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Operators and eigenstates

  • Thread starter einai
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27
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Operators and eigenstates (updated with new question)

Hi, I encountered the following HW problem which really confuses me. Could anyone please explain it to me? Thank you so much!

The result of applying a Hermitian operator B to a normalized vector |1> is generally of the form:

B|1> = b|1> + c|2>

where b and c are numerical coefficients and |2> is a normalized vector orthogonal to |1>.

My question is: Why B|1> must have the above form? Does it mean if |1> is an eigenstate of B, then b=!0 and c=0? But what if |1> is not an eigenstate of B?

I also need to find the expectation value of B (<1|B|1> ), but I think I got this part:

<1|B|1> = <1|b|1> + <1|c|2> = b<1|1> + c<1|2> = b

since |1> and |2> are orthogonal and they're both normalized. Does that look right?
 
Last edited:
95
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Originally posted by einai
My question is: Why B|1> must have the above form? Does it mean if |1> is an eigenstate of B, then b=!0 and c=0? But what if |1> is not an eigenstate of B?
an operator takes a vector in a space and turns it into another vector in (not necesarily) another space. if that space is spanned by the vectors |1> and |2>m then b|1> + c|2> is the general form a vector in that space. if you don't know how B looks like, then you have to assume this general form. If you find something oput about B, like |1> being an eigenstate, then c must be 0 as the two base vectors should be orthonormal.


I also need to find the expectation value of B (<1|B|1> ), but I think I got this part:

<1|B|1> = <1|b|1> + <1|c|2> = b<1|1> + c<1|2> = b

since |1> and |2> are orthogonal and they're both normalized. Does that look right?
you're absolutely right. trust yourself.
 

Hurkyl

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Why B|1> must have the above form?
Because we can "compute" b, c, and |2>.

You know how to compute b already; b = <1|B|1>. Can you hazard a guess as to how to compute c and |2>?
 
27
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Thank you very much, Sonty and Hurkyl. Now I understand the question. However, I have another one -

I'm trying to find <1|(B-<B>)2|1>. I broke it up like this:

<1|(B-<B>)2|1> = <1|B2|1> - <1|2bB|1> + <1|b2|1>, since we already found that <B> = b.

<1|2bB|1> = 2b <1|B|1> = 2b2
<1|b2|1> = b2

But I'm not too sure about <1|B2|1>....should I do something like this ? (Note: b, c are real constants)

<1|B2|1> = <1|B* B|1>
= (<1|b + <2|c)(b|1> + c|2>)
=<1|b2|1> + <1|bc|2> + <2|cb|1> + <2|c2|2>
= b2 + c2 (since the orthogonal terms cancel)

So <1|(B-<B>)2|1>
= b2 + c2 - 2b2 + b2
= c2
 

Hurkyl

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That looks right.

Incidentally, you don't need to use hermiticity for this one...

<1|(B^2)|1> = <1|BB|1> = <1|B(b|1>) = b<1|B|1> = b^2

so if you ever work with a nonhermitian transformation, you can still compute things like this.
 
27
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Originally posted by Hurkyl
<1|(B^2)|1> = <1|BB|1> = <1|B(b|1>) = b<1|B|1> = b^2
Wait....I think my answer doesn't look right. You got <1|(B^2)|1> = b^2, but I got

<1|B^2|1> = <1|B* B|1>
= (<1|b + <2|c)(b|1> + c|2> )
=<1|b^2|1> + <1|bc|2> + <2|cb|1> + <2|c^2|2>
= b^2 + c^2

Hm......what's wrong with my way of doing it then? Thanks again :smile:
 
95
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Originally posted by einai
Hm......what's wrong with my way of doing it then? Thanks again :smile:
nothing. it's just that Hurkyl is sometimes in a hurry and skips some terms :smile:
the thing I wonder about is when did we say that <B>=b=<1|B|1>? I have this strange feeling that <B>=b+c...
 

Hurkyl

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Somehow, when writing my post, I substituted into my head another problem where |1> was an eigenvector of B with eigenvalue b. :frown:

For your problem, hermiticity is needed because you don't know what B|2> is. Silly me!
 
27
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Originally posted by Sonty
nothing. it's just that Hurkyl is sometimes in a hurry and skips some terms :smile:
the thing I wonder about is when did we say that <B>=b=<1|B|1>? I have this strange feeling that <B>=b+c...
I got <B> like the following, and <B> is defined to be <1|B|1> in the problem :).

B|1> = b|1> + c|2>
<1|B|1> = <1|b|1> + <1|c|2> = b<1|1> + c<1|2> = b

I hope it's right......

Thanks :smile:!
 
27
0
Originally posted by Hurkyl
Somehow, when writing my post, I substituted into my head another problem where |1> was an eigenvector of B with eigenvalue b. :frown:

For your problem, hermiticity is needed because you don't know what B|2> is. Silly me!
Yeah, I did silly things all the time. But thanks though, now I understand this problem a lot better.
 

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