How Do You Calculate Zero Point Energy for a Half-Harmonic Oscillator?

[physgirl]

Homework Statement

find the zero point energy of particle of mass m in one dimension under:
V(x)=(1/2)kx^2 for x>0; infinity for x<0

Homework Equations

??

The Attempt at a Solution

I'm completely lost :( I took a similar approach to deriving eigenvalues for a particle in a box but that didn't make sense to me... Any starters?

 

[George Jones]

Do you recognize the potential for x>0?

 

[physgirl]

its the potential for a harmonic oscillator?

 

[physgirl]

hmm its a harmonic oscillator in a box with infinite length ?!

 

[George Jones]

its the potential for a harmonic oscillator?

Yes, so, for x>0, the wavefunction has to satisfy the same time-independent Schrodinger that the wavefunction for a harmonic oscillator satisfies. But now there is an extra boundary condition.

 

[physgirl]

so it's a harmonic oscillator that has no access to x values less than 0?

 

[physgirl]

So you would just use the operator H=-(hbar)/(2m)[d^2/dx^2]+(1/2)kx^2
Plug that into the Schrodinger equation... but when do you impose the new boundary condition...?

 

[George Jones]

so it's a harmonic oscillator that has no access to x values less than 0?

Exactly.

So you would just use the operator H=-(hbar)/(2m)[d^2/dx^2]+(1/2)kx^2
Plug that into the Schrodinger equation...

Yes, and you know the solution to [itex]H \psi = E \psi[/itex] for this potential.

but when do you impose the new boundary condition...?

Think about solving the Schrodinger equation for a particle in a box. For what region (in terms of the potential) is this equation solved? Why are boundary conditions imposed for a particle in a box?

 

[physgirl]

I thought for a box particle, really the only significant boundary condition was that the wavefunction has to be continuous. Therefore the cosine term has to disappear to leave with just the sine term that WILL go to 0 when x=0... I'm looking at the solution for the oscillator SE and I just see: (a_n)(H_n)(e^-(1/2)(alpha*x)^2)
I'm not really sure how to impose a boundary condition that says x cannot be below 0...?

 

[George Jones]

I thought for a box particle, really the only significant boundary condition was that the wavefunction has to be continuous. Therefore the cosine term has to disappear to leave with just the sine term that WILL go to 0 when x=0... I'm looking at the solution for the oscillator SE and I just see: (a_n)(H_n)(e^-(1/2)(alpha*x)^2)
I'm not really sure how to impose a boundary condition that says x cannot be below 0...?

For the particle in a box, the wavefuntion is set to zero everywhere the potential in infinite, since the particle has zero probability of being in this region. The Schrodinger is sovled where the potential is not inifinite (V = 0 in the case of the Harmonic oscillator), giving a linear combination of sin and cos. These solutions are joined by the boundary condition requirement that the the wavefunction is continuous at the transition from V non-infinite to V infinite, which means that wavefunction has to be zero where the transition in the potential occurs.

Follow the same strategy for your potential.