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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.

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

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  • $\begingroup$ 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 $A^\dagger$ $\endgroup$ – Ivan Jan 15 at 19:45
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    $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 15 at 20:29

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