Prove the operator d/dx is hermitian

[baldywaldy]

Hiya :) the title is meant to be prove it isn't hermitian

Homework Statement

Prove the operator d/dx is hermitian

Homework Equations

I know that an operator is hermitian if it satisfies the equation : <m|Ω|n> = <n|Ω|m>*

The Attempt at a Solution

Forgive the lack of latex , I have know idea how to use it and find it baffling.

the intergral of (fm* d/dx fn) dx = the intergral of fm* d fn
={fm* fn - the intergral of fn d fm*} between the limits x=infinity and - infinity.

This is where i get stuck. I just don't know where to go from here, like i said sorry for the lack of latex usage :(.

Thanks for the help :D

 

[ehild]

Think of integration by parts.

ehild

 

[baldywaldy]

I know intergration by parts but i just don't understand how to apply in this situation because there are two functions and an operator

 

[ehild]

d/dx f means that you differentiate f with respect to x. d/dx f = df/dx = f'

You have to show that [itex]\int{f_n f'_mdx}\neq (\int{f'_n f_mdx})^*[/itex]

ehild

 

[asteriatic]

Hello! I believe there is a mistake in the title of your question. The operator d/dx is actually hermitian, so there is no need to prove that it is not hermitian.

To prove that d/dx is hermitian, we need to show that it satisfies the equation <m|Ω|n> = <n|Ω|m>*. In this case, the operator Ω is d/dx and the states |m> and |n> are the functions fm(x) and fn(x).

Let's start with the left-hand side of the equation: <m|Ω|n>. This is equal to the integral of fm*(x)Ωfn(x)dx. Since Ω = d/dx, we can rewrite this as the integral of fm*(x)(d/dx)(fn(x))dx. Now, using integration by parts, we get:

∫fm*(x)(d/dx)fn(x)dx = fm*(x)fn(x) - ∫fn(x)(d/dx)fm*(x)dx

Note that the integral on the right-hand side is the same as the one we started with, but with the roles of fm(x) and fn(x) switched. Therefore, we can rewrite it as:

∫fm*(x)(d/dx)fn(x)dx = fm*(x)fn(x) - ∫fm(x)(d/dx)fn*(x)dx

Now, let's look at the right-hand side of the equation: <n|Ω|m>*. This is equal to the complex conjugate of <m|Ω|n>, which is the integral of fn*(x)Ωfm(x)dx. Following the same steps as before, we get:

∫fn*(x)(d/dx)fm(x)dx = fn*(x)fm(x) - ∫fn(x)(d/dx)fm*(x)dx

Notice that this is almost the same as what we got for the left-hand side, except for the order of fm and fn. However, since integration is commutative, the order doesn't matter and we can rewrite this as:

∫fn*(x)(d/dx)fm(x)dx = fm*(x)fn(x) - ∫fm(x)(d/dx)fn*(x)dx

And this is