# Time-evolution of a quantum system

1. May 26, 2014

### Laby

1. The problem statement, all variables and given/known data
Consider an electron bound in a hydrogen atom under the influence of a homogenous
magnetic field B = zˆB  . Ignore the electron spin. The Hamiltonian of the system is H = H0 −ωLz ,where
H0 is the Hamiltonian of the hydrogen atom with the usual eigenstates nlm and eigenenergies (0) En
(we use the superscript (0) to denote the unperturbed hydrogen atom), and ω =|e |B/(2mec).

At t = 0 the system is in the state: |ψ(t=0)>=1/$\sqrt{2}($(|2,1,-1> -|2,1,1>)
For each of the following states
calculate the probability of finding the system at some later time t > 0 in that state:

1(t)>=1/$\sqrt{2}(|2,1,-1> - |2,1,1>) 2(t)>=1/[itex]\sqrt{2}(|2,1,-1> + |2,1,1>) 3(t)>=1/[itex]\sqrt{2}(|2,1,0> 2. Relevant equations |ψ(t)>=U(t)*|ψ(t=0)> U(t)=exp(-i*H*t/\hbar) 3. The attempt at a solution So the time evolution of a system can be described by multiplying the wavefunction at t=0 by the function U(t) which in this case would be exp(-i*(En(o)-|e |B/(2mec)*t/\hbar). But then from there, I'm a bit confused. Do I need to transform the equation into a matrix form and then take the square of the coefficients of the superimposed wavefunctions to get the probabilities? If that was the case, then I'd be able to represent |ψ1(t)>=1/\sqrt{2}[-1,0,1], |ψ2(t)>=1/\sqrt{2}[1,0,1] and then |ψ3(t)>=[0,1,0] right? Thanks for your help. Last edited: May 26, 2014 2. May 26, 2014 ### Laby Ok, so I tried the matrix approach, but it doesn't seem quite right. Here is what I did: With U(t)=exp(-i*(En(o)-|e |B/(2mec)*t/\hbar) I split that up into U(t)=exp(-i*En(o)t/hbar)*exp(i*α*t/\hbar) where α=|e|B/(2mec) from there I used the equation exp(iθ[itex]\widehat{n}$$\bullet$$\vec{σ}$)=Icos(θ)+i$\widehat{n}$$\bullet$$\vec{σ}$sin(θ)

where 'I' is the identity matrix

which gave me Icos(αt)+iBLzsin(αt) where I used the matrix representation of Lz

Now multiplying this new expression for U(t) gives me
|ψ(t)>=exp(i*En(o)*(cos(αt)|ψ1> -i*B*hbar*sin(αt)|ψ2>)

And we can take the square of the coefficients of the wavefunctions as their probabilities. So for example, the probability of |ψ3> would be 0, because it doesn't appear in the equation. Now, apart from appear really tenuous, the squares of the probabilities don't add up to 1, which makes me really doubt this is correct. Maybe I'd need to integrate from t=0 --> infinity, but that's a rather hard integral, and not likely either. Am I at least close?