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McCoy13
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First order correction to particle-in-box eigenstates for Dirac perturbation
Calculate the first three nonzero terms in the expansion of the correction to the ground state [itex]\psi^{1}_{1}[/itex] for a Dirac delta perturbation of strength alpha at a/2 (box from 0 to a).
[itex]\psi^{1}_{n} = \sum_{m\neqn} \frac{\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle}{E^{0}_{n}-E^{0}_{m}}\psi^{0}_{m}[/tex]
I started with m=2 and tried to computer the integral [tex]\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle[/tex] by parts.
Calculations:
[tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{2 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]
integration by parts
[tex]\frac{2}{a}\alpha([sin(\frac{2 \pi x}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{2\pi}{a}\int^{a}_{0}cos(\frac{2 \pi x}{a})dx)[/tex]
This clearly equals 0. Also, this equation indicates that the correction will be 0 for any m even.
Trying m=3.
[tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]
integration by parts
[tex]\frac{2}{a}\alpha([sin(\frac{\pi x 3}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{3\pi}{a}\int^{a}_{0}cos(\frac{3 \pi x}{a})dx)[/tex]
However, this also is 0, and will be for any m odd.
This clearly cannot be the correct result.
Homework Statement
Calculate the first three nonzero terms in the expansion of the correction to the ground state [itex]\psi^{1}_{1}[/itex] for a Dirac delta perturbation of strength alpha at a/2 (box from 0 to a).
Homework Equations
[itex]\psi^{1}_{n} = \sum_{m\neqn} \frac{\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle}{E^{0}_{n}-E^{0}_{m}}\psi^{0}_{m}[/tex]
The Attempt at a Solution
I started with m=2 and tried to computer the integral [tex]\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle[/tex] by parts.
Calculations:
[tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{2 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]
integration by parts
[tex]\frac{2}{a}\alpha([sin(\frac{2 \pi x}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{2\pi}{a}\int^{a}_{0}cos(\frac{2 \pi x}{a})dx)[/tex]
This clearly equals 0. Also, this equation indicates that the correction will be 0 for any m even.
Trying m=3.
[tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]
integration by parts
[tex]\frac{2}{a}\alpha([sin(\frac{\pi x 3}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{3\pi}{a}\int^{a}_{0}cos(\frac{3 \pi x}{a})dx)[/tex]
However, this also is 0, and will be for any m odd.
This clearly cannot be the correct result.
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