Operators and eigenstates

einai

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?

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Sonty

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

Staff Emeritus
Gold Member
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>?

einai

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

Staff Emeritus
Gold Member
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.

einai

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 Sonty

Originally posted by einai
Hm......what's wrong with my way of doing it then? Thanks again nothing. it's just that Hurkyl is sometimes in a hurry and skips some terms 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

Staff Emeritus
Gold Member
Somehow, when writing my post, I substituted into my head another problem where |1> was an eigenvector of B with eigenvalue b. For your problem, hermiticity is needed because you don't know what B|2> is. Silly me!

einai

Originally posted by Sonty
nothing. it's just that Hurkyl is sometimes in a hurry and skips some terms 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 !

einai

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. 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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