Show that an operator is not hermitian. (quantum mechanics)

[leoflindall]

Homework Statement

Show that the operator O = i [tex]\frac{d2}{ dx2[/tex] (please not 2 a squared term, Latex not working. So i (d2/dx2)) is not hermitian operator for a particle in 1D with periodic boundary conditions.

Homework Equations

The Attempt at a Solution

I know to prove an operator is hermitian that T_nm must equal T _mn*.

To show this I would take the intergral [tex]\int \Psi m* T \Psi n[/tex]

What I don't understand is why the operator (in the same conditions) -[tex]\frac{h2}{2m}[/tex] [tex]\frac{d2}{dx2}[/tex] (Please note 2 is a squared term, so h2/2m . d2/dx2) is hermitian and the one above is not?

I assume it is either that the latter is negative, and that if the first operator is positive then it then the when you intergrate by parts it doesn't simplify to T_nm*?

Or that the first operator isn't hermitian as it is complex?

I get the feeling I'm missing something simple here, but any help would be greatly appreciated!

Leo

 

[nickjer]

The kinetic energy operator doesn't have an imaginary term in it "i". Try with the "i" and without and you will see.

You can see the same thing with the first derivative. If you have i d/dx (similar to momentum operator) it is Hermitian. Without the "i" it isn't.

 

[leoflindall]

Thank you for your help!

You can see the same thing with the first derivative. If you have i d/dx (similar to momentum operator) it is Hermitian. Without the "i" it isn't.

Do you mean that without the i is it hermitian, and with it isn't? As I'm told in the question that 'i d2/dx2 is not a hermitian (at least in 1D perodic boundary conditions).

What I don't understand is why the presense of the i term distinguishs between hermitian and non-hermitian?

Thank you for you help!

 

[SpiffyKavu]

One can define a hermitian operator by its effect on the inner product, given by the following.

Operator [tex] A [/tex] is said to be hermitian if:

[tex]
\int \psi_m^* A \psi_n dx = \int \left(A \psi_m \right)^* \psi_n dx
[/tex]

One can use the definition you are familiar with to derive the above definition (and vice versa). If we know that, for operator A, the matrix elements are given by:
[tex]
A_{mn} = \int \psi_m^* A \psi_n dx
[/tex]

we can then find :
[tex] A_{nm}^* = \left( \int \psi_n^* A \psi_m dx \right)^* = \int \left( A \psi_m \right)^* \psi_n dx
[/tex]

So using the first definition, one can simply evaluate the two integrals and see if they give the same answer.

 

[nickjer]

If you don't understand then just work it out. Do the math and you will see why.

 

[leoflindall]

Got it! I had forgotten that the operator was included in the compleax conjugate in the RHS of the intergral.

Silly I know!

Thank you for your help guys, much appreciated.

Leo