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Ene Dene
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Proove that position x and momentum p operators are hermitian.
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?
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?