Simple, $3 \times 3$ Hamiltonian with negative eigenvalues and $\langle H \rangle=0$

I have the following exercise:

Consider a three-dimensional system whose Hamiltonian is described by the following matrix: $$\begin{bmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$ a) What values are possible when the energy is measured?

b) A system is in the initial state $$|\Psi(t_0)\rangle = A\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$$ Normalize $| \Psi(t_0) \rangle$ and find $\langle H \rangle$.

So, for (a), I proceeded to determine the eigenvalues of $H$. I found (both manually, and then checked in Mathematica) $1,-1,5$. Here is my first problem:What does it mean to have a negative eigenvalue? I hadn't found that to be a big problem until I calculated $\langle H \rangle$ (Again, checked with Mathematica).

I got, surprisingly, $\langle H \rangle =0$. I'm having difficulty in analysing these results. How can i have a mean energy that equals $0$? Wouldn't this mean that there is no energy at all? I know about bound and scattering states, but shouldn't scattering states be continuous? I also found that in some cases the bound states were said to have positive energy (harmonic oscillator, for example).

Any

• The origin of energies is unphysical. Ergo, being negative vs. positive is unphyisical too. May 24, 2018 at 1:36
• Can i still do, say, a decomposition of the initial state in terms of the eigenvectors of this hamiltonian, and calculate probabilities? May 24, 2018 at 1:43

Eigenvalues can be negative: all the eigenvalues of the hydrogen atom are negative and given by $E_n=-\frac{13.6}{n^2}$eV.
As to your problem, the eigenstates are (if my algebra is not in error) $$\vert 5\rangle=\left(\begin{array}{c} 0\\ 0\\ 1\end{array}\right)\, ,\quad \vert -1\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c} i\\1\\ 0\end{array}\right)\, , \quad \vert 1\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c} -i\\1\\ 0\end{array}\right)$$ so your initial state is basically $$\vert\Psi(t_0)\rangle = \frac{(1-i)}{2}\vert -1\rangle + \frac{(1-i)}{2} \vert 1\rangle$$ so the average value will be $$\langle H\rangle = \frac{1}{{2}}(-1) + \frac{1}{{2}}(1)$$ i.e. your initial state can be found equally probably with energy $-1$ and with energy $+1$, so the average is $0$.
• Your eigenstates are indeed correct. I also found the average to be $0$ using the same method that you used. Thanks. May 24, 2018 at 2:07
• @VitorCGoergen Of course you can also compute directly $AA^*(1,1,0)\cdot H\cdot \left(\begin{array}{c} 1\\ 1 \\ 0\end{array}\right)$, which of course gives the same average value. The previous method illustrates better why you get $0$: it’s the average of +1 and -1 with equal probabilities. May 24, 2018 at 2:19