# 1-D harmonic oscillator problem

1. Jan 12, 2010

### 8614smith

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$E=(n+\frac{1}{2})\hbar\omega$$

3. 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?

Last edited by a moderator: Apr 24, 2017
2. Jan 12, 2010

### Staff: Mentor

3. Jan 12, 2010

### jdwood983

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".

Last edited by a moderator: Apr 24, 2017
4. Jan 12, 2010

### 8614smith

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 $$E_2$$

Last edited by a moderator: Apr 24, 2017
5. Jan 12, 2010

### jdwood983

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)

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

and you should also write

$$E^\prime_0=E_1=\left(1+\frac{1}{2}\right)\hbar\omega=\frac{3}{2}\hbar\omega$$

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

$$E^\prime_n=\left(2n+\frac{3}{2}\right)\hbar\omega$$