# Example of Susy QM and its partition function

1. Nov 19, 2015

### nrqed

I am reading an article by Tachikawa on the Nekrasov partition function ("A review on instanton counting and W-algebras"). The article is meant to be pedagogical but I have some trouble with what is supposed to be "baby" examples.

The first one involves susy QM on $\mathbb{C}^2$. He says the following:

"Let us first consider the quantum mechanics of a supersymmetric particle on $\mathbb{C}^2$, parametrized by $(z,w)$. Let the supersymmetry be such that z and w are invariant and $(\bar{z},\psi_{\bar{z}})$ and $(\bar{w}, \psi_{\bar{w}})$ are paired. This system has global symmetries $J_1$ and $J_2$ such that $(J_1,J_2) = (1,0)$ for z and $(J_1,J_2) = (0,1)$ for w.

Let us consider its supersymmetric partition function

$$Z ( \beta; \epsilon_1, \epsilon_2) = tr_{H} (-1)^F e^{\beta \epsilon_1 J_1 } e^{\beta \epsilon_2 J_2}$$

where H is the total Hilbert space. "

He then proceeds to evaluate this. I have further questions but for now, I am trying to understand these two global symmetries. They are just assumed, guessed or they follow from the theory? And what is the meaning of the epsilons? Usually, in a partition function we sum over the exponential of beta times the energy of each state, so here, it looks as if the states are labelled by the values of J1 and J2, with corresponding energies epsilon1 and epsilon2. But what does that mean, exactly? Why are global symmetries showing up in this way in the partition function?

Later on, he adds that the contribution from the the pairs $(\bar{z},\psi_{\bar{z}}$ and $(\bar{w}, \psi_{\bar{w}})$cancel out in the sum over states because they are susy partners. I can buy that. But then he adds that this leaves the sum over the supersymmetric states

$$H_{susy} \simeq \sum_{m,n \leq 0} \mathbb{C} z^m z^n$$

How is it that these are "states"?

I hope someone can give a hand. Thanks in advance.

2. Nov 20, 2015

### fzero

We can put the metric $ds^2 = |dz|^2 + |dw|^2$ on $\mathbb{C}^2$, so the sigma model for $(z,w)$ will be invariant under global phase changes of $z$ and $w$ independently. We can choose the generators of these symmetries to be $J_1 = z \partial_z - \bar{z} \partial_{\bar{z}}$ and similarly for $J_2$ with $w$.

These obviously commute, so we can use their eigenfunctions to describe the bosonic states, which take the form of monomials $c z^m w^n \bar{z}^\bar{m} w^\bar{n}$. I think the choice of $H = \epsilon_1 J_1 + \epsilon_2 J_2$ is convenient because if $\epsilon_{1,2}$ are incommensurate, then the bosonic states will have nondegenerate eigenvalues. I can't relate this to the actual Hamiltonian that one would derive from either the Lagrangian or the supercharge, but I could be missing something.

3. Nov 20, 2015

### nrqed

Very useful, thank you again fzero. Why is it natural to look at transformations that leave the metric invariant in order to guess a Hamiltonian? What is the formal connection?

Tachikawa then consider susy quantum mechanics on $\mathbb{C}P^1$ "under the influence of a magnetic flux of charge j=0,1/2,1...." . He then says

"The supersymmetric Hilbert space is then

$$H_{susy} \simeq \sum_{k=0}^{2j} \mathbb{C} z^k (\partial_z)^{\otimes j}$$

and is the spin j representation of SU(2) acting on $\mathbb{C}P^1$."
I don't know why there is an SU(2) there. Does it have to follow because there are magnetic charges or is it an ad hoc choice he is making? And what is $(\partial_z)^{\otimes j}$ supposed to mean?
Also, I am confused because in the previous example (of $\mathbb{C}^2$ he was talking about the Hilbert space as being the sum of $z^n w^n$. In the present case, what he gives looks like an operator, not a state. In any case, he then adds the mysterious sentence

"Let the global symmetry J to rotate z with charge 1. Then we have
$$Z = tr_{H_{susy}} e^{\beta \epsilon J} = e^{i \beta j \epsilon} + e^{i \beta \epsilon (j-1) } + \ldots e^{-i j\beta \epsilon}$$"

I see that he is taking the trace over states that are eigenstates of J with eigenvalues ranging from -j to j. So it looks like we must think of J as being a J_z of SU(2) (how is this a global symmetry to rotate z with charge 1?) and the supersymmetric states as being states of a given $J_{total}$. But I don't see what the hamiltonian is supposed to be, explicitly and what the states are, and how all this connects to the form he gives for the Hilbert space and for Z.

4. Nov 20, 2015

### fzero

As I mentioned, I can't think of a way to connect these operators to the Hamiltonian that we might compute canonically by considering the sigma model Lagrangian or some other method. So I would be conservative and think of them as symmetries that we use to classify the states. Since they leave the metric invariant, you can show that they leave the Lagrangian invariant and commute with the Hamiltonian (which should be the Laplacian on $\mathbb{C}^2$). We can therefore define a partition function using these operators in place of the Hamiltonian. We might even expect that there are certain quantities that we can compute that aren't sensitive to which operator that we chose, but are properties of the Hilbert space.

There is a natural action of $SU(2)$ on $\mathbb{CP}^1=S^2$ that can be seen in a few ways. In this case, it's easiest to just think of it as the rotations of the sphere, but for more general complex projective spaces other ways are better suited. I wrote a bit about it in a recent post. As in the earlier example, this is an isometry of the metric, Lagrangian, etc., so we should be able to use it in an analogous way.

Tachikawa is not explaining very much of the formalism here. There is a very readable explanation in this Witten paper (scanned version here.), starting with eq. (90) on page 308. The summary is that states with no fermions are described by functions $f(\phi^i)$ of the coordinates $\phi^i$ of the sigma model target space $M$. States with a single fermion are described by vector fields $f_i(\phi)$, and multifermion states are described by antisymmetric tensors. But each of these is an element of the appropriate exterior algebra of $M$, so we can think of the multifermion states as $p$-forms and then $Q$ has the interpretation of the exterior derivative. Then $Q^\dagger$ would be the dual coexterior derivative and $\{Q,Q^\dagger\} = H$ is $d*d + *d d = \Delta$, or the Laplacian on $M$. This is what I meant about the Hamiltonian being the Laplacian.

So in that description of the Hilbert space, the one fermion state is represented by a vector field, while the multifermion state is a tensor product of copies of the vector field. Tachikawa did not include that in the previous description because the combination of $\bar{z}^n$, $\bar{w}^n$ and $(\partial_{\bar{z}})^{\otimes n}$, $(\partial_{\bar{w}})^{\otimes n}$ canceled out in the sum. I didn't include the vector fields because I was lazy and figured you would ask soon enough!

The complex coordinate $z$ is related to the angles of the sphere by
$$z = \tan\tfrac{\theta}{2} \, e^{i\phi},$$
then you can work out that
$$ds^2 = \frac{ |dz|^2}{(1+|z|^2)^2} = \frac{1}{4} ( d\theta^2 + \sin^2\theta d\phi^2).$$
So $J_3$ acts on $z$ by shifting the phase, but the rest of $SU(2)$ has a more complicated action.

As in the previous case, we are using $J_3$ to classify the states in lieu of the canonical Hamiltonian. Here, with the magnetic field, I guess we have a better motivation for interpreting this as a Hamiltonian. Since the holomorphic functions on the sphere are also harmonic, they have zero energy unless there is a magnetic field. The magnetic field splits the energy levels precisely by the eigenvalues of $B J_3$.

5. Nov 20, 2015

### nrqed

I can't thank you enough, this is extremely instructive and interesting. I have to digest this and I will be back with more questions, I am sure. Thank you again for your time!

Patrick

6. Nov 21, 2015

### nrqed

I am trying to understand this line of reasoning (reading Witten's paper, thank you for the reference!). I have what is probably a stupid question: there are i scalar fields $\phi$ and the same number of Weyl spinors $\psi$ (which can be assembled into a Majorana spinor). But each spinor field had two components (let's say we are in four dimensions). A one fermion state is described by a vector field $f_i(\phi)$. I am trying to understand the label "i" here. I think that the "i" labels different vector fields (not the components of a given vector field), correct? But then, in 4 dimensions, a vector field would have four component whereas the spinor that it is supposed to represent has only two components. So what is the precise correspondence between the fermion state and the vector field representing it?

Thank you!

7. Nov 21, 2015

### fzero

The Hilbert space is infinite dimensional when $d>1$, so we can only make the direct geometrical connection between spinors and the exterior algebra in $d=1$. However, it is still worthwhile to figure out the counting.

In $d=4$, as you say, the irreducible spinors are either Weyl (2 complex components) or Majorana (4 real components), so the minimal $N=1$ SUSY has 4 supercharges. In $d=2$, we can impose a Majorana-Weyl condition on spinors. so this is 1 real component. The dimensional reduction of the $d=4$, $N=1$ SUSY to $d=4$ gives what we would call $N= (2,2)$ SUSY in 2d.

If just start in 2d, with real scalar fields $\phi^i$, then it's natural to let SUSY act on each of these to give a fermion $\psi_i$. However, if we choose null coordinates on 2d space, we can let SUSY act only on rightmovers $\phi^i(\bar{\sigma})$ but not leftmovers $\phi^i(\sigma)$. This is called (heterotic) $(1,0)$ SUSY and the version where we put SUSY on both right and leftmovers is $(1,1)$.

Spinors in 1d are constructed from those in 2d by dimensional reduction, so again the minimal spinor has 1 real component. From what I gather in the literature, the reduction of the 2d $(1,1)$ theory is called $N=1$ in 1d and pairs the scalar with a 2-component Majorana spinor with real components. It seems an abuse of notation The reduction of $(1,0)$ gives something called $N=\tfrac{1}{2}$ SUSY in 1d, which is equivalent to setting the components of the Majorana spinor of the $N=1$ theory equal. This is the minimanl SUSY and I would have called this $N=1$.

Witten's construction is consistent with the above. He allows the possibility that $\phi^i$ are complex, so he uses complex spinors $\psi_i$ subject to a Majorana condition. So there is one complex fermionic degree of freedom for each complex bosonic one.

So functions $\phi^i \in \bigwedge^0(M)$. The index $i$ can be properly thought of as being in the tangent space to $M$, so $\psi_i \in \bigwedge^1(M)$. If you wanted to include the spinor index, you would say that $\psi_{i\alpha} \in \bigwedge^1(M)\times S$, where $S$ is the vector space for the spinor index. But the Majorana condition projects $S$ to a one-dimensional space, so we can leave it off.

8. Nov 23, 2015

### nrqed

Thank you again so much! I am slowly digesting your response. Is there a good (i.e. pedagogical) reference for dimensional reduction?

A quick question (until I have absorbed more): by d=1 here, do you mean space-time or just space? On one hand, I would think that even in one space dimension the Hilbert space is infinite (e.g. the 1D harmonic oscillator) so it might be that you have in mind 0+1. On the other hand, Witten seems to be talking about quantum mechanics with the fields $\phi_i$ playing the role of coordinates $x_i$, so I am confused about what is going on. I am correct to interpret what Witten does as taking a quantum field theory, integrating over all the spatial coordinates and ending up with something that is essentially quantum mechanics?

Thanks again for your time. All your posts (not just in reply to me) are filled with great insights.

9. Nov 23, 2015

### fzero

I looked in a few places that I thought it would be explained and it really wasn't. The idea of dimensional reduction is that we start with a field theory with spacetime coordinates $X^M$ and Lorentz group $SO(D-1,1)$. The fields $\Phi(X)$ are in representations $\mathbf{R}$ of the Lorentz group. To dimensionally reduce, we split the coordinates into $X^M=(x^\mu, y^i)$ and restrict to fields that don't depend on the $y^i$. One could explicitly compactify those directions, expand in harmonics and restrict to the zero modes (perhaps by taking the small-volume limit), but it is not necessary to go through all the trouble.

After dimensional reduction, the symmetry of the originial Lorentz group is broken to $SO(d-1,1)\times SO(D-d)$, where the first factor is the new Lorentz symmetry and the second is like a flavor symmetry. The fields decompose into representations $\oplus \mathbf{r}\otimes \mathbf{r}'$. For instance a vector field $A_M\rightarrow A_\mu \oplus A_i$, . where the $A_i$ transform like a scalar under the new Lorentz group.

The key bit that we needed earlier was the corresponding decomposition of the spinors, so we need to know the spinor representations in any dimension. There is a convenient discussion in an appendix of Polchinski's String theory vol. 2. Otherwise a more convenient reference online are the BUSSTEP lectures of Figueroa-O'Farrill, specifically section 5 and the tables therein.

I wasn't clear enough. We can think of the classical fields as maps from "space-time" $B$ to a target space $M$; so $\phi^i(x^\mu): B \rightarrow M$ and $d$ is the dimension of $B$. So quantum mechanics has $d=1$, $x^0=t$ and $B=\mathbb{R}$. On the other hand, the $\phi^i$ are coordinates on $M$ and we've discussed examples where, as complex fields, $\phi^i = z$ or $\phi^i = z,w$.

But this is the Lagrangian description quantum mechanics of a particle on the target space $M$, where $\phi^i(t)$ are the positions and $\dot{\phi}^i$ the velocities. Adding SUSY and passing to the Hamilton which is the Laplacian gives us things like the Schrodinger equation acting on wavefunctions $f(\phi)$.

10. Nov 24, 2015

### nrqed

That's extremely useful and interesting. Thank you again for being so generous with your time! (and thank you for looking for reference and the one you gave). I have to set aside a couple of days to prepare a talk but will come back with more questions, I am afraid.

Regards,

Patrick