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That's all well and good, but how much wood would a woodchuck chuck if a woodchuck could chuck wood?
, \psi = A e^{ikx} + B e ^{-ikx} \;\;\;\;\;\; E = \frac{\hbar^2 k^2}{2m}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?
Le coeur a ses raisons que la raison ne connaît point. 
That's all well and good, but how much wood would a woodchuck chuck if a woodchuck could chuck wood?
That's all well and good, but how much wood would a woodchuck chuck if a woodchuck could chuck wood?
If a woodchuck could chuck wood, he would and should chuck wood. But if woodchucks can't chuck wood, they shouldn't and wouldn't chuck wood. Though were I a woodchuck, and I chucked wood, I would chuck wood with the best woodchucks that chucked wood
That's all well and good, but how much wood would a woodchuck chuck if a woodchuck could chuck wood?
If a woodchuck could chuck wood, he would and should chuck wood. But if woodchucks can't chuck wood, they shouldn't and wouldn't chuck wood. Though were I a woodchuck, and I chucked wood, I would chuck wood with the best woodchucks that chucked wood
Priceless!!
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.
What's the square root of 1? Answer WGAF!
What's the square root of 1? Answer WGAF!
Lol, the square root of -1 is far more interesting
imaginary! Le coeur a ses raisons que la raison ne connaît point. 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.
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