# I Schrödinger equation and interaction Hamiltonian

1. May 12, 2016

Given 1A.1 and 1A.2, I have been trying to apply the Schrödinger equation to reproduce 1A.3 and 1A.4 but have been struggling a bit. I was under the assumption that by applying $\hat{W} \rvert {\psi} \rangle= i\hbar \frac {d}{dt} \rvert{\psi} \rangle$ and then taking $\langle{k'} \lvert \hat{W} \rvert{\psi} \rangle$ and $\langle{i}\lvert \hat{W} \rvert{\psi} \rangle$ would allow me to produce 1A.3 and 1A.4. I may very well be incorrect in my methods, but did the following rough calculation and got a very different result. (In my calculation, I assumed $\hat{W} = H$.)

Any clarification on how to reproduce 1A.3 and 1A.4 would be greatly appreciated.

2. May 12, 2016

### blue_leaf77

What are $|i\rangle$, $|k\rangle$, and $w$? Moreover, how can you assume that $H=W$? The answer of all of these questions should be available in the resource you have there.

3. May 12, 2016

$\hat {W}$ is defined as the interaction Hamiltonian. $|i\rangle$ is the initial state the system is prepared in and $|k\rangle$ are all possible states it can evolve into where k is allowed to take all values between $- \infty$ to $\infty$ except $i$.

4. May 12, 2016

### Staff: Mentor

My guess is that $\hat{H} = \hat{H}_0 + \hat{W}$, and that $\varepsilon$ has something to do with the energy in the absence of the interaction (i.e., the eigenvalue of $\hat{H}_0$.

More details are needed. A reference would be nice.

5. May 12, 2016

The initial state $\rvert i \rangle$ has an energy equal to 0, and each state is separated in energy by a difference of $\varepsilon$. The energy difference between $\rvert i \rangle$ and $\rvert k \rangle$ is k$\varepsilon$.

Sure. This is Introduction to Quantum Optics by Grynberg and this problem begins on page 34.

6. May 12, 2016

### Staff: Mentor

So $\hat{H} | k \rangle = k \varepsilon | k \rangle + \hat{W} | k \rangle$. Try that in your method.

7. May 12, 2016

I got it, thank you!

Just one more question, is the statement $\langle{i} \lvert \hat{W} \rvert{k} \rangle = w_k$ necessarily true if $\langle{k} \lvert \hat{W} \rvert{i} \rangle = w_k$ ?

If $\hat{W}$ was an annihilation/creation operator, it seems like this definitely would not be true. Although I had to use both relations written above when reproducing 1A.3 and 1A.4. It might just be something regarding the density matrix that I missed in the text or might be implicitly assumed.

8. May 12, 2016

### George Jones

Staff Emeritus
I haven't done the calculation, but is the idea to calculate $\left<k'|H|\psi \right>$ twice, once using the Schrodinger equation, and once using $H = H_0 + W$ (?) together with the facts that 1) the kets are eigenstates (as in Dr. Claude's post) of $H_0$ and 2) the relations given in the original post.

Edit: TheCanadian posted while I was writing my post. This thread seems to have a surplus of Canadians.

9. May 12, 2016

### George Jones

Staff Emeritus
Is $\hat{W}$ Hermitian and

$$w_k = \frac{w}{\sqrt{1+\left(\frac{k \epsilon}{\Delta}\right)^2}}$$

real?