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I am stuck on this:

Deduce the eigenvalues of the operator $A^{\dagger} A$ are positive, where

$$A^{\dagger} = -\frac{\text{d}}{\text{d}x} + \tanh(x)$$ $$A = \frac{\text{d}}{\text{d}x} + \tanh(x)$$

The point is that those are not matrices, in the usual sense I understand them, so how shall I proceed?

I shall basically solve

$$(A^{\dagger}A)|\phi\rangle = a |\phi\rangle$$

That is

$$\text{det}(A^{\dagger}A - a\mathbb{1}) = 0$$

And here I somehow stopped.

I also managed to compute the commutator, and I found

$$[A^{\dagger}, A] = -2 \left[\frac{\text{d}}{\text{d}x}, \tanh(x)\right]$$

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Enrico M.
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2 Answers2

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You don't actually need to solve for either eigenvectors or eigenvalues, and in fact you don't need to know what $A$ is at all.

To prove the positivity, simply consider any eigenvector $\psi$ of $A^\dagger A$ with eigenvalue $\lambda$, $$ A^\dagger A \psi = \lambda \psi, $$ take the inner product of that equation with $\psi$, $$ ⟨\psi,A^\dagger A \psi⟩ = ⟨\psi,\lambda \psi⟩, $$ and use the defining property of the adjoint, $$ ⟨\psi,A^\dagger A \psi⟩ = ⟨A\psi,A \psi⟩, $$ to obtain a relationship that involves $\lambda$ and quantities (specifically, norms) which must always be non-negative.

If you want to prove that the eigenvalues are strictly positive, on the other hand, then you will need to work a good deal harder to rule out the possibility of a zero eigenvalue. With the information you've provided it's impossible to tell (as there are not enough boundary conditions), but you can start by showing that $\lambda=0$ requires that $$ ⟨A\psi,A \psi⟩=0 $$ and therefore that $A\psi$ itself be zero; that then gives you a solvable ODE that you can couple with the boundary conditions of the problem (you do have them, right?) to tell whether that $\psi$ is a reasonable state or not.

E.P.
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  • I don't have boundary conditions, since this exercise is the second point of a Quantum Mechanics exercise, the first point of which just told me to verify a solution of Schrödinger equation with a given potential and so on... – Enrico M. Aug 13 '18 at 21:06
  • Then you cannot prove strict positivity. – E.P. Aug 14 '18 at 09:01
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The point is that those are not matrices, in the usual sense I understand them, so how shall I proceed?

You might proceed by looking for eigenfunctions $\phi(x)$ such that

$$A^\dagger A\,\phi(x) = c\,\phi(x)$$

and show that all the $c$ are real and positive.

For example, consider the simpler case of $A = \frac{d}{dx}$ so that

$$A^\dagger A = -\frac{d^2}{dx^2}$$

and then the eigenvalue equation is

$$-\frac{d^2}{dx^2}\,\phi(x) = c\, \phi(x)$$

The eigenfunctions are well known to be

$$\phi_k(x) = A\sin(kx) + B\cos(kx)$$

with eigenvalues

$$c_k = k^2$$

Hal Hollis
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