# Are the factors of SUSY factorization the conjugate transpose of each other?

I am trying to learn about SUSY by reading Freed Cooper, et al.

I understand that a Hamiltonian with ground state energy equal to $$0$$ ($$E_0=0$$).

$$- \frac{\hbar^2}{2 m} \frac{d^2}{d x^2} \psi_0 + V_1(x) \psi_0(x) = 0$$

Can be factorized into:

$$H_1 = A^\dagger A$$

where

$$A = \frac{\hbar}{\sqrt{2m}} \frac{d}{dx} + W(x) \quad ; \quad A^\dagger = - \frac{\hbar}{\sqrt{2m}} \frac{d}{dx} + W(x) \qquad (1)$$

Another Hamiltonian can be obtained by reversing the order of the operators:

$$H_2 = A A^\dagger$$

Then the eigenfunctions and eigenvalues of $$H_1$$ and $$H_2$$ are related by:

$$H_1 \psi_n^1 = A^\dagger A \psi_n^1 = E_n^1 \psi_n^1 \rightarrow H_2 ( A \psi_n^1 ) = A A^\dagger A \psi_n^1 = E_n^1 ( A \psi_n^1 )$$

And

$$H_2 \psi_n^2 = A A^\dagger \psi_n^2 = E_n^2 \psi_n^2 \rightarrow H_1 ( A^\dagger \psi_n^2 ) = A^\dagger A A^\dagger \psi_n^2 = E_n^2 ( A \psi_n^2 )$$

(where the upper index is used to identify the Hamiltonian of the eigenfunctions)

These equations relate $$\psi_{n+1}^1$$ to $$\psi^2_n$$.

My question is if $$A$$ and $$A^\dagger$$ are the conjugate transpose of each other. If not, how could I calculate $$< \psi_{n+1}^1 |$$? In order to compute $$< \psi_{n+1}^1 | H_1 |\psi_{n+1}^1 > = E_{n+1}^1$$.

It makes sense because:

$$E_n^2 = <\psi_n^2 | H_2 | \psi_n^2 > = < \psi_{n+1}^1 A^\dagger | H_2 | A \psi_{n+1}^1 >$$

$$=< \psi_{n+1}^1 A^\dagger AA^\dagger A \psi_n^1 > = < \psi_n^1 H_1 H_1 \psi_{n+1}^1 > = (E_n^1)^2$$ However, equation (1) doesn't seem to be the conjugate transpose.

• FWIW: Yes, the operator $A^{\dagger}$ is the Hermitian adjoint of $A$. – Qmechanic Jan 14 at 14:08
• Then, why on equation (1) the sign is reversed where no imaginary number exists? – Ivan Jan 14 at 22:04

Yes, the operator $$A^{\dagger}$$ is the Hermitian adjoint of $$A$$: Integration by parts of $$\frac{d}{d x}$$ causes a minus sign: $$\int_{\mathbb{R}}\!dx~ \phi^{\ast}(x)\frac{d}{d x} \psi(x) ~=~-\int_{\mathbb{R}}\!dx~ \psi(x)\frac{d}{d x} \phi^{\ast}(x)$$ up to boundary terms. (A similar argument yields that the momentum operator $$\hat{p}=\frac{\hbar}{i}\frac{d}{d x}$$ is Hermitian.)
• Could you elaborate more on how the integration by parts causes the minus sign? I don’t really know where you need to integrate to obtain $A^\dagger$ – Ivan Jan 15 at 19:45