dummy variable in the operator's definition...
 
  • #4
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,133
160
... that is assuming that [tex]A|a'\rangle = a'|a'\rangle[/tex]

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.)
 
Last edited:
  • #5
338
0
You can't do this...[itex]a'[/itex] is essentially a dummy variable in the operator's definition...

Can I do this?


[tex]
\prod_{a'}\left(A-a'\right)\sum_{a'}\langle a'|\alpha\rangle|a'\rangle
[/tex]
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Can I do this?


[tex]
\prod_{a'}\left(A-a'\right)\sum_{a'}\langle a'|\alpha\rangle|a'\rangle
[/tex]

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

[tex]\prod_{a'}\left(A-a'\right)|\alpha\rangle=\prod_{a'}\left(A-a'\right)\sum_{a''}\langle a''|\alpha\rangle|a''\rangle
[/tex]
 
  • #7
338
0
[tex]\prod_{a'}\left(A-a'\right)\sum_{a''}\langle a''|\alpha\rangle|a''\rangle[/tex]
[tex]=\prod_{a'}\left(A\sum_{a''}\langle a''|\alpha\rangle|a''\rangle-a'\sum_{a''}\langle a''|\alpha\rangle|a''\rangle\right)[/tex]
[tex]=\prod_{a'}\left(\sum_{a''}\langle a''|\alpha\rangle A|a''\rangle-a'\sum_{a''}\langle a''|\alpha\rangle a'|a''\rangle\right)[/tex]
[tex]=\prod_{a'}\left(\sum_{a''}\langle a''|\alpha\rangle A|a''\rangle-\sum_{a''}\langle a''|\alpha\rangle \delta\left(a' - a''\right)\right)[/tex]
[tex]=\prod_{a'}\left(\sum_{a''}\langle a''|\alpha\rangle A|a''\rangle-\langle a'|\alpha\rangle \right)[/tex]
[tex]=\prod_{a'}\left(\sum_{a'}\langle a'|\alpha\rangle A|a'\rangle-\langle a'|\alpha\rangle \right)[/tex]
[tex]=\prod_{a'}\left(\langle a'|\alpha\rangle \left(\sum_{a'} A|a'\rangle-1\right) \right)[/tex]

If this is true, then this works out:

[tex]\sum_{a'} A|a'\rangle =1 [/tex]
 
  • #8
338
0
No, I can't take that inner product out of the sum. Damn it.
 
  • #9
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Just start by taking the product inside the sum (which you can you since they are over different indices), and then realize that

[tex]\prod_{a'}\left(A-a'\right)=(A-a_1)(A-a_2)\ldots(A-a'')\ldots[/tex]
 
  • #10
338
0
Just start by taking the product inside the sum (which you can you since they are over different indices), and then realize that

[tex]\prod_{a'}\left(A-a'\right)=(A-a_1)(A-a_2)\ldots(A-a'')\ldots[/tex]

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

[tex]\langle a''|\alpha\rangle A|a''\rangle - \langle a''|\alpha\rangle a''|a''\rangle[/tex]

Which is zero if

[tex]
A|a''\rangle = a''|a''\rangle
[/tex]
 
  • #11
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Which will give me a term in each element of the sum like this:

[tex]\langle a''|\alpha\rangle A|a''\rangle - \langle a''|\alpha\rangle a''|a''\rangle[/tex]

You need to be careful with this....is [itex](A-a'')[/itex] 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

[tex](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[/tex]
 
  • #12
338
0
You need to be careful with this....is [itex](A-a'')[/itex] 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

[tex](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[/tex]

I see. If it commutes, then my most recent "solution" is correct?
 
  • #13
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Yes, so just show that [itex]A-a'[/itex] commutes with [itex]A-a''[/itex] (for arbitrary scalars [itex]a'[/itex] and [itex]a''[/itex]), and you're done.
 
  • #14
338
0
Yyou're done.

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

What is the significance of [tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}[/tex]?

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

[tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}|\alpha\rangle[/tex]
[tex]=\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\sum_{n}\langle a_n|\alpha\rangle | a_n\rangle[/tex]
[tex]=\sum_{n}\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_n|\alpha\rangle | a_n\rangle[/tex]
[tex]=\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}[/tex]
[tex]=\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}[/tex]
[tex]=\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[/tex]
[tex]=\sum_{n}\prod_{m\ne n}\frac{ a_n - a_m}{a_n-a_m}|a_n\rangle \langle a_n|\alpha\rangle[/tex]
[tex]=\sum_{n}\prod_{m\ne n}|a_n\rangle \langle a_n|\alpha\rangle[/tex]
[tex]=\sum_{n}|a_n\rangle \langle a_n|\alpha\rangle[/tex]

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

[tex]\sum_{n}|a_n\rangle \langle a_n|=1[/tex]
 
  • #15
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Not quite. I've got parts (b) & (c) to do, which states:

What is the significance of [tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}[/tex]?

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

[tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}|\alpha\rangle[/tex]
[tex]=\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\sum_{n}\langle a_n|\alpha\rangle | a_n\rangle[/tex]

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 [itex]j[/itex] instead...
 
  • #16
338
0
[tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}|\alpha\rangle[/tex]
[tex]=\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\sum_{j}\langle a_j|\alpha\rangle | a_j\rangle[/tex]
[tex]=\sum_{j}\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_j|\alpha\rangle | a_j\rangle[/tex]
[tex]=\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}[/tex]
[tex]=\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}[/tex]
[tex]=\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[/tex]
[tex]=\sum_{j}\prod_{m\ne n}\frac{ a_j - a_m}{a_n-a_m}|a_j\rangle \langle a_j|\alpha\rangle[/tex]

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

[tex]\frac{ a_j - a_m}{a_n-a_m}[/tex]

...unless you know of another way to get rid of it, the product, and the sum.
 
  • #17
gabbagabbahey
Homework Helper
Gold Member
5,002
7
[tex]=\sum_{j}\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_j|\alpha\rangle | a_j\rangle[/tex]
[tex]=\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}[/tex]


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

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

[tex]\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}[/tex]

Do you see why?
 
Last edited:
  • #18
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,133
160
[tex]=\sum_{j}\prod_{m\ne n}\frac{ a_j - a_m}{a_n-a_m}|a_j\rangle \langle a_j|\alpha\rangle[/tex]

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
338
0
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 [tex]j=n[/tex].

That will leave me with the following:

[tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}\langle a_n|\alpha\rangle | a_n\rangle[/tex]

which I can write as:

[tex]= \prod_{m\ne n}\frac{A-a_m}{a_n-a_m}| a_n\rangle\langle a_n|\alpha\rangle[/tex]

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

[tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}[/tex]
 
  • #20
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Well, what is [itex](A-a_m)|a_n\rangle[/itex] for arbitrary [itex]m[/itex]?
 
  • #21
338
0
If I have this:
[tex]\prod_{m\ne n}\frac{A-a_m}{a_n-a_m}| a_n\rangle\langle a_n|\alpha\rangle[/tex]

I can write it like this:
[tex]\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[/tex]

Since those terms commute, I can multiply any of them by [tex]| a_n\rangle\langle a_n|\alpha\rangle[/tex]

Say, for example, the first one:

[tex]\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}[/tex]
[tex]=\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}[/tex]
[tex]=\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}[/tex]
[tex]=\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}[/tex]
[tex]=\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}[/tex]

Can I do that? Can I keep applying that [tex]|a_n\rangle\langle a_n|\alpha\rangle[/tex] to each term, converting it to "1"?
 
  • #22
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Yep....
 
  • #23
338
0
Then I'm done.

Spasibo
 
  • #24
284
3
You can't do this...[itex]a'[/itex] 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...
 
  • #25
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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 [itex]\prod_{a'}(A-Ia')=(A-Ia_1)(A-Ia_2)\ldots(A-Ia_n)[/itex], that is [itex]a'[/itex] is a dummy variable (the prime is a dummy index) being multiplied over. So, [tex]\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[/tex].
 

Related Threads on Sakurai 1.7

Replies
2
Views
1K
  • Last Post
Replies
24
Views
4K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
8
Views
2K
  • Last Post
2
Replies
28
Views
8K
  • Last Post
Replies
1
Views
3K
Top