Wave function properties.

LagrangeEuler

Wave function and its first derivative must be continuous becaus wave function is solution of Schroedinger equation:
Let's examine one dimensional case.
$\frac{d^2 \psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$
David J. Griffiths gives a problem in his quantum mechanics book
Here is very similar problem. That in infinite potential well function $\psi(x)=Ax$ for $0 \leq x \leq \frac{a}{2}$ and $\psi(x)=A(a-x)$ for $\frac{a}{2} \leq x \leq a$. Otherwise $\psi(x)=0$. This wave function do not have a derivative in point $x=\frac{a}{2}$. Is it possible that we choose wave function of this type?

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ZapperZ

Staff Emeritus
2018 Award
Why does the derivative have to be continuous?

If you have a delta function barrier as your potential, you will see a derivative jump. That isn't continuous.

Zz.

cgk

Wave function resulting from Coulomb potentials also have derivative discontinuities. For example, this applies if atomic cores are modelled as point charges in molecules (as in practice is usually done, it is an excellent approximation), but also for parts of the multi-electron wave function where two electrons come close to each other.

Those things are real. Derivatives do not have to be continuous.

LagrangeEuler

Because if derivative is not continuous then second derivetive can not be defined.

ZapperZ

Staff Emeritus
2018 Award
Because if derivative is not continuous then second derivetive can not be defined.
So? That doesn't dictate that you can't form a wavefunction.

See this:

http://quantummechanics.ucsd.edu/ph130a/130_notes/node154.html

Even with a discontinuity in the first derivative, you still get everything you need out of it.

So what is the problem?

Zz.

LagrangeEuler

Whenever you solve S. equation. For example, in case of potential barrier, in case of finite well... you define boundary condition of wave function and its first derivative. Why? My understanding of that is ok because S. equation is second order differential equation and so wave function and its derivative must be continuous.

This case is not the same. In the infinite well potential is zero. It is not delta function in the half of the well.

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ZapperZ

Staff Emeritus
2018 Award
Whenever you solve S. equation. For example, in case of potential barrier, in case of finite well... you define boundary condition of wave function and its first derivative. Why? My understanding of that is ok S. equation is second order differential equation and so wave function and its derivative must be continuous.

This case is not the same. In the infinite well potential is zero. It is not delta function in the half of the well.
Remember your original premise! You question on whether one can get a wavefunction when there's a discontinuity in the derivative. You have been given at least 2 different examples where this is possible. You never stated any kind restriction on what kind of a potential or over what boundary limits! Look at your original post.

Secondly, why is there a problem with the infinite square well? The ONLY requirement here is that the wavefunction is identically be zero at the boundaries. There's no restriction on the first derivative. There is no wavefuction leaking into the outside of the well, so how can you say there is a discontinuity there? Did you actually computed the first derivative of a wavefunction just outside of the well? How did you manage to do that?

Zz.

LagrangeEuler

In case of infinite potential well function is zero outside of the well, so first derivative of the function is also zero outside of the well. Right?

If it does not depend of the potential why then in the case of potential barrier we employ boundary condition for both wave function and its first derivative?

Thank you for discussion!

ZapperZ

Staff Emeritus
2018 Award
In case of infinite potential well function is zero outside of the well, so first derivative of the function is also zero outside of the well. Right?
No, where did you get such an idea? The wave function does not exist outside the well. How could you get a derivative on a non-existing entity?

There is a difference between {0} and {}. The latter is an empty set, and not the same as a set consisting of zeroes.

Zz.

LagrangeEuler

The wave function iz zero outside the well. How do you mean does not exist. It exist and it is zero!

ZapperZ

Staff Emeritus
2018 Award
The wave function iz zero outside the well. How do you mean does not exist. It exist and it is zero!
Look at your Schrodinger equation. Put V as infinity there. Now solve for the wave function outside the well. Tell me what it is.

Zz.

LagrangeEuler

Look from 14:40 to 15:50.

In one dimensional problems always
$\int^{\infty}_{-\infty}|\psi(x)|^2dx=1$.

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ZapperZ

Staff Emeritus
2018 Award
So? In this case, the normalization boundary is the width of the well because that is the EXTENT of the wave function!

Again, you started off being very particular about the mathematics, asking about the discontinuity and how there could be a wave function. Yet, without solving the wave function itself, you had zero problem claiming that the wave function is zero outside the well. Well then, use the math, solve the differential equation, and derive it for me for the solution outside the well.

I'm waiting....

Zz.

LagrangeEuler

I can solve Sch. eq. in that region in case of fine potential.
$\frac{d^2}{dx^2}\psi(x)+\frac{2m}{\hbar^2}(V-E)\psi(x)=0$
For $V>E$
$\frac{d^2}{dx^2}\psi(x)+\gamma^2\psi(x)=0$
solution is
$\psi(x)=Ce^{\gamma x}+De^{-\gamma x}$
$C=0$
$\psi(x)=De^{-\gamma x}=De^{-\frac{\sqrt{2m}}{\hbar}\sqrt{V-E}x}$
When $V \rightarrow \infty$, $\psi(x)=0$.

ZapperZ

Staff Emeritus
2018 Award
See? That wasn't too difficult, was it?

Now noticed what you did. You set C=0, which is correct. But there's nothing in the mathematics that actually tells you to do that, is there? What you did was you use the physics to indicate that the solution should not blow up! So at some point, you invoke the physics to arrive at the description that is accurate for the system you are solving for. It has nothing to do with the mathematics.

Thus, when you look at the wavefunction, the entire "world" or the system here might as well be only inside the well. The outside solution might as well not exist. I apologize for making the comparison between the empty set and set of zeroes. That was not correct. I wanted to delete it later, but you already replied, so I decided to just keep it.

But coming back to your original question, and as has been pointed out, even by you when you derived the wavefunction (which was why I wanted you to derive it), did you see any problem with the first derivative not being continuous at all in this case? Did the solution missed out on any description that the math might have missed due to such a discontinuity?

Please note that a discontinuity in the wavefunction itself might present a problem in terms of the physics, but not the first derivative.

Zz.

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