Calculating eigenvalues (Schrödinger stationary states)

I'm trying to determine the stationary states and the corresponding energies.

My system has an angular momentum that is given by quantum number $l=1$, and the eigenvectors to $L_z$ are given as $|+1\rangle,|0\rangle,|-1\rangle$ and form a base. (thank you for the input @ZeroTheHero)

$$|+1\rangle=\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix},|0\rangle=\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix},|-1\rangle=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$

The Hamiltonian is $$H = \frac {\hbar w}{\sqrt2} \begin{pmatrix} 0 & 1 &0\\ 1 & 0 & 1\\ 0 & 1 & 0\\ \end{pmatrix}$$

When calculating the eigenvalues (energies) of the stationary states, I first conclude that the hamiltonian eigenvalues should be identical to those of $L_z$. This would mean that I can write

$$|n\rangle=\psi _{n}$$ $$H \psi_n (x) = E_n\psi_n (x)$$

So for $\psi_{+1}$, I would get

$$H \psi_{+1}=\frac{\hbar w}{\sqrt 2} \begin{pmatrix} 0 & 1 &0\\ 1 & 0 & 1\\ 0 & 1 & 0\\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix} = \frac{\hbar w}{\sqrt 2} \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} = E_n \psi_{+1}$$

But this doesn't appear to give me the correct solution.

• Most likely $\vert -1\rangle$ is actually $(0,0,1)^T$ and not $-(1,0,0)^T$. To be complete: $$\vert 1\rangle \to (1,0,0)^T\, \quad \vert 0\rangle \to (0,1,0)^T\, ,\quad \vert -1\rangle \to (0,0,1)^T$$ as these are the three basis states in your space. – ZeroTheHero Oct 6 '17 at 14:07
• I will also point out that you can recognize your $H$ as one of the angular momentum operators. Thus, you can guess that the eigenvalues will be identical to those of $L_z$. – ZeroTheHero Oct 6 '17 at 14:25
• You need to start with a generic vector $\psi_{+1}=(a,b,c)^T$ and solve $H\psi_{m}=m\hbar \omega \psi_{m}$ – ZeroTheHero Oct 6 '17 at 15:43

I'm very confused about what you mean by all this $L_z$ stuff. I believe there is some confusion between a Hamiltonian (which is an operator) and the matrix representation of the Hamiltonian.

If your Hamiltonian is represented as a matrix (with respect to some basis) then the eigenvectors of the matrix are the eigenvectors of the Hamiltonian (stationary states) also written in that same basis. As in if we assume the matrix you give above is $H$ written in some arbitrary basis $\{|a\rangle,|b\rangle,|c\rangle \}$. The eigenvectors of this matrix are:

• $v_1 = \frac{1}{2}\{1,-\sqrt{2},1\} \quad$ corresponding eigenvalue: $-\hbar \omega$
• $v_2 = \frac{1}{2}\{1,\sqrt{2},1\} \ \ \ \quad$ corresponding eigenvalue: $\hbar \omega$
• $v_3 = \{1,0,-1\} \quad \ \ \ \ \ \$ corresponding eigenvalue: $0$

so the stationary states of the Hamiltonian would be:

• $|v_1\rangle = \frac{1}{2} \big( 1|a\rangle - \sqrt{2} |b\rangle + 1|c\rangle \big)$
• $|v_2\rangle = \frac{1}{2} \big( 1|a\rangle + \sqrt{2} |b\rangle + 1|c\rangle \big)$
• $|v_2\rangle = \frac{1}{\sqrt{2}} \big( 1|a\rangle - 1|c\rangle \big)$

So to answer your question, the states $\{|v_1\rangle,|v_2\rangle,|v_3\rangle\}$ are your stationary states once you replace the $\{|a\rangle,|b\rangle,|c\rangle \}$ with the basis that that Hamiltonian was given to you in.

If it was given to you in the $L_z$ basis $\{|+\rangle,|0\rangle,|-\rangle \}$ then your stationary states are:

• $|v_1\rangle = \frac{1}{2} \big( 1|+\rangle - \sqrt{2} |0\rangle + 1|-\rangle \big)$
• $|v_2\rangle = \frac{1}{2} \big( 1|+\rangle + \sqrt{2} |0\rangle + 1|-\rangle \big)$
• $|v_2\rangle = \frac{1}{\sqrt{2}} \big( 1|+\rangle - 1|-\rangle \big)$
• what are your $\vert a\rangle, \vert b\rangle$ and $\vert c\rangle$? Surely the Hamiltonian is an operator but can be written as a $3\times 3$ matrix in the space spanned by the three states with $\ell=1$. – ZeroTheHero Oct 6 '17 at 21:06
• thank you so much, this cleared a lot up concepts up for me! just one question, I see you've normalized your eigenvectors $|v_1 \rangle, |v_2 \rangle, |v_3 \rangle$, why is that? Is it necessary to normalize them? – armara Oct 7 '17 at 12:17
• Eigenvectors do not need to be normalized but when we are thinking of the eigenvector as being a quantum state then it technically should be. – Patrick Oct 12 '17 at 20:46