An electron is described by the Hamiltonian

$ H=\frac{e}{mc}\bar{S}\cdot\bar{B} $

where $\bar{S} =(S_x,S_y,S_z)$ is the spin operator and $\bar{B}$ the magnetic field.

For $t>0$ $\bar{B}=B_0\hat{x}$ and for $t=0$ the electron is in the state $|\psi\rangle=\frac{1}{\sqrt3}|+\rangle+\frac{\sqrt2}{\sqrt3}|-\rangle$, with $|+\rangle$ and $|-\rangle$ eigenvectors of the operator $S_z$ (with eigenvalues $\pm \frac{\hbar}{2}$).

I want to determine the time evolution of the state.

Since $B$ is along x, the dot product gives me $S_xB_0$ and I know that the operator $S_x$ is rapresented by the matrix $ \frac{\hbar}{2}\bigl(\begin{smallmatrix} 0 & 1 \\1 & 0\end{smallmatrix}\bigr)$.

For convenience $\frac{e\hbar B_0}{2mc}=\epsilon$.

Now, avoiding all the calculations (hoping I did them right), what I did is to find the eigenvalues and eigenvectors of my operator by setting the equation:

$ H(\alpha|+\rangle+\beta|-\rangle)=E(\alpha|+\rangle+\beta|-\rangle) $

From which the eigenvalues are $\epsilon$ and $-\epsilon$, and the respective eigenvectors (determining the relation between $\alpha$ and $\beta$ and normalizing) are:



Now I need to express the state $|\psi\rangle$ as a combination of the two eigenvectors:

$ |\psi\rangle=\frac{\sqrt2+2}{2\sqrt3}|\psi_1\rangle+\frac{\sqrt2-2}{2\sqrt3}|\psi_2\rangle $

And the time evolution is:

$ |\psi(t)\rangle=\frac{\sqrt2+2}{2\sqrt3}|\psi_1\rangle \exp[{-i\frac{\epsilon}{\hbar}t}]+\frac{\sqrt2-2}{2\sqrt3}|\psi_2\rangle \exp[{i\frac{\epsilon}{\hbar}t}] $

This is how I solved the problem but since I'm pretty new to the quantum mechanics I'd like to have some opinions.

Is this a reasonable procedure? Did I made some terrible mistakes or non-sense considerations?


1 Answer 1


Your method for solving this problem was just fine (though I didn't check the math). An alternative but equivalent technique would be to calculate the time evolution operator. This would allow you to calculate the evolution of any state. The time evolution operator is given by,

$\exp[-iHt] = \exp\left[- \frac{ie\hbar}{2mc}\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)\right]$

The matrix exponential for any Pauli matrix is easy enough to calculate by hand and there even exists a general formula (see Wikipedia). This procedure gives,

$\exp[-iHt] = I_{2\times2}\cos \epsilon - i \sigma_x \sin \epsilon$

where $\epsilon \equiv \frac{e\hbar}{2mc} B_0$. From here it is easy to act on any state with this matrix to give the state after a time $t$:

$\psi(t) = \exp[-iHt] \psi (0) $

  • $\begingroup$ Thanks, I didn't even thought about calculate the time evolution operator! You showed me a really efficient and convenient method. $\endgroup$
    – Charlie
    Commented Jan 6, 2014 at 9:10
  • $\begingroup$ And what if I want to calculate the probability of measuring $\pm\frac{\hbar}{2}$ for the operator $S_z$? $\endgroup$
    – Charlie
    Commented Jan 6, 2014 at 9:32
  • $\begingroup$ Is it $|\langle\pm|\psi\rangle|^2$? $\endgroup$
    – Charlie
    Commented Jan 6, 2014 at 9:46
  • $\begingroup$ Yes that's right. as long as by $\left|\psi\right\rangle$ you mean you mean the time evolved wavefunction. $\endgroup$
    – JeffDror
    Commented Jan 6, 2014 at 14:24

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