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Hypnotoad
Sep11-04, 07:10 PM
We are working on Dirac notation in my quantum class, and for the most part I see that it is a very easy way to do problems. But I am still getting stuck on how to deal with a few things on my current homework assignment. These come out of chapter 2 in the Cohen-Tannoudji book if you want to look them up.

#1. |\phi_{n}> are eigenstates of a Hermitian operator H and they form a discrete orthonormal basis. The operator U(m,n) is defined by U(m,n)=|\phi_{m}><\phi_{n}|.

b. Calculate the commutator [H,U(m,n)].

I'm not really sure how to deal with this in such a general case. I get to the first step: H|\phi_{m}><\phi_{n}|-|\phi_{m}><\phi_{n}|H

but I don't know where to go from there.

e. Let A be an operator, with matrix elements A_{mn}=<\phi_{m}|A|\phi_{n}>

Prove the relation:A=\Sigma A_{mn}U(m,n)

If I start with A_{mn}=<\phi_{m}|A|\phi_{n}>, is it legal to do this:

A_{mn}|\phi_{m}><\phi_{n}|=<\phi_{m}|\phi_{m}> A<\phi_{m}|\phi_{n}>

then, since the states are orthonormal:
<\phi_{m}|\phi_{m}>=<\phi_{n}|\phi_{n}>=1

If I can do that I get: A=A_{mn}|\phi_{m}><\phi_{n}|

but I'm not sure where the summation comes in.

#4. Let K be the operator defined by K=|\phi><\psi| where |\phi>, |\psi> are two vectors of the state space.

c. show that K can always be written in the form K=\lambda P_{1}P_{2} where \lambda is a constant to be calculated and P_{1}, P_{2} are projectors.

I'm not really sure where to get started on this one. Any hints would be appreciatted, especially since this is due monday morning and I won't have time to talk to my professor before hand.

Wong
Sep11-04, 11:30 PM
#1
b. If |\phi_{n}> is a eigenket of H, then what is the action of H on |\phi_{n}>? Similarly, what is <\phi_{n}|H? Also note that the communtator between two operators is in general an operator.

e. The identity operator is \sum_{n}|\phi_{n}><\phi_{n}|. Presumably you want to find A=?. Try to insert the identity operator in front and after A and exchange terms to see what you get.

#4

What is a projector? \frac{|\phi><\phi|}{<\phi|\phi>} and \frac{|\psi><\psi|}{<\psi|\psi>} would be projectors.

Hypnotoad
Sep12-04, 02:28 PM
So I figured out the first two questions, but I'm still stuck on that last one. I get to this point: K=\lambda \frac{|\phi><\psi|}{<\phi|\phi>} \frac{|\phi><\psi|}{<\psi|\psi>}

How do I get from that to K=\lambda \frac{|\phi><\phi|}{<\phi|\phi>} \frac{|\psi><\psi|}{<\psi|\psi>} ?

Hurkyl
Sep12-04, 02:46 PM
Remember that <\psi|\phi> is just a number...



Actually, how did you get to

K=\lambda \frac{|\phi><\psi|}{<\phi|\phi>} \frac{|\phi><\psi|}{<\psi|\psi>}

?

It seems to me that if you did things slightly different, you'd get the answer you seek.

Hypnotoad
Sep12-04, 03:19 PM
Remember that <\psi|\phi> is just a number...



Actually, how did you get to

K=\lambda \frac{|\phi><\psi|}{<\phi|\phi>} \frac{|\phi><\psi|}{<\psi|\psi>}

?

It seems to me that if you did things slightly different, you'd get the answer you seek.

I started with K=|\phi><\psi| and multiplied by \frac{<\phi|\phi>}{<\phi|\phi>} and \frac{<\psi|\psi>}{<\psi|\psi>}. That is how I got the \lambda=<\phi|\psi> term out front. Although, now that I look over it again, I'm not sure if I can rearrange the terms like that.

Hurkyl
Sep12-04, 04:00 PM
It's fine; numbers can always be moved around to wherever you want. Why did you opt to pull the <\phi|\psi> term out instead of the <\psi|\phi> term?

Hypnotoad
Sep12-04, 04:09 PM
It's fine; numbers can always be moved around to wherever you want. Why did you opt to pull the <\phi|\psi> term out instead of the <\psi|\phi> term?

I can't believe I didn't see that. Thanks for the help.