- #1
jaydnul
- 558
- 15
To start, we have:
[tex]ψ=e^{i(kx+wt)}[/tex]
[tex]ψ_x=ikψ[/tex]
[tex]ψ_{xx}=-k^2ψ[/tex]
[tex]λ=\frac{h}{p}[/tex]
[tex]k=\frac{2π}{λ}[/tex]
[tex]E=\frac{p^2}{2m}+V[/tex]
Then, multiply both sides of energy equation by ψ to get:
[tex]Eψ=\frac{p^2}{2m}ψ+Vψ[/tex]
And replace [itex]-k^2[/itex] in the wave equation with [itex]-\frac{p^2}{hbar^2}[/itex]. Then plug into energy equation for [itex]p^2[/itex] to get:
[tex]Eψ=-\frac{hbar^2}{2m}ψ_{xx}+Vψ[/tex]
1. So why do we just assume the the wave function will be a complex exponential when it could be anything. If we used a sine and cosine wave function, this derivation wouldn't work, but yet, after you get the final equation, you can get a sine and cosine wave function out of it.
2. This might be a dumb one, but why don't we just leave it at
[tex]ψ_{xx}=-\frac{p^2}{hbar^2}ψ[/tex]
and call it a day, we have our function because now we could plug in our value for p and find the wave function.
3. Also, since we used the De broglie equation for lambda, doesn't that mean that the De Broglie wavelength is really the wavelength of the particles wavefunction?
[tex]ψ=e^{i(kx+wt)}[/tex]
[tex]ψ_x=ikψ[/tex]
[tex]ψ_{xx}=-k^2ψ[/tex]
[tex]λ=\frac{h}{p}[/tex]
[tex]k=\frac{2π}{λ}[/tex]
[tex]E=\frac{p^2}{2m}+V[/tex]
Then, multiply both sides of energy equation by ψ to get:
[tex]Eψ=\frac{p^2}{2m}ψ+Vψ[/tex]
And replace [itex]-k^2[/itex] in the wave equation with [itex]-\frac{p^2}{hbar^2}[/itex]. Then plug into energy equation for [itex]p^2[/itex] to get:
[tex]Eψ=-\frac{hbar^2}{2m}ψ_{xx}+Vψ[/tex]
1. So why do we just assume the the wave function will be a complex exponential when it could be anything. If we used a sine and cosine wave function, this derivation wouldn't work, but yet, after you get the final equation, you can get a sine and cosine wave function out of it.
2. This might be a dumb one, but why don't we just leave it at
[tex]ψ_{xx}=-\frac{p^2}{hbar^2}ψ[/tex]
and call it a day, we have our function because now we could plug in our value for p and find the wave function.
3. Also, since we used the De broglie equation for lambda, doesn't that mean that the De Broglie wavelength is really the wavelength of the particles wavefunction?
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