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Quantum problem

  1. Aug 23, 2005 #1
    Quantum problem...:)

    let be the next quantum problem we have a Hamiltonian with a potential V so H=T+V we don,t know if V is real or complex the only thing we now is that if [tex]E_{n} [/tex] is an eigenvalue of the Hamiltonian also its conjugate [tex]E^*_{n}=E_{k} [/tex] will also be an eigenvalue we have necesarily that V is real.. ..my proof is that

    [tex] <\phi]p^{2}+V[\phi>=E_{n} [/tex] taking the conjugate:

    [tex] <\phi^*]p^{2}+V^{*}[\phi^*>=E^*_{n}=E_{k} [/tex] (2)but E_k is also an energy so we would have the equation:

    [tex] <\phi]p^{2}+V[\phi>=E_{k} [/tex]
    (3) ow equating down 2 and 3 we would have that

    [tex]<\phi^*]p^{2*}+V^{*}[\phi^*>= <\phi]p^{2}+V[\phi> [/tex]

    so we would have in the end that [tex]<\phi]V(x)[\phi>=<\phi^*]V^*(x)[\phi^*> [/tex] so the potential would be real....

    we have that [tex] <\phi][x>=\phi^*(x) [/tex] and that [tex] <x][\phi>=\phi(x) [/tex] we would also have that [tex]\phi^*(x,E_{n})=\phi^(x,E^*_{n})=\phi(x,E_{k}) [/tex]

    Another question is suppsed that p^2*=p^2 and that [tex]<p\phi][p\phi>=<\phi]p^{2}[\phi> [/tex]
    Last edited: Aug 23, 2005
  2. jcsd
  3. Aug 23, 2005 #2


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    Are you asking if your work is right? If so, I got lost at the third line. Is that a new wavefunction, or is it the same as the one from line 1? Your notation suggests it's the same, which doesn't follow from your work. And it's easy to prove p^2 is hermitian if p is.
  4. Aug 25, 2005 #3
    Another question how i would prove that [tex]<\phi|p^{2}|\phi>=<p\phi|p\phi> [/tex] ?..we know p is an observable so p=p+...
  5. Aug 25, 2005 #4
    [tex]<\phi|a\phi>=<a*\phi|\phi>[/tex] and since p*=p [tex]<\phi|p^{2}|\phi>=<p\phi|p\phi> [/tex]
  6. Aug 25, 2005 #5
    another question,let,s suppose we have the identity:

    [tex]\int_{-\infty}^{\infty}dx(|\phi_{n}(x)|^{2}+|\phi_{k}(x)|^{2})f(x)=0 [/tex]

    valid for all the k and n eigenfunctions of a given Hamiltonian, then necessarily we must have that f=0
  7. Aug 26, 2005 #6


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    That doesn't necessarily have to be so. You can construct a hamiltonian for which all wavefunctions are zero outside an interval (i.e. infinite sqaure well) and have f(x) take nonzero values outside that interval.
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