# first-order energy correction in case of 1-D delta-function

Recall that a 1-D delta-function potential well of the form V (x) =

âˆ’B delta(x) had exactly one bound state, with a double-tailed exponential wave function.

(a) Apply a harmonic oscillator perturbation of the form V â€²(x) = (m omega^2 x^2)/2. Calculate the

ground-state energy for this perturbed system to first order.

(b) Imagine instead that the harmonic oscillator is the original unperturbed system, and

the delta-function potential is the perturbation. Calculate the ground state energy for

this system to first order.

(c) Do your results agree with each other? Discuss why or why not.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Recall that a 1-D delta-function potential well of the form had exactly one bound state with a double-tailed exponential wave function.

(a) Apply a harmonic oscillator perturbation of the form Calculate the

ground-state energy for this perturbed system to first order.

(b) Imagine instead that the harmonic oscillator is the original unperturbed system, and

the delta-function potential is the perturbation. Calculate the ground state energy for

this system to first order.

(c) Do your results agree with each other? Discuss why or why not.

Solution.

1-D delta-function potential well is defined as and it is well known (see for example Ref. 1 and 2) that bound state wave function has next form:

,

where and energy of that bound state is .

a)

It is know from perturbation theory that first-order energy correction is

where is some perturbation (in our case ) and is known and unperturbed wave function (in our case that is bound state wave function of 1-D delta-function potential well).

So, first-order energy correction for will be:

It's easy to show that solution of integral is . Integral can be solved using integration by parts two times. You can also find the solution using http://www.wolframalpha.com/ (input is: x^2 e^(2kx) int from -inf to 0).

First-order energy correction for is

For , first-order energy correction is:

Solution of this integral is the same as the first one . It can be solved using integration by parts two times or website http://www.wolframalpha.com/.

So, for part we have same solution as for

Thus we have .

So, energy of the system corrected to the first-order is:

b)

Now we have harmonic oscillator as unperturbed system and delta-function potential as perturbation. Our tusk is to calculate first-order energy correction for the ground state of the system.

Wave function of quantum harmonic oscillator (see Ref. 3) and energy are

where are Hermite polynomials and

Here we consider ground state only

Perturbation is now .

First-order energy correction is

It is well known (see for example Ref. 4) that . In our case

First-order energy correction is now

.

Energy of system corrected to the first-order is:

.

c)

We can see that we have different energies in part a) and part b). Reason for this is that we have other unperturbed states in case b) and also other perturbations and corrections for these states. In other words when we have delta-function potential as perturbation, it couples nearest neighbour states of harmonic oscillator and we should also consider corrections for this states.

References

[1] https://en.wikipedia.org/wiki/Delta_potential

[2] http://www.physics.udel.edu/~msafrono/424-2011/Lecture%2012.pdf

[3] https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

[4] https://en.wikipedia.org/wiki/Dirac_delta_function

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