# Are there multiple equivalent ways of writing an eigenstate?

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?

• They differ just by a phase factor, notice that the second is $-i$ times the first one.
– Gold
Commented Oct 13, 2017 at 21:14
• But is there a preference for choosing one over the other? My professor's solution to this problem gave the second one. Commented Oct 13, 2017 at 21:17
• @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. Commented Oct 13, 2017 at 21:23
• 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. Commented Oct 14, 2017 at 1:01

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.
• 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. Commented Oct 14, 2017 at 2:01