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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?

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  • $\begingroup$ I don't understand the comment about $0$ of multiplicity $2$ for truncations: the sum of two degenerate matrices does not need to be even degenerate, forget about eigenvalue $0$ of multiplicity $2$ (just look at $N=2$). In general, truncation to large size and taking the limit for compact self-adjoint matrices allows you to faithfully recover all non-zero eigenvalues but can tell nothing about whether the eigenvalue $0$ is there (and I doubt if any finite size approximation is capable of that). Your case may be special, but it won't be pure "numerics" then anyway. $\endgroup$
    – fedja
    Commented Jan 19, 2023 at 13:55
  • $\begingroup$ @fedja, the truncated Hamiltonian has two degenerate matrices at the off-diagonal blocks, right? So the truncated Hamilltonian has a zero eigenvalue of geometric multiplicity $2$? $\endgroup$
    – António Borges Santos
    Commented Jan 19, 2023 at 14:00
  • $\begingroup$ @JochenGlueck thank you, that was embarassing... $\endgroup$
    – António Borges Santos
    Commented Jan 19, 2023 at 14:00
  • $\begingroup$ Wrong. Consider $N=2$, really. It takes 1 minute at most ;) $\endgroup$
    – fedja
    Commented Jan 19, 2023 at 14:01
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    $\begingroup$ @fedja maybe you are talking about a different object but what I see in this case is $$ H_N = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} $$ which seems to have rank $2$. $\endgroup$
    – António Borges Santos
    Commented Jan 19, 2023 at 14:05

2 Answers 2

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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.

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  • $\begingroup$ That is more due to the lucky fact that the eigenvector of $a^*a$ corresponding to the eigenvalue $0$ is finitely supported than to the proposed idea itself. If it weren't, you would never see $0$ in any truncation despite all words about chiral symmetry and such would remain perfectly valid for a general compact $a$, woudn't they? I mean, the method definitely works in this case, but the explanation of why doesn't hold water IMHO. Am I saying nonsense? :-) $\endgroup$
    – fedja
    Commented Jan 19, 2023 at 14:56
  • $\begingroup$ I am not quite sure what you mean; the property of $a$ I need is that $a^\ast$ does not annihilate any state in the Hilbert space ($a^\ast|\psi\rangle=0$ has no nonzero solution). $\endgroup$
    – Carlo Beenakker
    Commented Jan 19, 2023 at 15:10
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    $\begingroup$ I'm just an idiot: I'm looking at one operator and seeing another one. Don't know why. So, please, disregard everything I said and accept my apologies :-) $\endgroup$
    – fedja
    Commented Jan 22, 2023 at 22:28
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Arveson studied the general problem of numerically computing spectra in this paper. The take-home message is "numerical problems involving infinite dimensional operators require a reformulation in terms of C${}^*$- algebras. Indeed, it is only when the single operator $A$ is viewed as an element of an appropriate C${}^*$-algebra $\mathcal{A}$ that one can see the precise nature of the limit of the $n \times n$ eigenvalues distributions".

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  • $\begingroup$ this statement is so strong, it seems as if it invalidates the entire field of quantum many-body physics; would you know of an example of a Hermitian operator $H(a,a^\ast)$ where the "naive" truncation approach fails? $\endgroup$
    – Carlo Beenakker
    Commented Jan 20, 2023 at 7:19
  • $\begingroup$ @CarloBeenakker See Remark 3.13 of his paper for a simple class of examples. $\endgroup$
    – Nik Weaver
    Commented Jan 20, 2023 at 16:07
  • $\begingroup$ I don't think any of these examples are normally ordered Hermitian operators $H(a,a^\ast)$ (of the type $\sum_k c_k (a^\ast)^k a^k$ with $c_k\in\mathbb{R}$); I understood the question of the OP in that quantum mechanics context. $\endgroup$
    – Carlo Beenakker
    Commented Jan 20, 2023 at 16:49
  • $\begingroup$ Oh, I didn't realize $H(a,a^*)$ meant "normally ordered Hermitian operator" ... if your point is that this special class is tractable, I'm puzzled by the suggestion that Arveson's work "invalidates the entire field of quantum many-body physics". $\endgroup$
    – Nik Weaver
    Commented Jan 20, 2023 at 22:14
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    $\begingroup$ I think we agree, thanks for the clarification. $\endgroup$
    – Carlo Beenakker
    Commented Jan 21, 2023 at 9:07

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