# Why are Eigenvectors of a 1D quantum ising hamiltonian real

I was modelling the 1D transverse quantum Ising model and made a Kronecker product loop to find the Hamiltonian of the system, for a given magnetic field configuration. Now, my question is that when I outputted the eigenvectors of the Hamiltonian, they were always real. This highly contrasted with say a Linear Vibrational Coupling/ Linear+Quadratic Coupling model, as their Hamiltonians had complex eigenvectors.

I might be missing something here, why should the Eigenvectors of the 1D Transverse Quantum Ising be real, please help me out. I am a beginner in condensed matter physics and linear algebra so would appreciate it if someone explained from the basics.

• @JasonFunderberker Probably the OP did not write down the hamiltonain in the eigenbasis to begin with. So I guess their question is really about why the entries of the eigenvectors in the original basis ended up real. I think we need to know more about which basis OP started with to answer this.
– d_b
May 26 at 19:43
• @d_b Yeah, could be that I misread the question... I deleted my previous comment. Thanks. May 26 at 19:45

Why are Eigenvectors of a 1D quantum ising hamiltonian real

Is the Hamiltonian real and symmetric?

If so, then you can choose the eigenvectors to be real.

Of course, since the eigenvectors are only defined up to a phase, you can also force them to be complex if you want. But your numerical solver seems to have been nice enough to give you the real values.

The hamiltonian is a hermitian operator, so in a basis that diagonalizes it all its eigenvalues are real. This is exactly the eigenbasis. Here, to make it very explicit in component form, the hamiltonian takes in this basis the form $$\begin{bmatrix} E_0 & 0&0&0 \\ 0 & E_1&0&0 \\0&0&E_2&0 \\0&0&0&E_3 \\ \end{bmatrix}$$ and the ground state, for instance takes the form $$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ (there is an arbitrary constant).

• The question is about the eigenvectors. May 26 at 19:36
• The eigenvectors in the eigenbasis are constant multiples of unit vectors. @JasonFunderberker May 26 at 19:37
• @Diffycue Hi, if a matrix is Hermitian, cant we just conclude that the eigenvalues are real? My problem was that in LVC model I was getting complex eigenvectors, though the Hamiltonian is still Hermitian, and for quantum ising it was always real. Sorry if my question is dumb. May 26 at 19:50
• Should i share my code? May 26 at 19:51