# Eigenstates of spin-1 Hamiltonian involving $x,y,z$ components

I am trying to find the energy eigenvalues and eigenstates of the spin-1 system with Hamiltonian operator

$$H \enspace = \enspace a J_z^2 + b( J_x^2 - J_y^2 ) \quad , \qquad a, b \in \mathbb{R}$$

or in terms of ladder operators

$$H \enspace = \enspace a J_z^2 + \frac{b}{2} ( J_+^2 + J_-^2 )$$

My thinking is, that since this is a spin-1 system, the Hilbert space is spanned by the three states $$|1\rangle, |0\rangle, |-1\rangle$$. The ladder operators however work on those states as

$$J_{\pm} |m\rangle = \hbar \sqrt{2 - m(m\pm1)} |m+1\rangle$$

and so

$$J_+^2 |1\rangle = J_+^2 |0\rangle = J_-^2 |0 \rangle = J_-^2 |-1\rangle = 0$$

And

$$J_+^2 |-1\rangle = 2 \hbar^2 |1\rangle \quad , \qquad J_-^2 |1\rangle = 2\hbar^2 |-1\rangle$$

Applying the Hamiltonian to the (normalised) state $$| \psi \rangle = \tfrac{1}{\sqrt{2}}( |1\rangle + |-1\rangle )$$ yields

$$H | \psi \rangle = a J_z^2 | \psi \rangle + \frac{b}{2} ( J_+^2 + J_-^2 ) | \psi \rangle = a \hbar^2 | \psi \rangle + \frac{b}{2} \cdot 2\hbar^2 | \psi \rangle$$

and so

$$H | \psi \rangle = \hbar^2 (a + b) | \psi \rangle$$

However, this is only one eigenvector, but since $$H$$ is a spin-1 system it is 3-dimensional. So what would be the other eigenstates? I am not too experienced in quantum mechanics - is my reasoning here even okay or is it nonsense?

• One straightforward way you can do this is by looking at the matrix elements of the Hamiltonian. If you find the matrix $\langle i | H | j \rangle$ for $i,j = -1, 0, 1$ you can diagonalize this explicitly and find its Eigenvectors/Eigenvalues. Commented Apr 25 at 13:48

I am trying to find the energy eigenvalues and eigenstates of the spin-1 system with Hamiltonian operator

...

$$H \enspace = \enspace a J_z^2 + \frac{b}{2} ( J_+^2 + J_-^2 )$$

...

$$J_+^2 |1\rangle = J_+^2 |0\rangle = J_-^2 |0 \rangle = J_-^2 |-1\rangle = 0$$

And

$$J_+^2 |-1\rangle = 2 \hbar^2 |1\rangle \quad , \qquad J_-^2 |1\rangle = 2\hbar^2 |-1\rangle$$

... So what would be the other eigenstates?

You already have all the parts you need to construct the full matrix of the Hamiltonian, so you should do that and then diagonalize the Hamiltonian.

With the notation, and matrix element ordering $$H = \left( \begin{matrix} H_{-1,-1} & H_{-1,0} & H_{-1,1}\\ H_{0,-1} & H_{0,0} & H_{0,1}\\ H_{1,-1} & H_{1,0} & H_{1,1} \end{matrix} \right)\;,$$ we have: $$H = \left( \begin{matrix} a & 0 & b\\ 0 & 0 & 0\\ b & 0 & a \end{matrix} \right)\;.$$

Applying the Hamiltonian to the (normalised) state $$| \psi \rangle = \tfrac{1}{\sqrt{2}}( |1\rangle + |-1\rangle )$$ yields

...and so

$$H | \psi \rangle = \hbar^2 (a + b) | \psi \rangle$$

Above, you have found the eigenstate with eigenvalue $$a+b$$. Now you need to find the eigenstates with eigenvalues $$a-b$$ and $$0$$. The procedure is straightforward matrix diagonalization, as could be performed, for example, automatically by Wolfram Alpha.

• Ah, I see what I missed here. So the other eigenstates would be $|\psi_2 \rangle \propto ( |1\rangle - |-1\rangle)$ and $|\psi_3\rangle = |0\rangle$, correct? Commented Apr 25 at 13:57
• Looks right to me. Should be easy to check.
– hft
Commented Apr 25 at 14:15