5
$\begingroup$

I wish to evaluate the following commutator: $[\operatorname{sign}(X),\, \operatorname{sign}(P)]$. Is there a general method for evaluating $[\operatorname{f}(X), \operatorname{f}(P)]$? I thought of a Taylor expansion but $\operatorname{sign}(x)$ is discontinuous on $x=0$. How would you evaluate this commutator?

$\endgroup$
  • $\begingroup$ This seems better suited for Math.SE. $\endgroup$ – Brandon Enright Jan 5 '14 at 0:51
  • 3
    $\begingroup$ How are you defining the sign of an operator? $\endgroup$ – doetoe Jan 5 '14 at 1:12
  • 1
    $\begingroup$ @BrandonEnright I don't think so: we're meant to interpret it as finding a commutator given the QM CCR, not how to do it for a general commutation - at least I think so. $\endgroup$ – WetSavannaAnimal Jan 5 '14 at 8:31
  • 2
    $\begingroup$ @doetoe As the function $sgn$ is Borel measurable, the sign of a self-adjoint operator $A$ is an operator defined in terms of its spectral measure adopting the standard procedure of Borel functional calculus: $$sgn(A)= \int_{\sigma(A)} sgn(\lambda) dP^{(A)}(\lambda)\:.$$ Since $A$ is self-adjoint and thus densely-defined and closed, it is possible to prove that it is the partial isometry obtained from the polar decomposition of $A$: $$A =J |A|$$ where, in fact, $J = sgn(A)$. $\endgroup$ – Valter Moretti Jan 5 '14 at 16:16
  • $\begingroup$ Thanks VM9, I found a concise reference at planetmath.org/spectralmeasure. I guess this also points at an answer to the OP's question: you could evaluate the commutator $[f(A),f(B)]$ loosely speaking by integrating weighted commutators of projection operators over the spectra of $A$ and $B$. $\endgroup$ – doetoe Jan 5 '14 at 23:56
7
$\begingroup$

I) The CCR reads

$$\tag{1} [\hat{X},\hat{P}]~=~i\hbar ~{\bf 1}. $$

We can imitate the Dirac delta function and the signum function via the following integral representations$^1$

$$\tag{2} \delta(\hat{X})~=~ \int_{\mathbb{R}} \! \frac{{\rm d}p}{2\pi\hbar} \exp\left(\frac{p\hat{X}}{i\hbar}\right), \qquad \delta(\hat{P})~=~ \int_{\mathbb{R}} \! \frac{{\rm d}x}{2\pi\hbar} \exp\left(\frac{i x\hat{P}}{\hbar}\right), $$

$$\tag{3} {\rm sgn}(\hat{X})~=~ \int_{\mathbb{R}} \! \frac{i{\rm d}p}{\pi p} \exp\left(\frac{p\hat{X}}{i\hbar}\right), \qquad {\rm sgn}(\hat{P})~=~ \int_{\mathbb{R}} \! \frac{{\rm d}x}{i\pi x} \exp\left(\frac{ix\hat{P}}{\hbar}\right). $$

The sought-for commutator can e.g. be written in $\hat{X}\hat{P}$-ordered form

$$\tag{4} [{\rm sgn}(\hat{X}),{\rm sgn}(\hat{P})] ~=~ \iint_{\mathbb{R}^2} \! \frac{{\rm d}x~{\rm d}p}{\pi^2 xp} \left[1-\exp\left(\frac{px}{i\hbar}\right)\right] \exp\left(\frac{p\hat{X}}{i\hbar}\right) \exp\left(\frac{ix\hat{P}}{\hbar}\right). $$

In eq. (4) we have used the following truncated Baker-Campbell-Hausdorff formula

$$\tag{5} e^{\hat{A}}e^{\hat{B}} ~=~e^{[\hat{A},\hat{B}]}e^{\hat{B}}e^{\hat{A}}, $$

which holds if the commutator $[\hat{A},\hat{B}]$ commutes with both the operators $\hat{A}$ and $\hat{B}$.

II) On a wavefunction $\psi(x)=\langle x |\psi\rangle $ in the Schrödinger position representation,

$$\tag{6} \hat{X}~=~x, \qquad \hat{P}~=~\frac{\hbar}{i}\frac{\partial}{\partial x}, $$

we have

$$\tag{7} \langle x |{\rm sgn}(\hat{P}) |\psi\rangle~=~ \int_{\mathbb{R}} \! \frac{{\rm d}y}{i\pi y}\langle x+y |\psi\rangle, $$

and therefore the matrix element of the sought-for commutator becomes

$$\tag{8} \langle x |[{\rm sgn}(\hat{X}),{\rm sgn}(\hat{P})] |\psi\rangle ~=~ \int_{\mathbb{R}} \! \frac{{\rm d}y}{i\pi y}\left({\rm sgn}(x)-{\rm sgn}(x+y)\right)\psi(x+y).$$

--

$^1$ The Cauchy principal value is implicitly assumed in pertinent places.

$\endgroup$
  • 1
    $\begingroup$ Could you please explain where the interpretation of ${\rm sgn}(\hat{p})$ came from. I mean, I know you seem to have interpreted it as the multiplication operator ${\rm sgn}(\hat{p})$ in momentum co-ordinates and then Fourier-transformed back to the $P\, 1/x$ in position co-ordinates - am I right? So there seems to be a "known" or "standard" way to interpret ${\rm sgn}$ - it isn't obvious from the question - so what kind of applications does this come up in? If this is too complicated for one line, I'll ask it as a new question. $\endgroup$ – WetSavannaAnimal Jan 5 '14 at 8:23
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 5 '14 at 14:59
2
$\begingroup$

Here is my answer (I have had time to answer only today). First of all, in general, if $A$ is a self-adjoint operator in the Hilbert space $H$ with spectrum $\sigma(A)\subset R$ (actually a closed normal operator would be enough), and $f: \sigma(A) \to C$ is a Borel measurable function (so for instance continuous up to a finite number of points would be OK), $f(A)$ is defined as: $$f(A) := \int_{\sigma(A)} f(\lambda) dP^{(A)}(\lambda)\:,\qquad (1)$$ where $\{P^{(A)}_E\}_{E \in {\cal B}(R)}$ is the so-called spectral measure of $A$ (for instance see http://planetmath.org/spectralmeasure). The $P_E$ are orthogonal projectors labelled by Borel sets $E$. Actually it turns out that $P_E=0$ if $E\cap \sigma(A)= \emptyset$, in this sense the measure is concentrated on the spectrum of $A$. Physically speaking $\sigma(A)$ is the set of the values that the observable $A$ can assume.

The fact that $A$ is sefl-adjoint guarantees the existence of the above mentioned notions and the feasibility of the construction I go to summarize.

The integral in (1) is defined in a way similar to that used for Riemann or Lebesgue integrals, first defining the integral of a function $s$ attaining a finite number of values $s_1,\ldots, s_n$ on corresponding sets $E_1,\ldots, E_n$: $$S(A) := \int_{\sigma(A)} s(\lambda) dP^{(A)}(\lambda) := \sum_{i=1}^n s_i P_{E_i} \qquad (1)'$$ and then taking the limit over a sequence of such functions $S_j$ point-wise tending to $f$ as $j \to +\infty$: $$\int_{\sigma(A)} f(\lambda) dP^{(A)}(\lambda) \psi := \lim_{j\to +\infty} \int_{\sigma(A)} s(\lambda) dP^{(A)}(\lambda)\psi\:. \quad (2)$$ The notion of convergence is that of the Hilbert space of the theory. The given definition makes precise the domain of $f(A)$: It is given by the vectors $\psi \in H$ such that the limit in (2) exists.

It is worth stressing that:

(a) the operator $f(A)$ is bounded (that is continuous) and defined on the whole Hilbert space, if the function $f$ is bounded over $\sigma(A)$;

(b) it holds: $$A = \int_{\sigma(A)} \lambda dP^{(A)}(\lambda)$$ and this identity completely determines $\{P_E^{(A)}\}_{E\in {\cal B}(R)}$ for a given self-adjoint opertor $A$. It also arises taking (a) into accoutn, that $A$ is bounded if (and only if) the map $\lambda \to \lambda$ is bounded over $\sigma(A)$ that, in turn, means that $\sigma(A)$ is bounded.

To answer to the general question of the OP, the above given definition of $f(A)$ is that one has to use to compute things like $[f(X), f(P)]$.

If $f$ is not bounded the domains of $f(X)$ and $f(P)$ are not the whole Hilbert space, and thus great care has to be used in computing the commutator above, since it is defined only in a common invariant domain.

However, this is not the case for $f= sgn$, since it is bounded.

Let us pass to the computation of $[sgn(X), sgn(P)]$ that, consequently, is a bounded operator as well.

If $A=X$ (position operator in $L^2(R)$), its spectral measure is quite trivial: $$(P^{(X)}_E \psi)(x):= \chi_{E}(x)\psi(x)\:,\qquad (3)$$ where $\chi_E(x)=1$ if $x\in E$ and $\chi_E(x)=0$ if $x\not \in E$. Consequently, exploiting (1)' (because $sgn$ assumes only three values (respectively $-1$ in $E_1=(-\infty,0)$, $0$ in $E_2= \{0\}$, $1$ in $E_3= (0,+\infty)$), one immediately sees that: $$(sgn(X) \psi)(x) = sgn(x) \psi(x)\:. \qquad (4)$$

We have next to focus on the momentum operator $P$. In the following I will assume $\hbar=1$ for the sake of notational simplicity. Henceforth ${\cal F}: L^2(R) \to L^2(R)$ is the Fourier transform, defined on $L^1$ functions (and than extended by $L^2$ continuity into an unitary map on $L^2$) by the usual integral formula: $${\cal F}: \psi(x) \mapsto \hat{\psi}(p) := \frac{1}{\sqrt{2\pi}}\int_R e^{-ipx} \psi(x) dx\:.$$ With these definitions, it turns out that the spectral measure of $P$ is $$(P^{(P)}_E \psi)(x) := {\cal F}^{-1}\left( \chi_E \cdot \hat{\psi}\right)(x)$$ In other words: in momentum representation, the spectral measure of $P$ exactly coincides with that of $X$ in position representation.

As the spectral theory is "covariant" under unitary transformations, it implies in particular that $sgn(P)$ in momentum representation is again defined as:

$$\left(sgn(P)_{momentum} \hat{\psi}\right)(p) = sgn(p) \hat{\psi}(p)\:,$$ so that, coming back to the position representation: $$sgn(P)\psi = {\cal F}^{-1} \left(sgn(p) \hat{\psi}(p)\right)\:. \quad (5) $$

We are in a position to compute the wanted commutator. I will assume that $\psi \in {\cal S}(R)$ the Schwartz space, because, in this case the Fourier transform can be computed as the usual integral and since that space is dense in $L^2$ so that the final result can be achieved simply taking a limit (as the commutator being bounded as stressed above). With that choice of $\psi$ all integration can be safely swapped. I do not enter into details.

Exploiting (4) and (5) (and interchanging integrals) we have almost immediately:

$$\left(sgn(X)sgn(P) \psi\right)(x)= \int\int \frac{e^{ip(x-y)}}{2\pi} sgn(p) sgn(x) \psi(y) dy dp$$

and

$$\left(sgn(P)sgn(X) \psi\right)(x)= \int\int \frac{e^{ip(x-y)}}{2\pi} sgn(p)sgn(y) \psi(y) dy dp\:.$$

Taking the difference and inserting an $\epsilon$ prescription to separate integrals, we have:

$$\left([sgn(X),sgn(P)]\psi \right)(x) = \lim_{\epsilon \to 0^+}\int \left(\int \frac{e^{ip(x-y) -|p| \epsilon}}{2\pi} sgn(p) dp\right) (sgn(x) -sgn(y)) \psi(y) dy $$

Computing the integral (please check the values of the coefficients) we finally get:

$$\left([sgn(X),sgn(P)]\psi \right)(x) = \lim_{\epsilon \to 0^+}\frac{1}{i\pi}\int_R \frac{(x-y)(sgn(x) -sgn(y))}{(x-y)^2 + \epsilon^2} \psi(y) dy \:.$$

Formally, it is possible to introduce the so-called Cauchy principal value:

$$\frac{1}{2}\frac{(x-y)}{(x-y)^2 + 0^2} = Vp \frac{1}{x-y}$$

so the found identity can be re-arranged in terms of the Cauchy principal value as done in the other answer.

$\endgroup$
1
$\begingroup$

Why is this commutator needed? I would start by trying to evaluate $$\int\!dx'\,dp'\, \langle \phi |\operatorname{sgn}(x) |x'\rangle\langle x'|p'\rangle\langle p'|\operatorname{sgn}(p) |\psi\rangle$$ and then each integral gets split up, e.g.,

$$-\int_{-\infty}^0\!dx'\,dp'\, \langle \phi |x'\rangle\langle x'|p'\rangle\langle p'|\operatorname{sgn}(p) |\psi\rangle + \int_{0}^\infty\!dx'\,dp'\, \langle \phi |x'\rangle\langle x'|p'\rangle\langle p'|\operatorname{sgn}(p) |\psi\rangle$$ and similarly for $p'$, giving you 4 integrals. The other half of the commutator is $$- \int\!dx'\,dp'\, \langle \phi |\operatorname{sgn}(p) |p'\rangle\langle p'|x'\rangle\langle x'|\operatorname{sgn}(x) |\psi\rangle$$

Keep in mind that $\langle p'|x'\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{-i p x}$ so that when you subtract the last 4 integrals from the first 4, you will need to relabel a coordinate to be able to cancel things out.

$\endgroup$
  • $\begingroup$ This is just a suggestion. Don't waste your time with it if it doesn't seem to work out. $\endgroup$ – lionelbrits Jan 5 '14 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.