Infinite well problem, not normal probability function(?)

[sz0]

Homework Statement

A particle with mass m is trapped inside the infinite potential well:

0<x<L : U(x) = 0
otherwise: U(x) = ∞

ψ(x,t=0) = Ksin(3∏x/L)cos(∏x/L)

What energies can be measured from this system, what are the probabilities for these energies ?

Homework Equations

Schrödinger equation, |ψ(x,t)|^2 = 1

The Attempt at a Solution

I can solve this type of problem normally when ψ = Asin(...) + Bcos(...), but the look of the ψ function confuses me. My guess i should use the Schrödinger equation to give me E like normally.. but maybe that's not how I am supposed to do this. If it is, my math skills are not good enough. I've been trying quite hard to get something good out of (d/dx)^2(ψ) / ψ .

So thanks in advance for any help i can get. :)

 

[vela]

You want to express $\psi(x)$ as a linear combinations of the eigenstates. Try using a trig identity to do this.

 

[sz0]

Thank you, I will work on that this night and read up on how to do it, hopefully I will make it.

 

[sz0]

I haven't solved it, but these are my new trys, am I doing anything right ?

I start with what i think is a trigidentity,
ψ(X) = Ksin(\frac{3πx}{L})cos(\frac{πx}{L}) = \frac{K}{2}[sin(\frac{2πx}{L})+sin(\frac{4πx}{L})] , solving K from this should be trivial, but I will wait with it for now.

I have two approaches.
1: If there is a function ψ'_n = \frac{K}{2}sin(\frac{nπx}{L}) then ψ'₂ + ψ'₄ = ψ, but ψ_n + ψ_m ≠ ψ_n+m right? so what is all this good for either way?

2: \frac{-\hbar}{2m}\frac{dψ}{dx}² = Eψ with ψ =\frac{K}{2}[sin(\frac{2πx}{L})+sin(\frac{4πx}{L})] \Rightarrow \frac{dψ}{dx}² = -\frac{K}{2}[sin(\frac{2πx}{L})+4sin(\frac{4πx}{L})]( \frac{4π}{L})² \Rightarrow E = \frac{2\hbar}{mL^2}\frac{1+8cos(\frac{2πx}{L})}{1+2cos(\frac{2πx}{L})}

But what does this tell me about what values are allowed ? And was this a good way to go ?

Well if you vela or someone else could help me out a bit more I would appreciate it, thanks.

 

[vela]

Read up on the superposition of states.

 

[sz0]

Ok I did, there was nothing about it in my textbook i think so becuse of different notation online it was a bit slow but i hope i got this right.

ψ(X) = Ksin(3πx/L)cos(πx/L) = C[sin(2πx/L)+sin(4π/L)] is what I have after I solve 1= ∫|ψ|^2 dx [0,L] → K = 2/√L . For simplicity i introduce C = 1/√L = K/2 .

I want to write ψ as a linear combination of different ψ_n that is
ψ = Ʃ c_nψ_n = C(c₁sin(x/L)+c₂sin(2x/L)+c₃sin(3x/L)+c₄sin(4x/L)+...) = C[sin(2πx/L)+sin(4π/L)]

Another condition is Ʃc_n² = 1 \Rightarrow
c₂ = c₄ = 1/√2 \Rightarrow Cc₂=Cc₄ = √(2/L) \Rightarrow
ψ₂ = √(2/L)sin(2πx/L), ψ₄ = √(2/L)sin(4πx/L)

And now solving Schrödinger equation with ψ₂
E₂ = \frac{2\hbarπ^2}{mL^2}
with ψ₄
E₄ = \frac{8\hbarπ^2}{mL^2}

Is it correct to say now that these are the measurable energies, and that the constant c_n shows the probability for E_n beeing measured, in this case they are the same so it's 50% each ?

Thanks again.

 

[vela]

Yup, you got it.

 

[sz0]

Fantastic, thanks for your help.