1. Feb 7, 2007

### ultimateguy

1. The problem statement, all variables and given/known data
A is a non-Hermitian operator. Show that

$$i(A-A^t)$$

is a Hermitian operator.

2. Relevant equations
$$\int \psi_1^*\L\psi_2 d\tau=\int (\L\psi_1)^*\psi_2 d\tau$$
$$\int \psi_1^*A^t\psi_2 d\tau=\int (A\psi_1)^*\psi_2 d\tau$$

3. The attempt at a solution

$$\int \psi_1^*i(A-A^t)\psi_2 d\tau$$
$$=\int \psi_1^*iA\psi_2 d\tau + \int \psi_1^*(-iA^t)\psi_2 d\tau$$
$$=\int (iA^t\psi_1)^*\psi_2 d\tau + \int ((-iA)\psi_1)^*\psi_2 d\tau$$
$$=\int i((A-A^t)\psi_1)^*\psi_2 d\tau$$

Is this right? The signs are wrong in the third line, but taking the i out of the complex conjugate brackets fix them. Can I do that?

Last edited: Feb 7, 2007
2. Feb 7, 2007

### Tom Mattson

Staff Emeritus
I am confused. Quick question:

1.) Is $A^t$ supposed to be $A^{\dagger}$, the Hermetian conjugate of $A$?

3. Feb 7, 2007

### ultimateguy

Yes, I didn't know the Latex code for it.

4. Feb 7, 2007

### Tom Mattson

Staff Emeritus
Inside the TeX brackets it's \dagger.

Anyway, what you have here is an abstract operator, not a coordinate representation of an operator. So you shouldn't be throwing it into that integral relation. Just use the following:

Let $A$ and $B$ be operators and let $\lambda\in\mathbb{C}$.

$$(A+B)^{\dagger}=A^{\dagger}+B^{\dagger}$$

$$(\lambda A)^{\dagger}=\lambda^*A^{\dagger}$$

5. Feb 7, 2007

### ultimateguy

I just discovered that I wasn't taking the conjugate of the complex value when doing $$(\lambda A)^\dagger$$ in the method above. When I do this, everything actually works out fine.

6. Feb 7, 2007

### Tom Mattson

Staff Emeritus
No, there's a major problem with your solution, and I already told you what it is. You took an abstract operator and treated it as though it were the coordinate representation of an operator. Specifically, you treated it as though it were a first derivative operator (whether you realize it or not). That's the only way you could have mimicked the integration by parts step that you did with the momentum operator (which really is a first derivative operator). You should just use the abstract rules for operators that I quoted.

7. Feb 7, 2007

### ultimateguy

But I didn't mimic the integration by parts step, I used the definition of an adjoint operator. (Which is the second equation in the list) There's no integration in the method, only substitution.

8. Feb 8, 2007

### dextercioby

My solution

I'd say the problem's more involved than it first looks. One's not forced to reduce his analysis to bounded everywhere defined operators on an abstract complex Hilbert space, so the solution is really involved.

So $A:D(A)\rightarrow \mbox{Ran}(A)$ is a densly defined linear operator on an abstract complex Hilbert space $\mathcal{H}$, i.e.

$$\overline{D(A)}=\mathcal{H}$$ (1)

Therefore its adjoint exists and we also assume that the adjoint is densly defined and therefore its adjoint - $A^{\dagger\dagger}$ exists.

It it crucial to the proof that

$$\overline{D(A)\cap D\left(A^{\dagger}\right)}=\mathcal {H}$$ (2)

so that its adjoint exists so we can prove that

$$i\left(A-A^{\dagger}\right) \subseteq \left(i\left(A-A^{\dagger}\right)\right)^{\dagger}$$ (3)

First we tackle the domain question. The "i" multiplying the operators is inessential to the domain issues and therefore can be neglected.

$$D\left(A-A^{\dagger}\right)=D(A)\cap D\left(A^{\dagger}\right)\subseteq D\left(A^{\dagger\dagger}\right)\cap D\left(A^{\dagger}\right)$$ (4)

since a theorem insures us that, if the double adjoint exists, the double adjoint is an extension of the original operator, i.e.

$$A\subseteq A^{\dagger\dagger}$$ (5)

which means that

$$D(A)\subseteq D\left(A^{\dagger\dagger}\right)$$ (6) and

$$A\psi =A^{\dagger\dagger}\psi , \forall \psi\in D(A)$$ (7)

There's another theorem in Hilbert space that says

$$\left(A^{\dagger}-A^{\dagger\dagger}\right)\subseteq \left(A-A^{\dagger}\right)^{\dagger}$$ (8)

which means that

$$D\left(A^{\dagger}-A^{\dagger\dagger}\right)=D\left(A^{\dagger}\right)\cap D\left(A^{\dagger\dagger}\right)\subseteq D\left(\left(A-A^{\dagger}\right)^{\dagger}\right)$$ (9)

Compare now (9) and (4). It follows that

$$D\left(A-A^{\dagger}\right)\subseteq D\left(\left(A-A^{\dagger}\right)^{\dagger}\right)$$ (10)

from which trivially

$$D\left(i\left(A-A^{\dagger}\right)\right)\subseteq D\left(i\left(A-A^{\dagger}\right)^{\dagger}\right)$$ (11)

So the first part of the proof is done. We're tackling now the ranges/codomains issue. We need to prove that

$$i\left(A-A^{\dagger}\right) \chi =\left(i\left(A-A^{\dagger}\right)\right)^{\dagger} \chi , \ \forall \chi\in D\left(i\left(A-A^{\dagger}\right)\right)$$ (12)

Consider the scalar product:

$$\langle \psi, i\left(A-A^{\dagger}\right)\chi \rangle , \forall \psi \in D\left(A^{\dagger}\right)\cap D\left(A^{\dagger\dagger}\right)$$ (13)

We claim that:

$$\langle \psi, i\left(A-A^{\dagger}\right)\chi \rangle =\langle \psi, i\left(A\chi-A^{\dagger}\chi)\rangle=\langle -iA^{\dagger}\psi,\chi\rangle -\langle -iA^{\dagger\dagger}\psi ,\chi\rangle =\langle -iA^{\dagger}\psi,\chi\rangle -\langle -iA\psi ,\chi\rangle$$ (14)

by the virtue of (7). Then

$$\langle -iA^{\dagger}\psi,\chi\rangle -\langle -iA\psi ,\chi\rangle =\langle i\left(A-A^{\dagger}\right)\psi, \chi\rangle =\langle \psi, \left(i\left(A-A^{\dagger}\right)\right)^{\dagger}\chi\rangle$$ (15)

Compare (14) and (15) and we conclude that
$$i\left(A-A^{\dagger}\right) \chi =\left(i\left(A-A^{\dagger}\right)\right)^{\dagger} \chi$$

So the second part of the proof is completed. The whole proof is complete. QED.

Last edited: Feb 8, 2007
9. Feb 8, 2007

### Stephan Hoyer

ultimateguy: In the form of abstract operators, the definition of a Hermitian operator $A$ is $A = A^\dagger$. You can just use matrix algebra to get your answer. Your second equation in the list that you consider the definition of a Hermitian operator is the follows from the what a Hermitian operator looks like in integral form.

10. Feb 8, 2007

### Tom Mattson

Staff Emeritus
OK, I see that now. I still maintain that you don't have to use that integral relation though. If you want to find the adjoint of that operator, all you have to do is use the rules:

$$\left(i\left(A-A^{\dagger}\right)\right)^{\dagger}=-i\left(A^{\dagger}-A^{\dagger\dagger}\right)=i\left(A-A^{\dagger}\right)$$