12.9 Supersymmetric quantum mechanics

We present, in the next section, the physicists’ proof of the index theorem in its simplest setting. The proof is heavily based on path integral formulation of supersymmetric quantum mechanics (SUSYQM), which will be outlined in the present section.

We have studied the path integral quantization of bosons and fermions. If these particles are combined together, there appears a new symmetry called supersymmetry. We will introduce a special class of SUSYQM later, which turns out to be crucial in the proof of an index theorem.

This and the next sections may be read separately from the previous sections. The necessary tools are supplied to make these sections self-contained. Our exposition follows Alvarez (1995) and Nakahara (1998). Original references are Alvarez-Gaume L (1983) and Friedman and Windey (1984, 1985).

12.9.1 Clifford algebra and fermions

We restrict ourselves to a particle moving in ${\mathbb{R}}^3$ to start with. More general settings will be studied later. Let $\{{\psi }_i\}=\{{\psi }_1,{\psi }_2,{\psi }_3\}$ be real Grassmann variables, where $i=1,2,3$ labels the coordinate index. They satisfy the algebra

$$\{{\psi }_i,{\psi }_j\}=0$$

Let us consider the Lagrangian

$$L=\frac{\mathrm{i}}{2}{\psi }_i{\dot{\psi }}_i-\frac{\mathrm{i}}{2}{ϵ}_{ijk}{B}_i{\psi }_j{\psi }_k\text{\hspace{1em}(12.130)}$$

where $B_i$ is a real number. The canonical conjugate momentum for ${\psi }_i$ is

$${\pi }_i\equiv \frac{\partial L}{\partial {\dot{\psi }}_i}=-\frac{\mathrm{i}}{2}{\psi }_i.$$

$$H=-{\dot{\psi }}_i\frac{\mathrm{i}}{2}{\psi }_i-L=\frac{\mathrm{i}}{2}{ϵ}_{ijk}{B}_i{\psi }_j{\psi }_k.\text{\hspace{1em}(12.131)}$$

$$\theta =\frac{\mathrm{i}}{2}{\psi }_i\mathrm{d}{\psi }_i.\text{\hspace{1em}(12.132)}$$

$$\omega =\mathrm{d}\theta =\frac{\mathrm{i}}{2}\mathrm{d}{\psi }_i\wedge \mathrm{d}{\psi }_i\text{\hspace{1em}(12.133)}$$

$$[{\psi }_j,\mathrm{i}{\psi }_k{]}_{\mathrm{PB}}=\mathrm{i}{\delta }_{jk}.\text{\hspace{1em}(12.134)}$$

$$\left\{{\psi }_j,{\psi }_k\right\}={\delta }_{jk}.\text{\hspace{1em}(12.135)}$$

$$\{{\sigma }_j,{\sigma }_k\}=2{\delta }_{jk}$$

$$H=-\frac{1}{2}\mathbf{B}\cdot \mathbf{\sigma }.\text{\hspace{1em}(12.136)}$$

12.9.2 Supersymmetric quantum mechanics in flat space

$$L=\frac{1}{2}{\dot{x}}_k{\dot{x}}_k+\frac{\mathrm{i}}{2}{\psi }_k{\dot{\psi }}_k.\text{\hspace{1em}(12.137)}$$

Then the Hamiltonian is The Poincaré one-form of this system is The corresponding symplectic two-form is from which we obtain the Poisson bracket Quantization of the system is achieved by replacing this Poisson bracket by the anti-commutation relation This anti-commutation relation is called the Clifford algebra in ${\mathbb{R}}^3$ . Let ${\sigma }_i$ be the $i$ th component of the Pauli matrices. It is easily verified from the observation that ${\psi }_i={\sigma }_i/\sqrt{2}$ is the two-dimensional representation of the Clifford algebra. It is known that the finite-dimensional irreducible representation of the Clifford algebra is unique (modulo conjugate transformations). Thus, the Hilbert space of this system turns out to be $\mathcal{H}={\mathbb{C}}^2$ . The Hamiltonian is rewritten in terms of the Pauli matrices as This Hamiltonian is known as the Pauli Hamiltonian and describes a spin in a magnetic field. Similarly, the Clifford algebra defined in ${\mathbb{R}}^{2n}$ and ${\mathbb{R}}^{2n+1}$ acts on the Hilbert space $\mathcal{H}={\mathbb{C}}^{2^n}$ . The Pauli Hamiltonian is made only of the spin coordinates ${\psi }_i$ and is independent of the space coordinate $x_k$ . Accordingly, it cannot describe a travelling spin. Now the Hamiltonian is modified so that the spin may move around the space. This can be realized by adding a kinetic term to the Hamiltonian. Let us consider a spin in ${\mathbb{R}}^d$ and put $B=0$ to obtain the Hamiltonian

The coefficients of this Lagrangian have been chosen so that the system has a supersymmetry defined later. The canonically conjugate momenta are $p_k={\dot{x}}_k$ and ${\pi }_k=-\mathrm{i}{\psi }_k/2$ , from which we obtain the Poisson brackets of the system

$$[x_j,x_k{]}_{\mathrm{PB}}=[p_j,p_k{]}_{\mathrm{PB}}=0[x_j,p_k{]}_{\mathrm{PB}}=[{\psi }_j,{\psi }_k{]}_{\mathrm{PB}}={\delta }_{jk}.$$

It is easy to derive (anti)commutation relations from these Poisson brackets. The canonical (anti)commutation relations are

$$\left[x_j,x_k\right]=\left[p_j,p_k\right]=0\left[x_j,p_k\right]=\left\{{\psi }_j,{\psi }_k\right\}={\delta }_{jk}.\text{\hspace{1em}(12.138)}$$

The Hamiltonian is

$$H={\dot{x}}_j{p}_j-{\dot{\psi }}_j\frac{\mathrm{i}}{2}{\psi }_j-L=\frac{1}{2}{p}^2=-\frac{1}{2}\Delta \text{\hspace{1em}(12.139)}$$

where $\Delta ={\sum }_{k=1}^d{\partial }_k^2$ is the $d$ -dimensional Laplacian. The Hilbert space on which $H$ acts is $L^2({\mathbb{R}}^d)\otimes {\mathbb{C}}^{2^n}$ , where $L^2({\mathbb{R}}^d)$ stands for the set of square-integrable functions in ${\mathbb{R}}^d$ and $n\equiv [d/2]$ is the integer part of $d/2$ .

Variation of the Lagrangian yields

$$\delta L={\dot{x}}_j\frac{\mathrm{d}}{\mathrm{d}t}\delta x_j+\frac{\mathrm{i}}{2}\delta {\psi }_j{\dot{\psi }}_j+\frac{\mathrm{i}}{2}{\psi }_j\frac{\mathrm{d}}{\mathrm{d}t}\delta {\psi }_j.$$

Let us verify that the Lagrangian is invariant under the following supersymmetry transformation

$$\delta x_j=\mathrm{i}ϵ{\psi }_j\delta {\psi }_j=-ϵ{\dot{x}}_j\text{\hspace{1em}(12.140)}$$

where $ϵ$ is an ‘infinitesimal’ real Grassmann constant. In fact,

$$\begin{array}{rl}\delta L& =\mathrm{i}{\dot{x}}_jϵ{\dot{\psi }}_j-\frac{\mathrm{i}}{2}ϵ{\dot{x}}_j{\dot{\psi }}_j-\frac{\mathrm{i}}{2}{\psi }_jϵ{\ddot{x}}_j\\ & =\mathrm{i}{\dot{x}}_jϵ{\dot{\psi }}_j-\frac{\mathrm{i}}{2}ϵ{\dot{x}}_j{\dot{\psi }}_j-\frac{\mathrm{i}}{2}\frac{\mathrm{d}}{\mathrm{d}t}({\psi }_jϵ{\dot{x}}_j)+\frac{\mathrm{i}}{2}{\dot{\psi }}_jϵ{\dot{x}}_j\\ & =-\frac{\mathrm{i}}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left({\psi }_jϵ{\dot{x}}_j\right)\end{array}\text{\hspace{1em}(12.141)}$$

and the action $S=\int L\mathrm{d}t$ is left invariant. The corresponding charge (the generator) is called the supercharge and defined through the Noether’s theorem as⁴

$$ϵQ\equiv \mathrm{i}ϵp_j{\psi }_j=\mathrm{i}ϵ{\psi }_j{p}_j=\mathrm{i}ϵ{\psi }_j{\dot{x}}_j.\text{\hspace{1em}(12.142)}$$

Exercise 12.8

Show that

$$\delta x_j=\left[x_j,ϵQ\right]\text{\hspace{1em}(12.143)}$$

$$\delta {\psi }_j=\left\{{\psi }_j,ϵQ\right\}.\text{\hspace{1em}(12.144)}$$

4 Note that the mass of the particle is set to unity and hence we have $p_j={\dot{x}}_j$ .

These equations show that $Q$ is the generator of SUSY transformations.

Let us take $d=2n$ to be an even integer and quantize the system in the following. We introduce the matrix representation ${\psi }_j={\gamma }_j/\sqrt{2}$ , which is the generalization of the two-dimensional representation introduced in the previous subsection. Here ${\gamma }_j$ are the $d$ -dimensional Dirac matrices that satisfy the Clifford algebra

$$\left\{{\gamma }_j,{\gamma }_k\right\}=2{\delta }_{ij}.\text{\hspace{1em}(12.145)}$$

The Hamiltonian acts on the Hilbert space

$$\mathcal{H}=L_2({\mathbb{R}}^{2n})\otimes {\mathbb{C}}^{2^n}.$$

The supercharge takes the form, upon diagonalizing the coordinate,

$$Q=\mathrm{i}{\psi }_j{p}_j=\frac{1}{\sqrt{2}}{\gamma }_j\frac{\partial }{\partial x_j}.\text{\hspace{1em}(12.146)}$$

The operator

$$\partial /{\gamma }_j\frac{\partial }{\partial x_j}\text{\hspace{1em}(12.147)}$$

is nothing but the Dirac operator in Euclidean space ${\mathbb{R}}^{2n}$ and plays an important role in the proof of the index theorem.

The hypercharge $Q$ transforms in an interesting way under an SUSY transformation (12.140)

$$\begin{array}{rl}\delta Q& =\mathrm{i}(\delta {\psi }_j){\dot{x}}_j+i{\psi }_j\frac{\mathrm{d}}{\mathrm{d}t}\delta x_j=\mathrm{i}(-ϵ{\dot{x}}_j){\dot{x}}_j+\mathrm{i}{\psi }_j(\mathrm{i}ϵ{\dot{\psi }}_j)\\ & =-\mathrm{i}ϵ{\dot{x}}_j{\dot{x}}_j+ϵ{\psi }_j{\dot{\psi }}_j=-2\mathrm{i}ϵ\left(\frac{1}{2}{\dot{x}}_j{\dot{x}}_j+\frac{\mathrm{i}}{2}{\psi }_j{\dot{\psi }}_j\right)\\ & =-2\mathrm{i}ϵL.\end{array}\text{\hspace{1em}(12.148)}$$

Namely, the variation of the supercharge under an infinitesimal SUSY transformation is the Lagrangian!

We next consider the relation between the supercharge and the Hamiltonian of the system. Let us consider successive SUSY transformations with Grassmann parameters ${ϵ}_1$ and ${ϵ}_2$ . If a transformation with ${ϵ}_1$ is applied first and then ${ϵ}_2$ next, we obtain

$$x_j\stackrel{{ϵ}_1}{\to }{x}_j+\mathrm{i}{ϵ}_1{\psi }_j\stackrel{{ϵ}_2}{\to }{x}_j+\mathrm{i}({ϵ}_1+{ϵ}_2){\psi }_j-\mathrm{i}{ϵ}_1{ϵ}_2{\dot{x}}_j$$

$${\psi }_j\stackrel{{ϵ}_1}{\to }{\psi }_j-{ϵ}_1{\dot{x}}_j\stackrel{{ϵ}_2}{\to }{\psi }_j-({ϵ}_1+{ϵ}_2){\dot{x}}_j-\mathrm{i}{ϵ}_1{ϵ}_2{\dot{\psi }}_j$$

while if the order of the SUSY transformations is reversed,

$$\begin{array}{rl}& x_j\to x_j+\mathrm{i}({ϵ}_1+{ϵ}_2){\psi }_j-\mathrm{i}{ϵ}_2{ϵ}_1{\dot{x}}_j\\ & {\psi }_j\to {\psi }_j-({ϵ}_1+{ϵ}_2){\dot{x}}_j-\mathrm{i}{ϵ}_2{ϵ}_1{\dot{\psi }}_j.\end{array}$$

We find, from these results, the commutation relation of the SUSY variations:

$$\left[{\delta }_{{ϵ}_2},{\delta }_{{ϵ}_1}\right]={\delta }_{{ϵ}_2}{\delta }_{{ϵ}_1}-{\delta }_{{ϵ}_1}{\delta }_{{ϵ}_2}=-2\mathrm{i}{ϵ}_1{ϵ}_2\frac{\partial }{\partial t}.\text{\hspace{1em}(12.149)}$$

The observation that the commutation relation of two SUSY transformations is a time derivative, i.e. the Hamiltonian, suggests that the anti-commutation relation of the supercharge, the generator of the SUSY transformation, also yields the Hamiltonian. In fact,

$$\begin{array}{rl}\{Q,Q\}& =2Q^2=2(\mathrm{i}{p}_j{\psi }_j)(\mathrm{i}{p}_k{\psi }_k)\\ & =-p_j{p}_k\left({\psi }_j{\psi }_k+{\psi }_k{\psi }_j\right)=-p_j{p}_k{\delta }_{jk}\\ & =-2H.\end{array}$$

After all, the SUSY algebra reduces to

$$Q^2=-H.\text{\hspace{1em}(12.150)}$$

Since $Q$ is anti-Hermitian, the Hamiltonian is a Hermite operator with nonnegative spectrum.

In summary, we proved in equations (12.148) and (12.141) that

$$\delta Q=-2\mathrm{i}ϵL\delta L=\frac{1}{2}ϵ\frac{\mathrm{d}Q}{\mathrm{d}t}.\text{\hspace{1em}(12.151)}$$

If these equations are compared with the SUSY transformations (12.140) of the coordinates $x_j$ and ${\psi }_j$ , we readily notice that the roles played by bosonic quantities $(x_j$ and $L$ ) and the fermionic quantities $({\psi }_j$ and $Q$ ) are interchanged. Note that the variation of the supercharge $Q$ in (12.151) is always a time derivative of the Lagrangian $L$ . This observation is crucial in constructing a SUSY-invariant Lagrangian out of a supercharge $Q$ .

12.9.3 Supersymmetric quantum mechanics in a general manifold

Let $M$ be a Riemannian manifold with $\dim M=2n$ . The Riemannian metric is

$$\mathrm{d}{s}^2=g_{\mu \nu }\mathrm{d}{x}^{\mu }\mathrm{d}{x}^{\nu }$$

and the inner product of two vectors $X$ and $Y$ with respect to this metric is denoted as

$$⟨X,Y⟩=g_{\mu \nu }{X}^{\mu }{Y}^{\nu }.$$

The vector ${\psi }^{\mu }(t)$ belongs to $T{M}_{x(t)}$ at each instant of time $t$ . Therefore, ${\psi }^{\mu }(t)$ obeys the ordinary transformation rule for a vector under the coordinate transformation $x^{\mu }\to x^{\prime \mu }=x^{\prime \mu }(x^{\nu })$ :

$${\psi }^{\mu }\to {\psi }^{\prime \mu }=\frac{\partial x^{\prime \mu }}{\partial x^v}{\psi }^v.\text{\hspace{1em}(12.152)}$$

Then, under the SUSY transformation $\delta \equiv {\delta }_{ϵ}$ , the coordinates transform as

$$\delta x^{\prime \mu }=\frac{\partial x^{\prime \mu }}{\partial x^{\nu }}\delta x^{\nu }=\frac{\partial x^{\prime \mu }}{\partial x^{\nu }}\mathrm{i}ϵ{\psi }^{\nu }=\mathrm{i}ϵ{\psi }^{\prime \mu }$$

and

$$\begin{array}{rl}\delta {\psi }^{\prime \mu }& =\frac{{\partial }^2{x}^{\prime \mu }}{\partial x^{\nu }\partial x^{\lambda }}\delta x^{\lambda }{\psi }^{\nu }+\frac{\partial x^{\prime \mu }}{\partial x^{\nu }}\delta {\psi }^{\nu }\\ & =\frac{{\partial }^2{x}^{\prime \mu }}{\partial x^{\nu }\partial x^{\lambda }}\mathrm{i}ϵ{\psi }^{\lambda }{\psi }^{\nu }+\frac{\partial x^{\prime \mu }}{\partial x^{\nu }}(-\mathrm{i}ϵ{\dot{x}}^{\nu })=-ϵ{\dot{x}}^{\prime \mu }\end{array}$$

where the anti-commutativity of Grassmann numbers has been used to obtain the last equality. These transformation rules show that the SUSY transformation is covariant under the coordinate transformation $x^{\mu }\to x^{\prime \mu }$ .

The supercharge $Q$ introduced in the previous subsection should be generalized on the manifold $M$ as

$$Q=\mathrm{i}⟨\dot{x},\psi ⟩=\mathrm{i}{g}_{\mu v}(x){\dot{x}}^{\mu }{\psi }^v.\text{\hspace{1em}(12.153)}$$

The SUSY-invariant Lagrangian on $M$ is constructed from the SUSY variation of this $Q$ as

$$\begin{array}{rl}\delta Q& =\mathrm{i}{\partial }_{\lambda }{g}_{\mu \nu }\delta x^{\lambda }{\dot{x}}^{\mu }{\psi }^{\nu }+\mathrm{i}{g}_{\mu \nu }\delta {\dot{x}}^{\mu }{\psi }^{\nu }+\mathrm{i}{g}_{\mu \nu }{\dot{x}}^{\mu }\delta {\psi }^{\nu }\\ & =\mathrm{i}{\partial }_{\lambda }{g}_{\mu \nu }\mathrm{i}ϵ{\psi }^{\lambda }{\dot{x}}^{\mu }{\psi }^{\nu }+\mathrm{i}{g}_{\mu \nu }(\mathrm{i}ϵ{\dot{\psi }}^{\mu }){\psi }^{\nu }+\mathrm{i}{g}_{\mu \nu }{\dot{x}}^{\mu }(-ϵ{\dot{x}}^{\nu })\\ & =-2\mathrm{i}ϵ[\frac{1}{2}{g}_{\mu v}{\dot{x}}^{\mu }{\dot{x}}^v+\frac{\mathrm{i}}{2}{g}_{\mu v}{\psi }^v{\dot{\psi }}^{\mu }\\ & -\frac{\mathrm{i}}{2}{\dot{x}}^{\mu }\frac{1}{2}\left({\partial }_{\lambda }{g}_{\mu \nu }-{\partial }_{\nu }{g}_{\mu \lambda }-{\partial }_{\mu }{g}_{\lambda \nu }\right){\psi }^{\lambda }{\psi }^{\nu }]\\ & =-2\mathrm{i}ϵ\left(\frac{1}{2}{g}_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }+\frac{\mathrm{i}}{2}{g}_{\mu \nu }{\psi }^{\nu }{\dot{\psi }}^{\mu }+\frac{\mathrm{i}}{2}{\dot{x}}^{\mu }{g}_{\lambda \rho }{\Gamma }_{\mu \nu }^{\rho }{\psi }^{\lambda }{\psi }^{\nu }\right)\end{array}$$

where

$${\Gamma }_{\lambda \mu }^v=\frac{1}{2}{g}^{v\rho }\left({\partial }_{\lambda }{g}_{\rho \mu }+{\partial }_{\mu }{g}_{\lambda \rho }-{\partial }_{\rho }{g}_{\lambda v}\right)$$

is the Christoffel symbol associated with the Levi-Civita connection. Note the symmetry ${\Gamma }_{\mu \nu }^{\lambda }={\Gamma }_{\nu \mu }^{\lambda }$ . By comparing this $\delta Q$ with (12.151), we read off the Lagrangian,

$$\begin{array}{rl}L& =\frac{1}{2}{g}_{\mu \nu }(x){\dot{x}}^{\mu }{\dot{x}}^{\nu }+\frac{\mathrm{i}}{2}{g}_{\mu \nu }(x){\psi }^{\mu }\left(\frac{\mathrm{d}{\psi }^{\nu }}{\mathrm{d}t}+{\dot{x}}^{\lambda }{\Gamma }_{\lambda \kappa }^{\nu }(x){\psi }^{\kappa }\right)\\ & =\frac{1}{2}⟨\dot{x},\dot{x}⟩+\frac{\mathrm{i}}{2}⟨\psi ,\frac{D\psi }{Dt}⟩.\end{array}\text{\hspace{1em}(12.154)}$$

Here $D\psi /Dt$ is the covariant derivative of $\psi$ along the curve $x(t)$

Exercise 12.9

Show that the SUSY variation of the Lagrangian is proportional to the time derivative of the supercharge,

$$\delta L=\frac{1}{2}ϵ\frac{\mathrm{d}Q}{\mathrm{d}t}.\text{\hspace{1em}(12.155)}$$

The quantum version of the supercharge is

$$Q\sim g_{\mu v}{p}^{\mu }{\gamma }^v\text{\hspace{1em}(12.156)}$$

that is the Dirac operator $/\partial$ on $M$

Let us define some symbols that will be employed in the next section. The connection one-form is

$${\Gamma }_v^{\mu }=\mathrm{d}{x}^{\lambda }{\Gamma }_{\lambda v}^{\mu }\text{\hspace{1em}(12.157)}$$

while the Riemann curvature two-form is

$${\mathcal{R}}_{\nu }^{\mu }=\mathrm{d}{\Gamma }_{\nu }^{\mu }+{\Gamma }_{\sigma }^{\mu }\wedge {\Gamma }_{\nu }^{\sigma }.\text{\hspace{1em}(12.158)}$$

The Riemann curvature two-form is expanded in terms of ${\mathrm{dx}}^{\rho }\wedge {\mathrm{dx}}^{\sigma }$ to yield

$${\mathcal{R}}_{\nu }^{\mu }=\frac{1}{2}{R}_{\nu \rho \sigma }^{\mu }\mathrm{d}{x}^{\rho }\wedge \mathrm{d}{x}^{\sigma }\text{\hspace{1em}(12.159)}$$

the component of which is the ordinary Riemann curvature tensor. This component is also written in terms of the connection ${\nabla }_{\mu }$ as

$$\begin{array}{rl}{R}_{\lambda \mu \nu }^{\kappa }& =⟨\mathrm{d}{x}^{\kappa },{\nabla }_{\mu }{\nabla }_{\nu }\frac{\partial }{\partial x^{\lambda }}-{\nabla }_{\nu }{\nabla }_{\mu }\frac{\partial }{\partial x^{\lambda }}⟩\\ & ={\partial }_{\mu }{\Gamma }_{\nu \lambda }^{\kappa }-{\partial }_{\nu }{\Gamma }_{\mu \lambda }^{\kappa }+{\Gamma }_{\nu \lambda }^{\eta }{\Gamma }_{\mu \eta }^{\kappa }-{\Gamma }_{\mu \lambda }^{\eta }{\Gamma }_{\nu \eta }^{\kappa }.\end{array}\text{\hspace{1em}(12.160)}$$

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