# Hermitian operator

1. Jan 31, 2007

### alisa

show that if A and B are both Hermitian, AB is Hermitian only if [A,B]=0. where or how do io start?

2. Jan 31, 2007

### quasar987

From the property (AB)+=B+a+

+ denoting hermitian conjugation

3. Jan 31, 2007

### cristo

Staff Emeritus
Expand [A,B]=0, then use the hint above.

4. Jan 31, 2007

### dextercioby

Here's how i'd do it. Consider the scalar product

$$\langle x, AB y\rangle$$ (1)

for "x" unspecified yet and $y\in D(AB), \overline{D(AB)}=\mathcal{H}$ an arbitrary element.

$$\langle x, AB y\rangle = \langle x, A^{\dagger}B y\rangle$$ (2)

,where $D(A^{\dagger}B)=D(AB)$, since $D(A)\subseteq D\left(A^{\dagger}\right)$

$$\langle x, A^{\dagger}B y\rangle =\langle x, A^{\dagger}B^{\dagger} y\rangle$$ (3)

as $B\subseteq B^{\dagger}$. Therefore $y\in D\left(A^{\dagger}B^{\dagger}\right)$ and $D(AB)\subseteq D\left(A^{\dagger}B^{\dagger}\right)$.

$$\langle x, A^{\dagger}B^{\dagger} y\rangle =\langle Ax, B^{\dagger} y\rangle$$ (4),

if $x\in D(A)$.

$$\langle Ax, B^{\dagger} y\rangle =\langle BAx, y\rangle$$ (5),

if $x\in D(BA)\subseteq D(A)$.

$$\langle BAx, y\rangle=\langle ABx, y\rangle$$ (6),

since, by hypothesis $AB=BA$.
Finally

$$\langle ABx, y\rangle =\langle x, \left(AB)^{\dagger}y\rangle$$(7)

by the definition of the adjoint. Therefore $y\in D\left((AB)^{\dagger}\right)$ and

$$ABy=(AB)^{\dagger} y {} \wedge D(AB)\subseteq D\left((AB)^{\dagger}\right)$$ (8),

which means the operator AB is symmetric/hermitean.

QED.

Last edited: Jan 31, 2007
5. Jan 31, 2007

### quetzalcoatl9

wouldn't it be easier to just say that x,y are in the domain of the operators A, B (which i agree IS important to state, as is actually applying the operator to an element as opposed to treating it as some "algebraic" quantity as many books do)? in which case the proof would reduce to about 4 lines. in other words, when would the domain of A and its hermitian conjugate NOT be the same?

6. Jan 31, 2007

### wm

Dear alisa, Does it go like this?

Let |n> be an eigenstate of A and B. Then:

AB|n> = Abn|n> = bn.an|n>.

Therefore AB is hermitean because bn.an is real.

In a similar expansion you will find that: [A, B]|n> = 0.

So, given |n>, AB is hermitean and [A, B] = 0.

This should point you in the right direction; check it out, wm

Last edited: Feb 1, 2007
7. Feb 1, 2007

### dextercioby

I don't know about that. I don't claim that my proof is unique/the shortest possible.

By the Hellinger-Toeplitz theorem, iff the operator A is bounded. My proof accounts for the arbitrary character of the A and B operators.

8. Feb 1, 2007

### dextercioby

The spectrum of a hermitean operator is not necessarily real. Actually only the eigenvalues (points in the pure point spectrum) are real, however, one cannot guarantee that $|n\rangle$ is an element of the Hilbert space. Therefore one cannot guarantee that "a_{n}b_{n}" is real. And even if it was real, there are examples of nonhermitean operators with real spectral values.

9. Feb 1, 2007

### wm

Dear dextercioby, At alisa's level I thought it might go like this:

Let

(1) A|x> = a|x>. If A is hermitean then (by definition):

(2) <x|A|x> = <x|A|x>*. Substituting (1) into (2):

(3) <x|a|x> = <x|a|x>*; ie,

(4) a<x|x> = a*<x|x>; ie,

(5) a = a*; ie, a is real (zero being excluded as trivial).

Thus, in the example that I gave: bn.an is real (because each factor is real).

Would that be satisfactory at alisa's level? wm

10. Feb 1, 2007

### dextercioby

I don't know Alisa's level of education. And in this case i'd rather make no assumption, unlike you.

11. Feb 1, 2007

### wm

Dear dextercioby, With alisa's best interests in view, may I suggest that you alert her to the ''level of education'' at which my analysis fails?

Thanks; and best regards, wm

12. Feb 1, 2007

### dextercioby

Your analysis is mathematically faulty. At best it can be considered a heuristic approach with made with disputable arguments. But this is already off-topic, so i'd say "let's drop it".