# Commutator of hermitian operators

1. May 27, 2005

### Gideon

i searched the forum, but nothing came up. My question, how do you prove that [A,B] = iC if A and B are hermitian operators? I understand how C is hermitian as well, but i can't figure out how to prove the equation.

2. May 27, 2005

### seratend

What do you want to see? You have an equation AB -BA= iC

This equation tells you that ([A,B])^+=(AB-BA)^+ where + is the hermitian conjugate
=> ([A,B])^+= BA - AB=[B,A]=-[A,B]

=> the commutator of hermitian operators is an anti hermitian operator.
And an antihermitian operator is an hermitian operator times i.
[A,B] = iC just relates this fact nothing more.

Seratend.

Last edited: May 27, 2005
3. May 27, 2005

### dextercioby

Here's how i'd do it

Let's consider the densely defined linear operator

$$\hat{O}=:\left[\hat{A},\hat{B}\right]_{-}=:\hat{A}\hat{B}-\hat{B}\hat{A}$$ (1)

,where $\hat{A},\hat{B}$ are densly defined linear operators on the Hilbert space $\mathcal{H}$.

The domain of this operator is

$$\mathcal{D}_{\hat{O}} \subset \left(\mathcal{D}_{\hat{A}}\cap\mathcal{D}_{\hat{B}}\right)$$

We want to define the adjoint of $\hat{O}$.

-Let's consider a vector $\phi\in\mathcal{H}$.

-We define the action of a linear functional $F$
*continuous on the image of the linear operator $\hat{O}$
*associated to the linear operator $\hat{O}$ and to a vector $\phi\in\mathcal{H}$
on a vector $\psi\in\mathcal{D}_{\hat{O}}$ by the scalar product in $\mathcal{H}$

$$F_{\hat{O},\phi}\psi =:\langle \phi,\hat{O}\var\psi \rangle$$ (2)

If for a given $\phi\in\mathcal{H}$ the values of the linear functional (2) admit the representation given by the theorem of Riesz,i.e.

$$F_{\hat{O},\phi}\psi =:\langle \chi,\psi\rangle$$ (3)

,with $\chi\in\mathcal{H}$ uniquely determined by $\hat{O}$ and $\phi$,

then there exists a densly defined linear operator denoted by $\hat{O}^{\dagger}$ called the adjoint of the linear operator $\hat{O}$

which obeys

$$\chi=:\hat{O}^{\dagger}\phi$$ (4)

Therefore

$$\phi\in\mathcal{D}_{\hat{O}^{\dagger}}$$ (5)

$$\chi\in\mathcal{I}_{\hat{O}^{\dagger}}$$ (6)

We thus obtain

$$\left\langle \hat{O}^{\dagger}\phi,\psi\right\rangle =\left\langle\phi,\hat{O}\psi\right\rangle$$ (6)

To be continued.

Daniel.

Last edited: May 27, 2005
4. May 27, 2005

### HackaB

WTF are you talking about? It appears that you have been reading too many math books, and that it has affected your brain. Or perhaps you are just trying to show off.

5. May 27, 2005

Staff Emeritus

There speaks a man ever so proud of his own ignorance (or if you are a lady, switch the appropriate pronouns).

6. May 27, 2005

### HackaB

7. May 27, 2005

### dextercioby

So that's how the adjoint of the commutator is defined.

$$F_{-\left[\hat{A},\hat{B}\right]_{-},\phi} =\left\langle\phi,-\left[\hat{A},\hat{B}\right]_{-}\psi\right\rangle =\left\langle\phi,\hat{B}\hat{A}\psi\right\rangle-\left\langle\phi,\hat{A}\hat{B}\psi\right\rangle$$ (7)

$$\psi\in\mathcal{D}_{\left[\hat{A},\hat{B}\right]_{-}} \Rightarrow \psi\in\mathcal{D}_{\hat{A}} \ \mbox{and} \ \psi\in\mathcal{D}_{\hat{B}}$$ (8)

By hypothesis

$$\hat{A}\subset\hat{A}^{\dagger}$$ (9)

$$\hat{B}\subset\hat{B}^{\dagger}$$ (10)

From (8),(9) and (10) we conclude that

$$\psi\in\mathcal{D}_{\hat{A}^{\dagger}} \ \mbox{and} \ \psi\in\mathcal{D}_{\hat{B}^{\dagger}}$$ (11)

Moreover

$$\hat{A}\hat{B}\psi=\hat{A}^{\dagger}\hat{B}^{\dagger} \psi$$ (12)

$$\hat{B}\hat{A}\psi=\hat{B}^{\dagger}\hat{A}^{\dagger} \psi$$ (13)

Using (7),(12) and (13),one gets

$$F_{-\left[\hat{A},\hat{B}\right]_{-},\phi} =\left\langle\phi,\left(\left(B^{\dagger}\hat{A}^{\dagger}-\hat{A}^{\dagger}\hat{B}^{\dagger}\right)\psi\right\rangle=\left\langle\phi,\left(\left(\hat{A}\hat{B}\right)^{\dagger}-\left(\hat{B}\hat{A}\right)^{\dagger}\right)\psi\right\rangle$$ (14)

,where i used the operatorial inclusion

$$\left(\hat{A}\hat{B}\right)^{\dagger} \supseteq \hat{B}^{\dagger}\hat{A}^{\dagger}$$ (15)

Using in (14) another operatorial inclusion

$$\left(\hat{A}-\hat{B}\right)^{\dagger}\supseteq \hat{A}^{\dagger}-\hat{B}^{\dagger}$$ (16)

and the definition of the commutator,one gets

$$F_{-\left[\hat{A},\hat{B}\right]_{-},\phi} =F_{\left[\hat{A},\hat{B}\right]_{-}^{\dagger},\phi}$$ (17)

Q.e.d.

Daniel.

8. May 27, 2005

### dextercioby

You call it "show off",i call it "quantum mechanics in its splendor"

Daniel.

Last edited: May 27, 2005
9. May 27, 2005

### HackaB

OK, call it whatever you want. While you're here, how does that "proof that group multiplication is associative" go again?

10. May 27, 2005

### dextercioby

I hope everyone sees that (17) implies

$$-\left[\hat{A},\hat{B}\right]_{-}\subset \left[\hat{A},\hat{B}\right]_{-}^{\dagger}$$

,even though i've clearly proven only the equality of the 2 images.I've only hinted the way to prove the inclusion of the domains.

Daniel.

11. May 27, 2005

### dextercioby

You'll descover through those 7000+ posts many more mistakes.The important thing is that i've learned from each and everyone of them...

Daniel.