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Quantum problem

  • Thread starter eljose
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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]
 
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Answers and Replies

StatusX
Homework Helper
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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.
 
492
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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+...
 
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eljose said:
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+...
[tex]<\phi|a\phi>=<a*\phi|\phi>[/tex] and since p*=p [tex]<\phi|p^{2}|\phi>=<p\phi|p\phi> [/tex]
 
492
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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
 
Galileo
Science Advisor
Homework Helper
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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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