evildarklord1985
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How Does Symmetry Simplify Quantum Harmonic Oscillator Integrals?
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Show work done and the formulas /relations you know, then someone will try to help you.
evildarklord1985
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My first thought about this problem is trying to set up a boundary condition for the wavefunction. For 0<x<A, the wavefunction takes the form of simple harmonic oscillators, i.e. N exp(-1/2 alpha x^2) . With boundary condition where psi(x) = 0 at x=0, I'm stuck with the expression of N=0. Maybe, I didn't think it right...
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N is just a normalization factor, leave that now for the moment.
But you also have Hermite polynomials in the harmonoc wave functions right?
But you also have Hermite polynomials in the harmonoc wave functions right?
evildarklord1985
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so, the harmonic wave function can be written as : N H(y) exp(-1/2 alpha x^2) . With the mentioned boundary condition, do I get N*H(y) = 0 ...how would I go from here on to solve the problem?
eys_physics
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The energy is calculated as the expectation value of the Hamiltonian i.e.:
[tex]E_{n}=\int{{\phi_{n}{(x)H\phi_{n}(x)dx}}[/tex]
where [tex]\phi_{n}(x)[/tex] is the eigenfunction for the energy-level [tex]E_{n}[/tex]
In the case of an ordinary one-dimensional harmonic oscillator
[tex]H=\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}[/tex]
And therefore the energy is given by
[tex]E_{n}=\int_{-\infty}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}[/tex]
In your case you instead integrate from 0 to [tex]\infty[/tex].
So the energy is:
[tex]E_{n}=\int_{0}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}[/tex]
[tex]E_{n}=\int{{\phi_{n}{(x)H\phi_{n}(x)dx}}[/tex]
where [tex]\phi_{n}(x)[/tex] is the eigenfunction for the energy-level [tex]E_{n}[/tex]
In the case of an ordinary one-dimensional harmonic oscillator
[tex]H=\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}[/tex]
And therefore the energy is given by
[tex]E_{n}=\int_{-\infty}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}[/tex]
In your case you instead integrate from 0 to [tex]\infty[/tex].
So the energy is:
[tex]E_{n}=\int_{0}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}[/tex]
evildarklord1985
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but then how do you actually carry out the intergration?
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have you tried? Where do you get stuck?
eys_physics
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Hi again
I did only give you a clue so you could continue by yourself. But maybe you need some more help. As you know the integral over the even interval [tex]-\infty[/tex] to [tex]\infty[/tex] you can use the fact that the integrand is an even function (why?), i.e.
[tex]f(-x)=f(x)[/tex]
Know you want to integrate over the half interval and therefore use the rule (I suppose you have seen it in some course in mathematics):
[tex]\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx[/tex]
I will give you a detailed solution because I want you to try by yourself, but kmow I suppose you have pathway at least.
I did only give you a clue so you could continue by yourself. But maybe you need some more help. As you know the integral over the even interval [tex]-\infty[/tex] to [tex]\infty[/tex] you can use the fact that the integrand is an even function (why?), i.e.
[tex]f(-x)=f(x)[/tex]
Know you want to integrate over the half interval and therefore use the rule (I suppose you have seen it in some course in mathematics):
[tex]\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx[/tex]
I will give you a detailed solution because I want you to try by yourself, but kmow I suppose you have pathway at least.