Hermiticity of AB where A and B are Hermitian operator?

[Haynes Kwon]

Trying to prove Hermiticity of the operator AB is not guaranteed with Hermitian operators A and B and this is what I got:
$$<\Psi|AB|\Phi> = <\Psi|AB\Phi> = ab<\Psi|\Phi>=<B^+A^+\Psi|\Phi>=<BA\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$
but since A and B are Hermitian eigenvalues a and b are real,
Therefore we have
$$ ab<\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$ since multiplication of the numbers commutes, this is same as
$$b^*a^*<\Psi|\Phi>=a^*b^*<\Psi|\Phi> = <AB\Psi|\Phi>$$
So Hermiticity is guaranteed?

[Moderator's note: Moved from a technical forum and thus no template.]

 

[PeroK]

Trying to prove Hermiticity of the operator AB is not guaranteed with Hermitian operators A and B and this is what I got:
$$<\Psi|AB|\Phi> = <\Psi|AB\Phi> = ab<\Psi|\Phi>=<B^+A^+\Psi|\Phi>=<BA\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$
but since A and B are Hermitian eigenvalues a and b are real,
Therefore we have
$$ ab<\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$ since multiplication of the numbers commutes, this is same as
$$b^*a^*<\Psi|\Phi>=a^*b^*<\Psi|\Phi> = <AB\Psi|\Phi>$$
So Hermiticity is guaranteed?

First, you cannot assume that $\Phi, \Psi$ are eigenvectors of $A, B$.

What is the definition of Hermicity?

Do you know any results for taking the Hermitian conjugate of a product?

 

[Haynes Kwon]

First, you cannot assume that $\Phi, \Psi$ are eigenvectors of $A, B$.

What is the definition of Hermicity?

Do you know any results for taking the Hermitian conjugate of a product?

Definition of Hermicity is $$<\Psi|A|\Phi> = <\Psi|A\Phi> = <A\Psi|\Phi>$$ so, $$A^+ = A $$ Taking the Hermitian conjugate of a product AB yields $B^+A^+$.

If I cannot assume that the two wavefunctions are not eignvectors of A,B, how should I approach this proof?

 

[PeroK]

If I cannot assume that the two wavefunctions are not eignvectors of A,B, how should I approach this proof?

This is a good start:

Taking the Hermitian conjugate of a product AB yields $B^+A^+$.

 

[Haynes Kwon]

$$(AB)^+ = B^+A^+ = BA $$ since A and B are Hermitian operators. Now I have to prove the commutator $[A,B] = AB - BA$ may be non-zero. I will try to compute $[A,B]<\Psi|\Phi>$

 

[PeroK]

$$(AB)^+ = B^+A^+ = BA $$ since A and B are Hermitian operators. Now I have to prove the commutator $[A,B] = AB - BA$ may be non-zero. I will try to compute $[A,B]<\Psi|\Phi>$

You could find an example of two Hermitian operators that do not commute. You might already know of such an example.

 

[Haynes Kwon]

You could find an example of two Hermitian operators that do not commute. You might already know of such an example.

I know $[x,p]$ is not zero, so this could be the counter example. But is there any way to generalize this proof?

 

[PeroK]

I know $[x,p]$ is not zero, so this could be the counter example. But is there any way to generalize this proof?

Generalise in what way? What you have shown is that the product is Hermitian if and only if the operators commute. That is a complete statement of the answer.

In general, therefore, the product is not Hermitian. Although you might think that perhaps all Hermitian operators commute, so you need to find an example of two that don't commute.

Position and momentum operators are one example. But, also, you could look at Hermitian 2x2 matrices - that's the simplest system to look at. All you have to do is find two Hermitian 2x2 matrices that do not commute. One counterexample is enough.

 

[Demystifier]

So Hermiticity is guaranteed?

No, if $A$ and $B$ are hermitian, then $AB$ does not need to be hermitian. However, the symmetric product
$$\frac{1}{2}(AB+BA)$$
is always hermitian.

 

[MatthewKrup]

Does this mean that (AB-BA) is also hermitian, and if you might by "i" such as i(AB-BA) and since A and B are hermitian, this product is no longer real, therefore not hermitian?

 

[PeroK]

Does this mean that (AB-BA) is also hermitian

You should be able to prove or disprove this yourself.

and if you might by "i" such as i(AB-BA) and since A and B are hermitian, this product is no longer real, therefore not hermitian?

This is very confused. If $A$ is Hermitian, then $iA$ is skew-Hermitian. Hermitian is not the same as "real".

You can, however, put these two things together to show that if $A, B$ are Hermitian, then so is:
$$\frac{1}{2i}(AB - BA)$$

 

[vanhees71]

and
$$\frac{1}{2}(AB+BA)$$
too.