Dirac Notation Help

1. Sep 11, 2004

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.

Last edited: Sep 11, 2004
2. Sep 11, 2004

Wong

#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.

Last edited: Sep 11, 2004
3. Sep 12, 2004

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

4. Sep 12, 2004

Hurkyl

Staff Emeritus
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.

Last edited: Sep 12, 2004
5. Sep 12, 2004

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.

6. Sep 12, 2004

Hurkyl

Staff Emeritus
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?

7. Sep 12, 2004