So my problem here is that I'm confused about how to solve for the eigenstates corresponding to certain eigenvalues.

For my problem I have the Hamiltonian $$ H=E_0 \begin{pmatrix} 3 & 5i \\ -5i & 3 \\ \end{pmatrix} $$ which yields the eigenvalues $E_1=8E_0$ and $E_2=-2E_0$

Now I plug these eigenvalues back into the eigenvalue equation for the problem $(H-E_nI) |E_n\rangle = 0 $ which gives \begin{align} a_n(3E_0-E_n)+b_n(5iE_0)&=0 \tag{1}\\ -a_n(5iE_0)+b_n(3E_0-E_n)&=0 \tag{2} \end{align} where I've represented $|E_n\rangle$ as $|E_n\rangle = \begin{pmatrix} a_n \\ b_n \\ \end{pmatrix}\, .$ I use equation (2) and arrive at $$ a_n=-ib_n \frac{(3E_0-E_n)}{5E_0} \qquad a_1=ib_1 \tag{3} $$ Now here is where my confusion comes in. I can either plug $a_1$ into the normalization condition $$ |a_1|^2 + |b_1|^2=1 $$ as follows $$ |ib_1|^2+|b_1|^2 = 1 \qquad |b_1|^2 = \frac{1}{2} \qquad\Rightarrow \qquad a_1 = \frac{i}{\sqrt{2}} $$ and $$ |E_1\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} i \\ 1 \\ \end{pmatrix} $$ or I can rearrange equation (3) as $$ b_1=-ia_1 $$ and plug $b_1$ into the normalization condition to get $$ |a_1|^2=\frac{1}{2} \qquad b_1=-\frac{i}{\sqrt{2}} $$ which gives $$ |E_1\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i \\ \end{pmatrix} $$

So are these eigenstates equivalent or did I make a mistake somewhere? Is there a particular variable ($a_1$ or $b_1$ in this case) that we're supposed to substitute into the normalization condition first?

  • $\begingroup$ They differ just by a phase factor, notice that the second is $-i$ times the first one. $\endgroup$
    – Gold
    Commented Oct 13, 2017 at 21:14
  • $\begingroup$ But is there a preference for choosing one over the other? My professor's solution to this problem gave the second one. $\endgroup$
    – Elvis
    Commented Oct 13, 2017 at 21:17
  • $\begingroup$ @Elvis There is generally no preference, especially not in a problem like this. For some classes of problems there are phase conventions but in those cases you'll be told what they are. $\endgroup$
    – knzhou
    Commented Oct 13, 2017 at 21:23
  • 1
    $\begingroup$ Note that a ket is not identical to a state - a state is a ray in Hilbert space where the kets $e^{i\phi}|\psi\rangle$ belong to the same ray. $\endgroup$ Commented Oct 14, 2017 at 1:01

1 Answer 1


If $\vert \psi\rangle$ is an eigenstate of $\hat \Lambda$, then so is $\alpha\vert\psi\rangle$ for any complex $\alpha$ since $$ \hat \Lambda \alpha \vert\psi\rangle = \alpha \hat \Lambda\vert\psi\rangle = \alpha \lambda \vert\psi\rangle = \lambda (\alpha\vert\psi\rangle)\, . $$

Normalization pins down the $\alpha$’s to be of the form $e^{i\varphi}$, but you cannot do better than this, i.e. if $\vert \psi\rangle$ is a normalized eigenvector, then $e^{i\varphi}\vert\psi\rangle$ is an equally valid normalized eigenvector.

Physical quantities such as average values of the type $\langle \psi\vert \hat{\cal O}\vert\psi\rangle$ do not depend on the $e^{i\varphi}$ phase, so how you choose this overall factor is unimportant.

In your case, your two eigenvectors differ by an overall factor of $-i$, which is precisely of the form $e^{i\varphi}$. You can work out $\varphi$ by yourself.

  • $\begingroup$ To be sure, it's not that $e^{i\phi}|\psi\rangle$ is an equally valid state, it's that $e^{i\phi}|\psi\rangle$ belong to the same ray, i.e., the same state. $\endgroup$ Commented Oct 14, 2017 at 2:01
  • $\begingroup$ @AlfredCentauri Nice observation. I changed “eigenstate” to “eigenvector” to accommodate your comment. $\endgroup$ Commented Oct 14, 2017 at 2:05

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