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Proving E must exceed the min potential

  • Thread starter emob2p
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  • #1
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I've come across Problem 2.2 out of Griffiths' Intro to Quantum book second edition. The problem says to show that E must exceed the minimum value of V(x) for all normalizable solutions to the Schroed. eq. Naturally I started with the normalization condition: int(|phi|^2)=1 and started taking derivatives on this. However, I cannot arrive at a contradiction. Any thoughts? Or any other ways to show the same result? Thanks.
 

Answers and Replies

  • #2
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hi, emob2p.

I think I would start from the time independent schroedinger equation, [tex]H\phi=E\phi[/tex], where H is the Hamiltonian, [tex]\phi[/tex] the wavefunction. Normally, H=p^2/(2m)+V(x), where p is the momentum operator. First multiply both sides with [tex]\phi^{*}[/tex], then note that p is hermitian (and the [tex]\phi^{*}p^{2}\phi[/tex] term can be made into the square of modulus of a function), and then compare both sides. You may arrive at a contradiction if E is smaller than the minimum of V.
 
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  • #3
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Here's my reasoning. Solve for ψ'' as Griffiths suggests. Now notice that if E<V, then ψ>0 and ψ''>0, or ψ<0 and ψ''<0. From calculus recall that if the second derivative is positive, then you have a local minimum. Therefore as x→∞ so does ψ→∞. In order to be normalizable, ψ must go to 0. So V must be greater than E so that ψ'' and ψ have different signs.
 

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