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Probability of the lowest eigenenergy, quantum theory



So I have my solutions of part b, but I'm not sure how to go about c.

What eigenfunction am I using, exactly?

For the region ∣x∣<a|x|<a I have the eigenfunction ϕn(x)=Acosnx+Bsinnx\phi_n(x) = A cosnx + B sinnx. The lowest eigenenergy should be at n=0 I think, but doing this leaves just the constant A. Do we work this constant out using a boundary condition?

Edit: Oh of course I'm using cn=<ϕn,ψ>c_n = <\phi_n , \psi> and P(E=En)=∣cn∣2P(E=E_n)=|c_n|^2
(edited 12 years ago)
quantumq.png58.6KB
Original post by wanderlust.xx


So I have my solutions of part b, but I'm not sure how to go about c.

What eigenfunction am I using, exactly?

For the region ∣x∣<a|x|<a I have the eigenfunction ϕn(x)=Acosnx+Bsinnx\phi_n(x) = A cosnx + B sinnx. The lowest eigenenergy should be at n=0 I think, but doing this leaves just the constant A. Do we work this constant out using a boundary condition?

Edit: Oh of course I'm using cn=<ϕn,ψ>c_n = <\phi_n , \psi> and P(E=En)=∣cn∣2P(E=E_n)=|c_n|^2


Normalising your wavefunction should yield A and, yes, n=0 is the lowest energy. This can be seen by the fact that n clearly determines the frequency of the eigenstates and E=h*f
(edited 12 years ago)
Original post by ben-smith
Normalising your wavefunction should yield A and, yes, n=0 is the lowest energy. This can be seen by the fact that n clearly determines the frequency of the eigenstates and E=h*f


Oh yeah.

So after normalising I got ∣A∣2=12a|A|^2 = \frac{1}{2a}.

Thus

1a2∫−aasin(πxa)dx=2π\dfrac{1}{a\sqrt{2}} \displaystyle \int_{-a}^{a} sin(\dfrac{\pi x}{a}) dx = \dfrac{\sqrt{2}}{\pi}

Hence

∣c0∣2=P(En=E0)=4π2|c_0|^2 = P(E_n = E_0) = \dfrac{4}{\pi^2}

Does this look right? Seems fine to me, but I'd appreciate a correction if needed!

Thanks for the help. :smile:
Original post by wanderlust.xx
Oh yeah.

So after normalising I got ∣A∣2=12a|A|^2 = \frac{1}{2a}.

Thus

1a2∫−aasin(πxa)dx=2π\dfrac{1}{a\sqrt{2}} \displaystyle \int_{-a}^{a} sin(\dfrac{\pi x}{a}) dx = \dfrac{\sqrt{2}}{\pi}

Hence

∣c0∣2=P(En=E0)=4π2|c_0|^2 = P(E_n = E_0) = \dfrac{4}{\pi^2}

Does this look right? Seems fine to me, but I'd appreciate a correction if needed!

Thanks for the help. :smile:


Actually, having done a bit of reading, I have feeling n=0 is not actually the lowest energy eigenstate as it makes the wave function discontinuous at the sides of the box. sorry about that.
Original post by ben-smith
Actually, having done a bit of reading, I have feeling n=0 is not actually the lowest energy eigenstate as it makes the wave function discontinuous at the sides of the box. sorry about that.


The problem now is that normalising becomes a pain in the arse, since we'd have ϕ1(x)=Acosx+Bsinx, ∣x∣<a\phi_1(x) = Acosx+Bsinx, \ |x|<a
(edited 12 years ago)
Okay, here's another idea. Since the potential is symmetric, I can split our solutions into even and odd states, yielding

Even:
Unparseable latex formula:

\psi_n^{even}(x)=\dfrac{1}{\sqrt{a}}cos\dfrac{(n-\frac{1}{2})\pix}{a}, \ n= 1,2,... \ |x| < a



Eneven=ℏ2π22ma2(n−12)2E_n^{even} = \dfrac{\hbar^2 \pi^2}{2ma^2}(n-\dfrac{1}{2})^2

Odd:

ψnodd(x)=1asinnπxa, n=1,2,... ∣x∣<a\psi_n^{odd}(x)=\dfrac{1}{\sqrt{a}}sin\dfrac{n \pi x}{a}, \ n= 1,2,... \ |x| < a


Enodd=n2ℏ2π22ma2E_n^{odd} = \dfrac{n^2 \hbar^2 \pi^2}{2ma^2}

Hence our lowest eigenenergies for Even and Odd parity states respectively will be

E1even=ℏ2π28ma2E_1^{even} = \dfrac{\hbar^2 \pi^2}{8ma^2}

E1odd=ℏ2π22ma2E_1^{odd} = \dfrac{\hbar^2 \pi^2}{2ma^2}

but now clearly we have E1even<E1oddE_1^{even} < E_1^{odd}, so we use the even eigenfunction ψ1even(x)=1acosπx2a\psi_1^{even}(x)=\dfrac{1}{\sqrt{a}}cos\dfrac{\pi x}{2a}

in

c1=<ψ1even,ψ>c_1 = <\psi_1^{even}, \psi>

Does this work? If not, why not?

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