Solving Semantic Question: Infinite Square with Kinetic Energy

[Void123]

Homework Statement

If I have an electron in an infinite square with kinetic energy greater than the potential moving along a barrier until it reaches a turning point (after, which, is just zero ad infinitum) and I am asked to calculate the probability that it continues along its 'initial path' on the barrier after the turning point.

Would this just be the probability that the potential rises above zero after the turning point?

Otherwise, I'm not sure I understand the question exactly.

Thanks.

Homework Equations

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The Attempt at a Solution

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[Void123]

Correction: its potential step.

 

[jdwood983]

So you have a wave function outside the potential step ($\psi_I$) and a wave function inside the potential step ($\psi_{II}$), correct? It looks to me like you need to calculate $|\psi_{II}|^2$, given that E>V.

 

[Void123]

So you have a wave function outside the potential step ($\psi_I$) and a wave function inside the potential step ($\psi_{II}$), correct? It looks to me like you need to calculate $|\psi_{II}|^2$, given that E>V.

Yes, so the conditions are:

$$\Psi_{1}, x < 0$$

$$\Psi_{2}, x > 0$$

Now, for $$x < 0$$ the potential step is set at 6 eV, where the energy of the electron is 8 eV. At $$x > 0$$ the potential is zero.

So, the probability density after zero is going to be constant (correct?) and since $$E >> V_{0}$$ the particle will be free and unaffected by the potential. However, wouldn't that also make $$\Psi_{2}$$ non-normalizable? How would I evaluate the probability density from 0 to infinity?

 

[Void123]

Am I going to have to develop a wave packet, in terms of the Fourier transform, to solve this problem?

 

[jdwood983]

So, the probability density after zero is going to be constant (correct?) and since $$E >> V_{0}$$ the particle will be free and unaffected by the potential. However, wouldn't that also make $$\Psi_{2}$$ non-normalizable? How would I evaluate the probability density from 0 to infinity?

I'd be careful in saying that $E\gg V_0$ because, given the numbers there, it isn't much greater than, it's just plain greater than.

From what I gather of your problem, this is the picture of what you have:

where you need to find the wave function inside the brown portion. You should know that the basic form of the wave function is an exponential (rather the sum of two exponentials: $\psi(x)= A_r\exp[ikx]+A_l\exp[-ikx]$).

Out of curiosity, what textbook are you using?

 

[Void123]

But how am I going to calculate the probability when the standard wave function (for x > 0) is not normalizable? Or am I missing something here.

The text I'm using is authored by Zettili.

 

[jdwood983]

But how am I going to calculate the probability when the standard wave function (for x > 0) is not normalizable? Or am I missing something here.

The text I'm using is authored by Zettili.

How is it not normalizable?

 

[Void123]

How is it not normalizable?

Well, I'm under the assumption that its going to be a free electron for x > 0, since the potential is zero in that region. And if that is so, I thought free particles produce non-normalizable wave functions in which case it is necessary to construct a wave packet instead.

 

[jdwood983]

Well, I'm under the assumption that its going to be a free electron for x > 0, since the potential is zero in that region. And if that is so, I thought free particles produce non-normalizable wave functions in which case it is necessary to construct a wave packet instead.

I don't have Zettili's textbook, but I do have Griffiths which states clearly (all emphasis are his own):

"We must choose a multiplicative factor so as to ensure that

$$\int_{-\infty}^\infty\left|\Psi(x,t)\right|^2dx=1$$

is satisfied. This process is called normalizing the wave function. For some solutions to the Schrodinger equation, the integral is infinite; in that case no multiplicative factor is going to make it 1. The same goes for the trivial solution $\Psi=0$. Such non-normalizable solutions cannot represent particles, and must be rejected. Physically realizable states correspond to the "square-integrable" solutions to Schrodinger's equation"

So the wave function must be normalizable for it to represent a real particle.

 

[jdwood983]

Well, I'm under the assumption that its going to be a free electron for x > 0, since the potential is zero in that region. And if that is so, I thought free particles produce non-normalizable wave functions in which case it is necessary to construct a wave packet instead.

Also, just because the potential is zero doesn't imply anything about the wavefunction except the following:

$$-\frac{\hbar^2}{2m}\,\frac{\partial^2\Psi(x)}{\partial x^2}+V(x)\Psi(x)=E\Psi(x)\longrightarrow-\frac{\hbar^2}{2m}\,\frac{\partial^2\Psi(x)}{\partial x^2}=E\Psi(x)$$

which you should be able to solve the latter part with ease.