QM: Questions on spin

1. Mar 21, 2009

Niles

1. The problem statement, all variables and given/known data
Hi all.

I am looking at the spin-state (i.e. we neglect other degrees of freedom for this system) of two particles 1 and 2 given by:

$$\left| {\psi (t = 0)} \right\rangle = \frac{1}{{\sqrt 3 }}\left( {\left| \uparrow \right\rangle _1 \left| \uparrow \right\rangle _2 + \left| \uparrow \right\rangle _1 \left| \downarrow \right\rangle _2 + \left| \downarrow \right\rangle _1 \left| \downarrow \right\rangle _2 } \right)$$

where the subscript denotes the particle which has a spin-direction given by the arrow (up or down).

Question #1: Say I conduct a measurement of the spin for particle 1, and I get spin up. This means that there are two possible states the system is now in, more specifically the first and second state. Now I measure the spin of particle 2, and I can get either up or down, but what is the probability of this?

Attempt for #1: I believe it is simply ½, since the particles are now in a state, which is a linear combination of the two first states in $\psi(t=0)$. Can you confirm this?

******

Question #2: The two particles are in the "original" state $\psi(t=0)$. We let them evovle by according to the Hamiltonian given by:

$$\hat H = \omega_1 S_{1,z} + \omega_2 S_{2,z},$$

where the omega's are just positive constants and the operatores are spin in the z-direction for particle 1 and 2. I have to find $\psi(t)$ at some random time t.

Attempt for #2: I find the eigenenergies of each of the three possible states at time t=0, and then I just multiply each of the three states in $\psi(t=0)$ with the time-constant exp(-iEt/hbar), with the respective energy for each state.

If this is correct, then why can we just multiply by the exponential time factor? I mean, this came when the solved the S.E., but that was for a different Hamiltonian.

Best regards,
Niles.

2. Mar 21, 2009

xepma

As for the second question, the idea is as follows. Any wavefunction $$\Psi(x,t)$$ can be expanded in terms of energy eigenstates $$\psi_n(x)$$. At t=0, this expansion looks like:

$$\Psi(x,0) = \sum_n c_n \psi_n(x)$$

Now you can prove, on very general grounds, that the evolution of an energy-eigenstate through time is quite simple, namely, we only have an energy and time-dependent phase factor: $$e^{iE_nt/\hbar}$$. So the complete decomposition is:

$$\Psi(x,t) = \sum_n c_n \psi_n(x)e^{iE_nt/\hbar}$$

This is precisely the answer you gave, only now for a specific Hamiltonian (i.e. set of energy states and energy eigenvalues). It is in fact a very general result.

3. Mar 21, 2009

Niles

Thank you for reading through my post, and taking the time to reply. I really appreciate it.

4. Mar 22, 2009

Niles

If the states $\psi_n(x)$ are not energy eigenstates (i.e. not eigenfunctions of the Hamiltonian), then how does the time-dependence look like?

Is it still exp(-iEt/hbar)? Or is it not the energy E, but the relevant eigenvalue that should be used instead?

Last edited: Mar 22, 2009
5. Mar 23, 2009

xepma

No, the time-dependence is always given in terms of the energy-eigenstates. If you use basis states which are not eigenstates of the Hamiltonian, then their evolution gets a lot more complicated. Probably the only way to know what the exact expression is, is to write these basis states in terms of energy eigenstates.

This is why energy-eigenstates are so important: their time evolution is quite simple.

6. Mar 23, 2009

Niles

Ahh, ok, very interesting. Thanks!