Proove that position x and momentum p operators are hermitian.(adsbygoogle = window.adsbygoogle || []).push({});

Now, more generaly the proof that operator of some opservable must be hermitian would go something like this:

[tex]A\psi_{n}=a_{n}\psi_{n}[/tex]

Where A operator of some opservable, [tex]\psi_{n}[/tex] eigenfunction of that operator and [tex]a_{n}[/tex] are the eingenvalues of that operator, which are real because that is what we messure.

So:

[tex]<\psi_{n}|A|\psi_{n}>=<\psi_{n}|a_{n}|\psi_{n}>=<a_{n}\psi_{n}|\psi_{n}>=<\psi_{n}|a_{n}\psi_{n}>[/tex]

Since [tex]a_{n}[/tex] is real. And then:

[tex]<A^+\psi_{n}|\psi_{n}>=<\psi_{n}|A\psi_{n}> => <A\psi_{n}|\psi_{n}>=<\psi_{n}|A\psi_{n}> => A^+=A[/tex]

But how do I apply this to concrete problem, for example on operator [tex]p_{x}=-ih\frac{d}{dx}[/tex] (I used h for h/2Pi). Would this be a good analogy:

[tex]-ih\frac{d}{dx}u(x)=p_{x}u(x)[/tex]

[tex]u(x)=Cexp(\frac{i}{h}(p_{x}x))[/tex]

[tex]<Cexp(\frac{i}{h}(p_{x}x))|-ih\frac{d}{dx}|Cexp(\frac{i}{h}(p_{x}x))>=<Cexp(\frac{i}{h}(p_{x}x))|p_{x}|Cexp(\frac{i}{h}(p_{x}x))>=<p_{x}^*Cexp(\frac{i}{h}(p_{x}x))|Cexp(\frac{i}{h}(p_{x}x))>=[/tex]

[tex]=<Cexp(\frac{i}{h}(p_{x}x))|p_{x}Cexp(\frac{i}{h}(p_{x}x))>[/tex]

But since [tex]p_{x}[/tex] is real:

[tex]<-ih\frac{d}{dx}(Cexp(\frac{i}{h}(p_{x}x)))|Cexp(\frac{i}{h}(p_{x}x))>=<Cexp(\frac{i}{h}(p_{x}x))|-ih\frac{d}{dx}(Cexp(\frac{i}{h}(p_{x}x)))>[/tex]

Thus operator of p is hermitian.

Is this correct?

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# Homework Help: Proof that x and p are hermitian

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