FORUMula1.com - F1 Forum

[welshie]

[Jabberwocky]

I can not see the image

[bud]

mui comico

[texasmr2]

I dont get it??

[madbrad]

An OK joke, bit of a groaner really.

[darwin dali]

Too nerdy and not funny enough.

[texasmr2]

welshie recieve's a 'non-official' warning for stale humor .

[darwin dali]

I can not see the image

[Jabberwocky]

I was blind but now I can see...


well the picture anyway. wished I had not bothered mind. 2/10

[Gaz]

the fact he made a whole new thead about such a s*** joke is terrible.

[madbrad]

... then what does it say that we are making this much volume in this thread about that joke?

[Amanda]

I got all excited when I read the topic title, I though we we're going to have a genuine Quantum Mechanics discussion

[darwin dali]

I got all excited when I read the topic title, I though we we're going to have a genuine Quantum Mechanics discussion

The particle in a 1-dimensional potential energy box is the most simple example where restraints lead to the quantization of energy levels. The box is defined as zero potential energy inside a certain interval and infinite everywhere outside that interval. For the 1-dimensional case in the x direction, the time-independent Schrödinger equation can be written as:[38]

- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} = E \psi.

The general solutions are:

\psi = A e^{ikx} + B e ^{-ikx} \;\;\;\;\;\; E = \frac{\hbar^2 k^2}{2m}

or

\psi = C \sin kx + D \cos kx \; (by Euler's formula).

The presence of the walls of the box restricts the acceptable solutions of the wavefunction. At each wall:

\psi = 0 \; \mathrm{at} \;\; x = 0,\; x = L.

Consider x = 0

* sin 0 = 0, cos 0 = 1. To satisfy \scriptstyle \psi = 0 \; the cos term has to be removed. Hence D = 0.

Now consider: \scriptstyle \psi = C \sin kx\;

* at x = L, \scriptstyle \psi = C \sin kL =0\;
* If C = 0 then \scriptstyle \psi =0 \; for all x. This would conflict with the Born interpretation
* therefore sin kL = 0 must be satisfied, yielding the condition.

kL = n \pi \;\;\;\; n = 1,2,3,4,5,... \;

In this situation, n must be an integer showing the quantization of the energy levels.

[Leo-The_Red]

I got all excited when I read the topic title, I though we we're going to have a genuine Quantum Mechanics discussion

The particle in a 1-dimensional potential energy box is the most simple example where restraints lead to the quantization of energy levels. The box is defined as zero potential energy inside a certain interval and infinite everywhere outside that interval. For the 1-dimensional case in the x direction, the time-independent Schrödinger equation can be written as:[38]

- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} = E \psi.

The general solutions are:

\psi = A e^{ikx} + B e ^{-ikx} \;\;\;\;\;\; E = \frac{\hbar^2 k^2}{2m}

or

\psi = C \sin kx + D \cos kx \; (by Euler's formula).

The presence of the walls of the box restricts the acceptable solutions of the wavefunction. At each wall:

\psi = 0 \; \mathrm{at} \;\; x = 0,\; x = L.

Consider x = 0

* sin 0 = 0, cos 0 = 1. To satisfy \scriptstyle \psi = 0 \; the cos term has to be removed. Hence D = 0.

Now consider: \scriptstyle \psi = C \sin kx\;

* at x = L, \scriptstyle \psi = C \sin kL =0\;
* If C = 0 then \scriptstyle \psi =0 \; for all x. This would conflict with the Born interpretation
* therefore sin kL = 0 must be satisfied, yielding the condition.

kL = n \pi \;\;\;\; n = 1,2,3,4,5,... \;

In this situation, n must be an integer showing the quantization of the energy levels.

erm.......

[scotty]

I got all excited when I read the topic title, I though we we're going to have a genuine Quantum Mechanics discussion

The particle in a 1-dimensional potential energy box is the most simple example where restraints lead to the quantization of energy levels. The box is defined as zero potential energy inside a certain interval and infinite everywhere outside that interval. For the 1-dimensional case in the x direction, the time-independent Schrödinger equation can be written as:[38]

- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} = E \psi.

The general solutions are:

\psi = A e^{ikx} + B e ^{-ikx} \;\;\;\;\;\; E = \frac{\hbar^2 k^2}{2m}

or

\psi = C \sin kx + D \cos kx \; (by Euler's formula).

The presence of the walls of the box restricts the acceptable solutions of the wavefunction. At each wall:

\psi = 0 \; \mathrm{at} \;\; x = 0,\; x = L.

Consider x = 0

* sin 0 = 0, cos 0 = 1. To satisfy \scriptstyle \psi = 0 \; the cos term has to be removed. Hence D = 0.

Now consider: \scriptstyle \psi = C \sin kx\;

* at x = L, \scriptstyle \psi = C \sin kL =0\;
* If C = 0 then \scriptstyle \psi =0 \; for all x. This would conflict with the Born interpretation
* therefore sin kL = 0 must be satisfied, yielding the condition.

kL = n \pi \;\;\;\; n = 1,2,3,4,5,... \;

In this situation, n must be an integer showing the quantization of the energy levels.

That's all well and good, but how much wood would a woodchuck chuck if a woodchuck could chuck wood?