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Proof that x and p are hermitian

  1. Aug 16, 2007 #1
    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?
     
  2. jcsd
  3. Aug 16, 2007 #2
    If you want to prove, that eigenvalues of Hermitian operator are real, then you should do something like you have done now. But if you instead want to prove that some given operator is Hermitian, then you should not use eigenstates in your calculation. Operator A being Hermitian means that for two arbitrary physical states [itex]\psi[/itex] and [itex]\phi[/itex] an equation

    [tex]
    \langle A\psi |\phi\rangle = \langle \psi| A\phi\rangle
    [/tex]

    is true. You are not allowed to assume that these states would be eigenstates.

    If the operator is defined in position representation in terms of derivative operators, like the momentum operator is, this proof can be carried out using integration by parts. First write down the inner product in the position representation as an integral, and see what you can do.
     
  4. Aug 17, 2007 #3
    I understand what you're saying but I haven't been able to proove it.
     
  5. Aug 17, 2007 #4

    CompuChip

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    Homework Helper

    Work out the inner product.
    I'll do the x one for you, try the p yourself (it's harder, because you have to find a way to throw the derivative over to the other side ).

    I'll work in "real" space:
    [tex]\langle x\phi | \psi \rangle = \int x^* \phi^*(x) \psi(x) \, \mathrm{d}x [/tex]
    but since in this base the x-operator just multiplies by x, and x* = x as the position is real,
    [tex]\cdots = \int \phi^*(x) x^* \psi(x) \, \mathrm{d}x = \int \phi^*(x) x \psi(x) \, \mathrm{d}x = \langle \phi | x \psi \rangle.[/tex]

    I know, it looks trivial, but as I said, the momentum is harder. Try it (hint: partial integration)
     
  6. Aug 19, 2007 #5
    I did it, thanks. Now that I have seen your example it was not hard for momentum.
     
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