1-D harmonic oscillator problem

[8614smith]

Homework Statement

Consider a particle of mass m moving in a one-dimensional potential,

V(x)=\infty for x\leq0

V(x)=\frac{1}{2}m{\omega^2}{x^2} for x>0

This potential describes an elastic spring (with spring constant K = m\omega^2) that can be extended but not compressed.

By reference to the solution of the 1-D harmonic oscillator potential sketch and state the form of the valid eigenfunctions, and state the corresponding eigenvalues, for the ground and first excited state.

Homework Equations

E=(n+\frac{1}{2})\hbar\omega

The Attempt at a Solution

https://www.physicsforums.com/attachment.php?attachmentid=23025&stc=1&d=1263339354
Here is the potential sketch

Since the potential V(x)=\infty for x\leq0, The potential at x=0 is infinity. So the wave function at x=0 must equal 0. If it were anything else the particle would have infinite energy. Therefore eigenfunctions must pass through the origin.

http://131.104.156.23/Lectures/CHEM_207/CHEM_207_Pictures/p75a_72gif

From the graph, n=1 and n=3 are the lowest eigenfunctions to pass through the origin.

Ground state eigenfunction n=1 {E_1}=(1+\frac{1}{2})\hbar\omega=\frac{3}{2}\hbar\omega

1st excited state eigenfunction n=3 {E_2}=(3+\frac{1}{2})\hbar\omega=\frac{7}{2}\hbar\omega

Can someone tell me if this is right?

 

[Mark44]

Please don't double post!

 

[jdwood983]

1st excited state eigenfunction n=3
<br /> {E_2}=(3+\frac{1}{2})\hbar\omega=\frac{7}{2}\hbar\omega<br />

I believe you mean E_3=\cdots and not what you wrote. But otherwise, this is correct. This problem is a good example of http://en.wikipedia.org/wiki/Parity_(physics)#Quantum_mechanics".

 

[8614smith]

I believe you mean E_3=\cdots and not what you wrote. But otherwise, this is correct. This problem is a good example of http://en.wikipedia.org/wiki/Parity_(physics)#Quantum_mechanics".

No, i meant E_2 as i thought it is the 2nd energy level i.e. the 1st excited state, can you tell me why this its not E_2

 

[jdwood983]

No, i meant E_2 as i thought it is the 2nd energy level i.e. the 1st excited state, can you tell me why this its not E_2

The reasoning is purely semantics. Since the formula is E_n=\left(n+\frac{1}{2}\right)\hbar\omega, if you have n=3 in the parenthesis, you should have n=3 as the subscript on E. What you should say is (I made a mistake too)

<br /> E^\prime_1=E_3=\left(3+\frac{1}{2}\right)\hbar\omega=\frac{7}{2}\hbar\omega<br />

and you should also write

<br /> E^\prime_0=E_1=\left(1+\frac{1}{2}\right)\hbar\omega=\frac{3}{2}\hbar\omega<br />

because the first equation is the new first excited state and the second equation is the new ground state. If you calculate more levels, you should find that

<br /> E^\prime_n=\left(2n+\frac{3}{2}\right)\hbar\omega<br />