Quantum Mechanics problem

[Erasmus00]

At infinite potential the probability density becomes zero. However we can still introduce a constant in the exponential part of the solution. e^(-ax+:) thereby introducing a finite probability density at any arbitary boundary...

 

Right, AT INFINITE POTENTIAL THE PROBABILITY DENSITY BECOMES 0. Thats the key point.

 

If you solve for the finite well, you find finite probability at the wall (and even IN the wall), but as you take the limit that the potential at the wall goes to 0, the probability at the wall also goes to 0.

-Will

[dkv]

If the potential at well tends to zero the attenuation constant doesnt tend to be zero. psi(at the wall)= C e^(-alpha*L/2)

U -> 0 we get alpha ->0 psi ->C

U->infinity we get alpha -> infintiy psi -> 0

However given the intial condition the solutions are not unique.

if psi=C e^(-alpha*L/2 + const) then also we get the same the solutions...

 

(we observe only energy and momentum...using which we can calculate psi)

But it is a clear that there are infinite solutions of wave function which satisfy the given energy and momentum .therefore it is impossible to know the precise wave function. That sums up the discussion.

 

There can be an uncertainity in the psi without changing the Energy and Momentum of the system.

[Erasmus00]

However given the intial condition the solutions are not unique. .

 

This is not true, and is your main sticking point. You are simply wrong. In a finite square well, there are three regions we must solve for. Since Schroedinger's equation is a second order diff-eq, we get six constants that must be specified. Agreed?

 

Our first condition is that our probability function be normalizable, which sets two of these constants to 0 (the ones outside the well that grow exponentially). This leaves us with four.

 

The conditions that the wavefunction and its derivative be continuous at the walls gives us three new equations (one is redundant), and reduces us to only 1 constant.

 

Finally, the requirement of normalization (total integrated area of psi^2 is unity) sets this last constant.

 

The solution IS unique. Work through it.

-Will

[dkv]

Our first condition is that our probability function be normalizable, which sets two of these constants to 0 (the ones outside the well that grow exponentially). This leaves us with four.

Without any in the normalizability we can assume finite constants...

psi = A sin(kx) + d

 

At the boudary we get

psi = B e^(-a.L/2 +c) = A sin(kL/2)+d

demanding differential conitnuity we get

-a.Be^(-a.L/2+c) = A.k cos(kL/2)

by dividing eqn 2 by eqn 1 we get

-a=k cos(kL/2)/(sin(kL/2) +2

which can represent the situation equally well for any infinite values of d and c.

Equation 1 and 2 are normalizable.

 

The point is we can choose any aribtary value of d and c and solve the equation to find the wave function, wavelength and attenuation constant.

In other words there are infinite number of solutions of the wave function...

[Erasmus00]

Without any in the normalizability we can assume finite constants...

psi = A sin(kx) + d

 

Without any what in the normalizability? A wavefunction must be normalized for its interpretation to be meaningful!

 

In other words there are infinite number of solutions of the wave function...

 

For a given energy level in one dimension, the solution is unique as is shown in any intro quantum book. Normalization sets the constant you want to add!

-Will

[dkv]

without any loss in the normalizability we can write the equation as

psi = A sin(kx) +d

 

The energy level remains unique for infinite number constants.

Doesnt that say something? The probability wave function doesnt exsit on its own. The probability function depends on the observer and the initial conditions he wishes to use.(not only the measurement but also the initial conditions ... an obeserver can observe the frequency of the events which he wants to see!!)

[Erasmus00]

without any loss in the normalizability we can write the equation as

psi = A sin(kx) +d

 

The energy level remains unique for infinite number constants.

Doesnt that say something? The probability wave function doesnt exsit on its own. The probability function depends on the observer and the initial conditions he wishes to use.(not only the measurement but also the initial conditions ... an obeserver can observe the frequency of the events which he wants to see!!)

 

As I've tried to explain to you over and over, you cannot add that constant d arbitrarily. Only certain choices allow for this to be true, for the infinite well is must be 0. Consider plugging your solution into the infinite square well.

 

[math] -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi = E \psi[/math]

 

[math] \frac{\hbar^2k^2}{2m} A \sin\left(kx\right) = E A\sin\left(kx\right) +d E [/math]

 

So d must be 0. Is this more clear?

-Will

[dkv]

The boundary condition is arbitary.

[Erasmus00]

The boundary condition is arbitary.

 

I didn't invoke a boundary condition above! Only Schroedinger's equation! What, specifically, are you having trouble understanding?

-Will

[dkv]

your solution adds dE why should dE be zero ?

Can you tell me please ?

Arent you reframing the problem by manipulating the equations and creating the madness... Cant you see the standard method of solution on the web ? Cnat you see what I am trying to explain ? You bastards teach stupid physics in schools and colleges .... You will go to the HELL...

[Erasmus00]

your solution adds dE why should dE be zero ?

 

All I did was plug psi into the equation. When you do this, E multiplies through on to psi. For the left side to match the right side of the equation, d must be 0. Its the only solution.

Can you tell me please ?

 

Cant you see the standard method of solution on the web ? Cnat you see what I am trying to explain ?

 

I understand what you are saying- but its simply not correct. The boundary conditions AREN'T arbitrary.

-Will