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I'm interested in calculating the operator norm of a Hermitian operator, say $B$, acting on the Hilbert space of square integrable functions. The context is I have an optical system in all its infinite dimensional glory. To be clear, by operator norm I mean the largest eigenvalue, or $||B||_{\infty}=\sup \{ \langle \psi \vert B \vert \psi \rangle | \langle \psi \vert \psi \rangle =1 \}$.

I have a neat expression for $B$ in terms of its Characteristic function / Wigner function. My question is whether anyone knows a neat and easy way of calculating the operator norm within the phase space picture. Of course, I could truncate the operator to small Fock numbers (assuming that this is a good approximation), work out the matrix in the Fock basis and crunch the numbers. However, I'm just curious whether there is a more elegant solution. I have in mind such equalities as $tr ( A ) = \int W_{A}(r) dr $ and $tr ( A^{\dagger} B ) = \pi \int W_{A}(r)W_{B}(r)dr$, which give a useful correspondence between Wigner functions and the operators themselves.... are these any such expressions for norms?

Being pessimistic I guess the answer might simply be NO! However, I thought it would be worth consulting the collective wisdom of TPSE.

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For sure you have $||B||_{\infty} \leq \pi \ \sup_{r} |W_B(r)|$. – Piotr Migdal Nov 15 '11 at 17:09
@PiotrMigdal I did actually spot that bound myself after I posted, but I would want something much tighter. e.g. tight enough to give the right answer for a thermal Gaussian. – Earl Nov 16 '11 at 10:57
up vote 7 down vote accepted

It is not an complete answer, but (hopefully) it may help:
$$\text{tr} \left( A B \right) = \pi \int W_A(r) W_B(r) dr = \pi \int P_A(r) Q_B(r) dr,$$ where $A$ and $B$ are self-adjoint operators, $r=(p,q)$ is a point of the phase space, $W(r)$ is a Winger function, $P(r)$ - Glauber-Sudarshan P-representation and $Q(r)$ - a Husimi Q representation.

The second equality holds, as $Q(r)$ is a Gaussian-smeared $W(r)$, and $P(r)$ is a Gaussian-sharpened $W(r)$ (to see it explicitly, use the convolution theorem).

Then you can use advantages of each of the mentioned representations:

  • $P_A(r)$: is sometimes easier to handle than its Wigner counterpart,
  • $Q_B(r)$: is easily obtainable numerically (or sometimes - even analytically) from $W_B(r)$; for a non-negatively definite operator it is nonnegative, i.e. $Q_B(r) \geq 0$.

In particular for $P_A(r)\geq 0$ and $Q_B(r)\geq 0$ one gets a bound $$ \int P_A(r) Q_B(r) dr \leq \left( \sup_r\ Q_b(r) \right) \int P_A(r)dr = \sup_r\ Q_b(r).$$ In other words, when one restricts to finding maximum over a mixture of coherent states, then $\pi\ \sup_r\ Q_b(r)$ is the upper bound, which is tight.

However, the state achieving maximum need not to be coherent, so in general one gets $$||B||_\infty \geq \pi\ \sup_\alpha \langle \alpha | B | \alpha \rangle = \pi \ \sup_r\ Q_B(r).$$

Additionally, for some applications it may help that: $$\sup_r\ Q_B(r)\leq \sup_r\ W_B(r)\leq \sup_r\ |W_B(r)|.$$

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Thanks for your great answer. I didn't think to use the Husimi Q representation, but it seems to give quite a good bound! Though the bound is not always tight, it is tight for some important cases. For example, your bound would be tight for a thermal Gaussian, and displacements of a thermal Gaussian. I think this might be tight enough to be useful to me. – Earl Nov 16 '11 at 11:01
Actually, I'm not quite sure I understand why this is an upper bound on $||B||_{\infty}$ and not lower, and why there isn't a $\pi$ involved. So, via wikipedia, we have a $Q$ function: $Q_{B}(\alpha) = \frac{1}{\pi} \langle \alpha \vert B \vert \alpha \rangle $. Taking the sup gives, $\sup_{\alpha} Q_{B} ( \alpha ) = \sup_{\alpha} \langle \alpha \vert B \vert \alpha \rangle / \pi \leq \sup_{\psi} \langle \psi \vert B \vert \psi \rangle / \pi = ||B||_{\infty} / \pi$. Have I misunderstood something here? – Earl Nov 16 '11 at 11:14
@Earl: Haste makes waste - sorry for my omission. I lost $\pi$ and it wasn't on purpose. When it comes to the inequality (this one with $||B||_\infty$) - now I see it is not right. I wanted to use $\int P(r) Q(r) dr \leq \int P(r) (\sup_{R} Q(R)) dr$ but I've forgotten that $P(r)$ is not positive. I am fixing my post in a sec. – Piotr Migdal Nov 16 '11 at 12:23
Cool. But maybe we can also have both bounds such that $\pi \sup_{r} |Q_{B}(r)| \leq ||B||_{\infty} \leq \pi \sup_{r} |P_{B}(r)|$. For the following reason. $||B||_{\infty} = \pi | \sup_{\psi} \int P_{B}(r) Q_{\psi}(r) dr | \leq \sup_{\psi} \int | P_{B}(r) |.| Q_{\psi}(r) | dr $, where $Q_{\psi}$ is for some $\vert \psi \rangle$. Which is simply $||B||_{\infty} \leq \pi (\sup_{r}|P(r)|)(\int |Q_{\psi}|(r))$, but since Q is positive and integrates to 1, we have $||B||_{\infty}\leq \pi \sup_{r}|P(r)| $. This gives the other bound, which happens to be the one I prefer to have. – Earl Nov 16 '11 at 13:35
Ignore my last comment, as I think the Wigner function gives a better upper bound than the P-function. – Earl Nov 16 '11 at 14:18

In the case when $B$ is a compact operator, a limiting process can be applied to get the operator's norm to any arbitrarily small precision. This process is a Hilbert space generalization of the linear algebra relation for a Hermitian matrix $A$.

$ ||A||_{\infty} = \lim_{n\rightarrow \infty}|tr(A^n)|^{\frac{1}{n}}$.

For the data given in the question the operator composition is to be performed by means of the star product:

$ W_{AB}(r) =(W_A \star W_B)(r) =\int W_A(r_1)W_B(r_2) k(r, r_1, r_2) dr$

Thus one needs to know the star product kernel $k$. (Of course, in the ordinary phase space case, the star product is the Moyal star product).

Thus formally for the operator case:

$ ||B||_{\infty} = \lim_{n\rightarrow \infty}|tr(\star^{n} B)|^{\frac{1}{n}}$.

The limit exists when $B$ is compact.

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A small tweak in the transcription of Cayley's method: $ ||B||_{\infty} = \lim_{n\rightarrow \infty}|\int dr (\star B)^{n}|^{\frac{1}{n}}$. It may be easier to parse out this way. – Cosmas Zachos Sep 16 '15 at 17:04
Actually, the abs value in the above power iteration algorithm produces the spectral radius, but it is assumed the reader appreciates the trivial connection to the norm sought. – Cosmas Zachos Sep 16 '15 at 19:40
@Cosmas Zachos Thank you for your remarks Prof. Zachos. It is a great pleasure to have you on physics S.E. – David Bar Moshe Sep 17 '15 at 7:32
Thanks. Actually a practical and instructive illustration of this method is using [the Weyl transform of a Gaussian of operators][1] to compute the norm of the bounded Boltzmann operator $B\equiv\exp(-\frac{\hat{p}^2+\hat{x}^2}{2kT})$. [1]:… – Cosmas Zachos Sep 23 '15 at 19:51

There is an obvious property of the Wigner function when one takes the limit $p\rightarrow\infty$. By the Riemann's lemma, the Wigner function is expected to go to zero in this limit but it does driven by the largest eigenvalue. Your question can be stated in the following way. The Wigner function for a generic operator $A$ can be defined as

$$ W_A(x,p)=\frac{1}{2\pi\hbar}\int_{-\infty}^{+\infty}dy e^{-\frac{i}{\hbar}py}\langle x-y|A|x+y\rangle. $$

Now, assume that you have diagonalized $A$ so that $A|a_n\rangle=a_n|a_n\rangle$, you can write immediately

$$W_A(x,p)=\frac{1}{2\pi\hbar}\sum_n a_n\int_{-\infty}^{+\infty}dy e^{-\frac{i}{\hbar}py}\phi_n^*(x-y)\phi_n(x+y)=\frac{1}{2\pi\hbar}\sum_n a_n W_n(x,p) $$

being $\phi_n(x)=\langle a_n|x\rangle$. Of course, if the spectrum of $A$ will not run to infinity, taking the limit of increasing $p$ should grant the work done. This can also be argued from the unbounded case of the Hamiltonian of the harmonic oscillator. In this case you will get

$$W_A(x,p)=\sum_n E_n W_n^{H.O.}(x,p)$$

being now

$$ W_n^{H.O.}(x,p)=\frac{(-1)^n}{\pi\hbar}e^{-\frac{p^2}{\hbar^2\kappa^2}-\kappa^2x^2}L_n\left(2\frac{p^2}{\hbar^2\kappa^2}+2\kappa^2x^2\right)$$

and $L_n(x)$ the Laguerre polynomials and $\kappa^2=m\omega/\hbar$. In the limit $p\rightarrow\infty$, being $x$ fixed, the Laguerre polynomial will give the driving contribution $p^n$. In this particular case, it is the "energy" of the oscillator to drive to zero with the power of $n$.

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