The ladder operator in quantum mechanics are the operators


$$a^\dagger \ = \ \frac{1}{\sqrt{2}} \left(-\frac{d}{dq} + q\right)$$
and
$$a \ \ = \ \frac{1}{\sqrt{2}} \left(\ \ \ \!\frac{d}{dq} + q\right).$$


They are differential operators on $\mathbb R.$ If one writes them in the Hermite basis, then


$$a^\dagger = \begin{pmatrix}
0 & 0 & 0 & 0 & \dots & 0 & \dots \\
\sqrt{1} & 0 & 0 & 0 & \dots & 0 & \dots \\
0 & \sqrt{2} & 0 & 0 & \dots & 0 & \dots \\
0 & 0 & \sqrt{3} & 0 & \dots & 0 & \dots \\
\vdots & \vdots & \vdots & \ddots & \ddots & \dots & \dots \\
0 & 0 & 0 & \dots & \sqrt{n} & 0 & \dots & \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix}$$ and


$$a =\begin{pmatrix}
0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\
0 & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\
0 & 0 & 0 & \sqrt{3} & \dots & 0 & \dots \\
0 & 0 & 0 & 0 & \ddots & \vdots & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\
0 & 0 & 0 & 0 & \dots & 0 & \ddots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.$$


Now assume I was interested in numerically computing the spectrum of


$$H = \begin{pmatrix} 0& a\\a^\* & 0\end{pmatrix}.$$


I absolutely know that this can be computed by hand, but I wonder about how to do this numerically.


A naive idea would be to truncate the above matrices at a large size $N$, but this leads to the wrong spectrum as both matrices then have a non-zero nullspace once they are truncated (it is clear since 0 is then an eigenvalue of geometric multiplicity $1$ for both matrices). Hence, the truncated numerics would predict that the Hamiltonian $H$ has an eigenvalue $0$ of multiplicity 2 rather than 1, which is correct.


Does anybody know how to numerically overcome this pseudospectral effect?



Spectrum of operator involving ladder operators
**Q:** *Does anybody know how to numerically overcome this pseudospectral effect?*


The key idea is "normal ordering". Rewrite the problem in such a way that annihilation operators $a$ appear to the right of creation operators $a^\ast$. In this particular case, first notice that $H$ has chiral symmetry, if $\lambda$ is an eigenvalue then also $-\lambda$ is an eigenvalue. We can thus reconstruct the full spectrum of $H$ without sign ambiguities from the spectrum of $H^2$,
$$H^2=\begin{pmatrix}
1+a^\ast a&0\\
0&a^\ast a
\end{pmatrix}.$$
We can now safely truncate each block to $N$ states. The spectrum of $H\_N^2$ contains 0 as eigenvalue with multiplicity 1.


