Quantum mechanics, zero point energy

[fluidistic]

Homework Statement

Estimate the zero point energy of an electron confined into a region of 10^{-14}m which is around the magnitude of the nucleus. Compare this energy with the potential energy between a proton and an electron and the gravitational potential energy with the same distance 10^{-14}m. Discuss on the possibility that an electron can be found inside a nucleus.

Homework Equations

E_p_1=-\frac{ke^2}{r}
E_p_2=-\frac{Gm_em_p}{r}.
\int _{0}^{10^{-14}} |\Psi (x)|^2dx=1.

The Attempt at a Solution

I guess I must find that the zero point energy of the electron is greater than both the potential energy gravitational and Coulombian and so there's a possibility to find it inside a nucleus.

What I've done is to write the time independent Schrodinger's equation with a potential function V(x)=0. Because I assumed that there were walls of potential surrounding the electron and inside the region its potential would be 0.
The solution to Schrodinger's equation gave me \Psi (x)=c_1 e^{-\frac{2m_eEx}{\hbar}} + c_2. Just to be sure, the equation I "solved" is -\frac{\hbar}{2m_e} \frac{d^2 \Psi}{dx^2}=E\Psi.
But I'm not given any initial conditions, maybe only contour conditions...
I need some help in order to find c_1 and c_2.

Then I guess I must see if |\Psi |^2 is normalized by evaluating \int _{0}^{10^{-14}} |\Psi (x)| ^2dx and it must equal to 1?
If it's not, I just multiply \Psi by a constant?

 

[fzero]

If the electron is confined to the box, what must be the value of the wavefunction evaluated on the sides of the box?

Also, this is technically a 3D problem, you should make some assumptions about the geometry and set up the proper integrals to compute normalizations and expectation values. You're also missing something in your solution since it's not periodic.

 

[fluidistic]

Thanks a lot for your help.

If the electron is confined to the box, what must be the value of the wavefunction evaluated on the sides of the box?

Like at any other point: zero I believe.

Also, this is technically a 3D problem, you should make some assumptions about the geometry and set up the proper integrals to compute normalizations and expectation values.

I agree the problem should be a 3D one, but since they give the value 10^{-14}m, I think they meant a 1D problem.

You're also missing something in your solution since it's not periodic.

Oh damn it...
Is the equation to solve equivalent to \Psi ''-\frac{E}{K} \Psi =0 with K=-\frac{\hbar}{2m_e}? If so I'll recheck what I've done.

 

[fzero]

Thanks a lot for your help.
Like at any other point: zero I believe.

OK, so that's your boundary condition.

I agree the problem should be a 3D o
ne, but since they give the value 10^{-14}m, I think they meant a 1D problem.

It's possible, especially if you haven't covered the hydrogen atom yet.

Oh damn it...
Is the equation to solve equivalent to \Psi ''-\frac{E}{K} \Psi =0 with K=-\frac{\hbar}{2m_e}? If so I'll recheck what I've done.

Writing it that way will only get you in trouble. The equation is

\Psi ''= -\omega^2 \Psi with \omega= \sqrt{\frac{2m_eE}{\hbar}},

which has periodic solutions.

 

[fluidistic]

OK, so that's your boundary condition.



It's possible, especially if you haven't covered the hydrogen atom yet.



Writing it that way will only get you in trouble. The equation is

\Psi ''= -\omega^2 \Psi =0 with \omega= \sqrt{\frac{2m_eE}{\hbar}},

which has periodic solutions.

Thanks once again.
How did you know that \Psi '' =0 and further, that it's also worth -\omega ^2 \Psi?

Ok I'll work on the solutions.

 

[fzero]

Thanks once again.
How did you know that \Psi '' =0 and further, that it's also worth -\omega ^2 \Psi?

Sorry that was a typo that I've now fixed. I was editing one of your equations and forgot to delete the =0 part.

 

[fluidistic]

Sorry that was a typo that I've now fixed. I was editing one of your equations and forgot to delete the =0 part.

Ok I see.
I reached \Psi (x)=c_1 sin (\omega x) + c_2 \cos (\omega x).
With the boundary condition \Psi (0)=0, I get that c_2=0.

My problem is with the other boundary condition, namely \Psi (10^{-14})=0.
It gives c_1 \sin (\omega \cdot 10^{-14})=0 \Rightarrow c_1=0 and thus \Psi (x)=0 \forall x.

 

[fzero]

Ok I see.
I reached \Psi (x)=c_1 sin (\omega x) + c_2 \cos (\omega x).
With the boundary condition \Psi (0)=0, I get that c_2=0.

My problem is with the other boundary condition, namely \Psi (10^{-14})=0.
It gives c_1 \sin (\omega \cdot 10^{-14})=0 \Rightarrow c_1=0 and thus \Psi (x)=0 \forall x.

You're forgetting that \sin(\omega L) can vanish for certain values of \omega L. Instead of concluding c_1=0, we conclude that \omega and hence the energy E is quantized.

 

[fluidistic]

You're forgetting that \sin(\omega L) can vanish for certain values of \omega L. Instead of concluding c_1=0, we conclude that \omega and hence the energy E is quantized.

You're absolutely right.
I get 10^{-14}\omega =n \pi with n\in \mathbb{Z}. Taking n=1 for ground state, I get E_0=\frac{10^{24}\pi ^2 \hbar}{2 m_e}\approx 7.7 \times 10 ^{24}J which is obviously way too large. I'll recheck my arithmetics.

 

[fluidistic]

Hmm can someone help me fixing my error? I've redone the arithmetics and still find this ridiculous answer.
Edit: I think I made a mistake writing down Schrodinger's equation. There should be a hbar squared term instead of just hbar...

Edit 2: I found out that the zero point energy is around 2.37 \times 10 ^{-8}J (still enormous to me, what do you say?), while the Coulombian potential energy is of the order of 10^{-14}J and the gravitational potential energy if of the order of 10^{-53}J. Now how does this imply that an electron can be found inside a nucleus, I have no idea.