# N=2 Supersymmetric Quantum Mechanics

1. Jan 30, 2007

### Tom Mattson

Staff Emeritus
The title says it all. I'd like to have an ongoing discussion about Witten's SUSY QM for N=2. At the moment I'm going through a book called Supersymmetric Methods in Quantum and Statistical Physics by Georg Junker, and I've got some issues that I'd like to discuss.

The first, and most basic, is: Can anyone out there recommend another good book, but one that has exercises in it? This book is meant to be read by grad students, but it has no problems to cut your teeth on.

Here's the next question. Junker cites Witten's definition of the supercharges thusly.

$$Q_1:=\frac{1}{\sqrt{2}}\left(\frac{p}{\sqrt{2m}}\otimes\sigma_1+\Phi(x)\otimes\sigma_2\right)$$
$$Q_2:=\frac{1}{\sqrt{2}}\left(\frac{p}{\sqrt{2m}}\otimes\sigma_2-\Phi(x)\otimes\sigma_1\right)$$

He then goes on to say that you can get the N=2 Hamiltonian by $H=2Q_1^2=2Q_2^2$. Now when I go ahead and compute $2Q_1^2$, I get the following.

$$H=\frac{p^2}{2m}\otimes\sigma_1^2+\Phi^2(x)\otimes\sigma_2^2+\frac{1}{2m}\{p,\Phi(x)\}\otimes\{\sigma_1,\sigma_2\}$$

This is wrong. Since $\sigma_1^2=\sigma_2^2=1$, first two terms are OK. But the anticommutators in the third term should in fact be commutators. Why doesn't the normal distributive law work here?

Last edited: Jan 30, 2007
2. Jan 30, 2007

### George Jones

Staff Emeritus
I get

$$\frac{p^{2}}{2m}\otimes \sigma _{1}^{2}+\phi ^{2}\otimes \sigma _{2}^{2}+\frac{\phi p}{\sqrt{2m}}\otimes \sigma _{2}\sigma _{1}+\frac{p\phi }{\sqrt{2m}}\otimes \sigma _{1}\sigma _{2}$$

when I compute $H=2Q_1^2.$

Now use $\sigma_2 \sigma_1 = - \sigma_1 \sigma_2$ (Is this true?), giving

$$\frac{p^{2}}{2m}\otimes \sigma _{1}^{2}+\phi ^{2}\otimes \sigma _{2}^{2}+\left( \frac{p\phi }{\sqrt{2m}}-\frac{\phi p}{\sqrt{2m}}\right) \otimes \sigma _{1}\sigma _{2}.$$

Finally, use $\sigma_1 \sigma_2 - \sigma_2 \sigma_1 = 2\sigma_1 \sigma_2.$

Does this give the right answer?

3. Jan 31, 2007

### dextercioby

Yes, it's either that, or that the momentum and the superpotential anticommute. U can compute $H=2Q_1^2$ to convince yourself.

It's correct, as far as we know.

4. Feb 3, 2007

### Tom Mattson

Staff Emeritus
Yes, it does. I actually found my dumb mistake after posting the thread. To finish this off, we would use $[p,\Phi(x)]=-i\hbar\frac{d\Phi(x)}{dx}$ and $\sigma_1\sigma_2=i\sigma_3$. This leads us to the Hamiltonian as presented in the book:

$$H=\left( \frac{p^2}{2m} + \Phi^2(x) \right) \otimes 1 + \frac{\hbar}{\sqrt{2m}}\frac{d\Phi(x)}{dx} \otimes \sigma_3$$

5. Mar 3, 2007

### Tom Mattson

Staff Emeritus
Here's another thing regarding SUSY QM that I am curious about. It involves a contrast between N=1 and N=2 SUSY QM.

N=1 SUSY QM
It has just one supercharge, $Q_1$ given by:

$$Q_1:=\frac{1}{\sqrt{4m}}\left(\vec{p}-\frac{e}{c}\vec{A}\right)\cdot\vec{\sigma}$$

So when we compute the Hamiltionian from $H=2Q_1^2$, we get the usual Pauli Hamiltonian.

$$H=\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{B}\cdot\vec{\sigma}$$

First, this Hamiltonian explicitly involves the so-called minimal substitution for the EM interaction. That makes sense, but why doesn't the N=2 SUSY QM mention the EM potential? I can only assume that it's buried in the SUSY potential $\Phi(x)$ that I mentioned earlier in the thread.

And second, after Junker states the N=1 Hamiltonian, he writes, "It is an amusing and interesting observation that supersymmetry suggests g=2." Well DUH! It looks like the supercharge was specifically written down to produce that result. Why would anyone think this is so special?

Last edited: Mar 3, 2007
6. Mar 7, 2007

### nrqed

Hi Tom.

Maybe I am missing the point, but it does sound a bit surprising to me.
If you impose that the coefficient of the (p-e A/c)^2 term must be correct, this fixes completely the coefficient of the supercharge. There is no freedom left. So the fact that g comes out to be equal to two is intriguing, no?

(it's kind of interesting that the correct value of g comes out of a nonrelativistic but supersymmetric treatment, whereas the correct value of g usually falls out of the relativistic treatment. I wonder if it's purely a coincidence or there is a deeper reason, maybe having to do with the connection between supersymmetry and Poincare invariance...I am just thinking out aloud here)

7. Mar 10, 2007

### Tom Mattson

Staff Emeritus
But the coefficient of the $(p-eA/c)^2$ term is the coefficient of the supercharge. It looks to me like "imposing that the coefficient...must be correct" was exercising some freedom, and that the supercharge was engineered precisely to reproduce the result $g=2$.

The superalgebra that defines SUSY QM is $\{Q_i,Q,j\}H\delta_{ij}$. For N=1, we have only one supercharge $Q$, so $2Q^2=H$. The coefficient of $Q$ is as I posted only if you assume that $H$ is the Pauli Hamiltonian. That is, you get $g=2$ only if you first assume that $g=2$.

Unless, of course, I'm not understanding it correctly.

8. Mar 10, 2007

### nrqed

Hi Tom. By the way, I read some entries in your blog and we are in a similar situation (I did my PhD on the use of effective field theories, did a postdoc, then started teaching at a college and now want to get back doing research in particle physics/string theory).

It seems that we are not on the same wavelength It's very possible that I am missing completely your point so forgive me if that's the case. Here's the way I see it: the supercharge must contain the Pauli matrices and when squared, it must yield the $\frac{(\vec{p} - e \vec{A})^2}{2m}$ term.

So if we write the supercharge as $c_1 (\vec{p} - e \vec{A}) \cdot \vec{\sigma}$ and we square this, we find that the coefficient c_1 is completely fixed by the requirement that we want to recover $\frac{(\vec{p} - e \vec{A})^2}{2m}$. There is no freedom left. Now, we do get a sigma dot B term, but there was no guarantee that g would come out to be equal to 2. Do you see what I mean?

Last edited: Mar 10, 2007
9. Mar 10, 2007

### Tom Mattson

Staff Emeritus
I do see what you mean, but I still don't see why anyone would be impressed by it. You have a supercharge satisfying $H=2Q^2$, and you write down $Q=(4m)^{-1/2}(\vec{p}-e\vec{A}/c)$ and compute $H$. You then observe that enforcing the coefficient on the first term automatically generates the correct coefficients on both terms of the Pauli Hamiltonian. Fine, I understand that.

But there's no reason that you couldn't turn the problem around: Start with the Pauli Hamiltonian and solve the operator equation $H=2Q^2$ for $Q$. You cannot come up with anything other than the stated supercharge (modulo some gauge transformation).

I just don't see why that is so remarkable. It seems no more remarkable than saying that $x=1$ satisfies the equation $2x^2=2$.

10. Mar 10, 2007

### nrqed

Ok. This is what I meant.
But there is no guarantee that it could have worked (unless I am missing something). [/quote]
It's different. The analogy would be more that we have the same situation as starting from, say, 1-2x + x^2 and realizing that this can be written as (1-x)^2. It's lucky that it can be written as a perfect square. If we have 1-2x + 2x^2, then it does not work anymore.

But maybe I am missing something. So let's say that g was equal to 5 instead of 2. You are saying that we can always easily write down a supercharge yielding the correct Hamiltonian no matter what g is. I think that we can't. So tell me what the supercharge would be if g was 5.

Regards

Patrick

11. Mar 10, 2007

### Tom Mattson

Staff Emeritus
Aha. The different g would force us to abandon the superalgebra that defines SUSY in the first place. OK, I get it now.

I think that Junker's admiration of this fact is lost on me because I already knew that the Pauli Hamiltonian could be "factored" like that. It was pointed out in a footnote in Sakurai's Advanced Quantum Mechanics (in fact, he came very close to writing down this supercharge without knowing it, IIRC). So to me it looks like the superalgebra was reverse-engineered to produce the correct answer. In other words, it doesn't look like "SUSY suggests g=2" (as Junker put it). It looks more like "g=2 suggests SUSY".

I think I am missing something big by starting with SUSY QM, and not SUSY QFT, which is where all of this came from.

12. Mar 10, 2007

### nrqed

Is it a good book? I want to learn SUSY (ultimately with the goal to understand supersymmetric field theories, superstring theory and things like the AdS/CFT correspondence). It would be very useful for me to work my way trough a textbook or some good review papers in parallel with someone else because there is nobody within 100 km from where I am who does that type of stuff.

Patrick

13. Mar 10, 2007

### Tom Mattson

Staff Emeritus
The book is OK. There are no exercises, so I fill in the missing lines as exercises. I am doing this as an independent study research course, with the hopes of getting into SUSY QFT in the fall. I thought I would start slow with SUSY QM first and then review QFT over the summer before getting into "the real deal".

14. Mar 13, 2007

### Tom Mattson

Staff Emeritus
By the way, as part of this independent study I've been asked to go through the following paper.

Algebraic Approach to Shape Invariance, by A. B. Balantekin. It was published in Phys. Rev. A, and may look different there. Anyway, if you get the book, the quickest route to understanding this paper would be to read Chapters 2 and 3, then 5. Chapter 5 covers shape invariant potentials.

But I'm getting ahead of myself here. I'll post more on N=2 later today, now that my N=1 questions have been cleared up.

15. Mar 27, 2007

### Tom Mattson

Staff Emeritus
I have a couple of questions about Witten's N=2 SUSY QM, as presented in Junker's book. I'll start with one of them. In Figure 2.1 in Section 2.2.3 of the book, Junker depicts what he calls a "typical" spectrum for good SUSY. He shows 3 degenerate ground states, 2 of even Witten parity and 1 of odd Witten parity. His point is that the states of even and odd W-parity need not pair up in the ground state when SUSY is good. OK, that's fine. But then in section 3.3.1, he gives the ground state for good SUSY in the very same model, or so I would think.

The ground state is:

$$\phi_0^-(x)=\phi_0^-(x_0) \exp \left[ -\frac{\sqrt{2m}}{\hbar}\int_{x_0}^x dz \Phi (z) \right]$$.

Junker then specifically makes the point that "...the eigenvalue $E_0$ is not degenerate."

So to make a long story short, in one chapter he says that in a typical spectrum for good SUSY, the ground state is degenerate. Then in the very next chapter he says that it's not. What gives?