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September 16, 2012


Update (10/18/2012) — Mea Culpa:

Sonia pointed out to me that my (mis)interpretation of Ozawa was too charitable. We ended up (largely due to Steve Weinberg’s encouragement) writing a paper. So… where does one publish simple-minded (but, apparently, hitherto unappreciated) remarks about elementary Quantum Mechanics?

Sonia was chatting with me about this PRL (arXiv version), which seems to have made a splash in the news media and in the blogosphere. She couldn’t make heads or tails of it and (as you will see), I didn’t do much better. But I thought that I would take the opportunity to lay out a few relevant remarks.

Since we’re going to be talking about the Uncertainty Principle, and measurements, it behoves us to formulate our discussion in terms of density matrices.

A quantum system is described in terms of a density matrix, ρ\rho, which is a self-adjoint, positive-semidefinite trace-class operator, satisfying Tr(ρ)=1,Tr(ρ 2)1 Tr(\rho)=1,\qquad Tr(\rho^2) \leq 1 In the Schrödinger picture (which we will use), it evolves unitarily in time

(1)ρ(t 2)=U(t 2,t 1)ρ(t 1)U(t 2,t 1) 1\rho(t_2) = U(t_2,t_1) \rho(t_1) {U(t_2,t_1)}^{-1}

except when a measurement is made.

Consider a self-adjoint operator AA (an “observable”). We will assume that AA has a pure point spectrum, and let P i (A)P^{(A)}_i be the projection onto the i thi^{\text{th}} eigenspace of AA.

When we measure AA, quantum mechanics computes for us

  1. A classical probability distribution for the values on the readout panel of the measuring apparatus. The moments of this probability distribution are computed by taking traces. The n thn^{\text{th}} moment is A n=Tr(A nρ) {\langle A^n\rangle} = Tr(A^n\rho) In particular, the variance is (ΔA) 2=Tr(A 2ρ)(Tr(Aρ)) 2 {(\Delta A)}^2 = Tr(A^2\rho) - {\left(Tr(A\rho)\right)}^2
  2. A change (which, under the assumptions stated, can be approximated as occurring instantaneously) in the density matrix,
(2)ρ afterρ^(ρ,A)= iP i (A)ρP i (A)\rho_{\text{after}}\equiv \hat{\rho}(\rho,A) = \sum_i P^{(A)}_i \rho P^{(A)}_i

Thereafter, the system, described by the new density matrix, ρ^\hat{\rho}, again evolves unitarily, according to (1).

The new density matrix, ρ^\hat{\rho}, after the measurement1, can be completely characterized by two properties

  1. All of the moments of AA are the same as before A n=Tr(A nρ^(ρ,A))=Tr(A nρ) \langle A^n\rangle = Tr(A^n \hat{\rho}(\rho,A))= Tr(A^n \rho) (In particular, ΔA\Delta A is unchanged.) Moreover, for any observable, CC, which commutes with AA ( [C,A]=0[C,A]=0 ), C n=Tr(C nρ^(ρ,A))=Tr(C nρ) \langle C^n\rangle = Tr(C^n \hat{\rho}(\rho,A))= Tr(C^n \rho)
  2. However, the measurement has destroyed all interference between the different eigenspaces of AA Tr([A,B]ρ^(ρ,A))=0,B Tr([A,B] \hat{\rho}(\rho,A)) = 0, \qquad \forall\, B

Note that it is really important that I have assumed a pure point spectrum. If AA has a continuous spectrum, then you have to deal with complications both physical and mathematical. Mathematically, you need to deal with the complications of the Spectral Theorem; physically, you have to put in finite detector resolutions, in order to make proper sense of what a “measurement” does. I’ll explain, later, how to deal with those complications

Now consider two such observables, AA and BB. The Uncertainty Principle gives a lower bound on the product

(3)(ΔA) ρ(ΔB) ρ12|Tr(i[A,B]ρ)|{(\Delta A)}_\rho {(\Delta B)}_\rho \geq \tfrac{1}{2}\left|Tr(-i[A,B]\rho)\right|

in any state, ρ\rho. (Exercise: Generalize the usual proof, presented for “pure states” to the case of density matrices.)

As stated, (3) is not a statement about the uncertainties in any actual sequence of measurements. After all, once you measure AA, in state ρ\rho, the density matrix changes, according to (2), to

(4)ρ^(ρ,A)= iP i (A)ρP i (A)\hat{\rho}(\rho,A) = \sum_i P^{(A)}_i \rho P^{(A)}_i

so a subsequent measurement of BB is made in a different state from the initial one.

The obvious next thing to try is to note that, since the uncertainty of AA in the state ρ^(ρ,A)\hat{\rho}(\rho,A) is the same as in the state ρ\rho, and since we are measuring BB in the state ρ^(ρ,A)\hat{\rho}(\rho,A), we can apply the Uncertainty Relation, (3) in the state ρ^\hat{\rho}, instead of in the state, ρ\rho. Unfortunately, Tr([A,B]ρ^(ρ,A))=0Tr([A,B]\hat{\rho}(\rho,A))=0, so this leads to an uninteresting lower bound on the product the uncertainties

(5)(ΔA)(ΔB)=(ΔA) ρ^(ρ,A)(ΔB) ρ^(ρ,A)0(\Delta A) (\Delta B) = {(\Delta A)}_{\hat{\rho}(\rho,A)} {(\Delta B)}_{\hat{\rho}(\rho,A)} \geq 0

for a measurement of AA immediately followed by a measurement of BB.

It is, apparently, possible to derive a better lower bound on the product of the uncertainties of successive measurements (which is still, of course, weaker than the “naïve” 12|Tr(i[A,B]ρ)|\tfrac{1}{2}\left|Tr(-i[A,B]\rho)\right|, which is what you might have guessed for the lower bound, had you not thought about what (3) means). But I don’t know how to even state that result at the level of generality of the above discussion

Instead, I’d like to discuss how one treats measurements, when AA doesn’t have a pure point spectrum. When it’s discussed at all, it’s treated very poorly in the textbooks.

Measuring Unbounded Operators

Let’s go straight to the worst-case, of an unbounded operator, with Spec(A)=Spec(A)=\mathbb{R}. Such an operator has no eigenvectors at all. What happens when we measure such an observable? Clearly, the two conditions which characterized the change in the density matrix, in the case of a pure point spectrum,

  1. Tr(A nρ^(ρ,A))=Tr(A nρ)Tr(A^n \hat{\rho}(\rho,A))= Tr(A^n \rho)
  2. Tr([A,B]ρ^(ρ,A))=0,BTr([A,B] \hat{\rho}(\rho,A)) = 0, \qquad \forall\, B

are going to have to be modified. The second condition clearly can’t hold for all choice of BB, in the unbounded case (think A=xA=\mathbf{x} and B=pB=\mathbf{p}). As to the first condition, we might hope that the moments of the classical probability distribution for the observed measurements of AA would be calculated by taking traces with the density matrix, ρ^\hat{\rho}. But that probability distribution depends on the resolution of the detector, something which the density matrix, ρ\rho, knows nothing about.

To keep things simple, let’s specialize to = 2()\mathcal{H}=\mathcal{L}^2(\mathbb{R}) and A=xA=\mathbf{x}. Let’s imagine a detector which can measure the particle’s position with a resolution, σ\sigma. Let’s define a projection operator P x 0:ψ(x)dyσπe ((xx 0) 2+(yx 0) 2)/2σ 2ψ(y) P_{x_0}:\quad \psi(x) \mapsto \int\frac{d y}{\sigma\sqrt{\pi}} e^{-\left( (x-x_0)^2+(y-x_0)^2\right)/2\sigma^2}\psi(y) which reflects the notion that our detector has measured the position to be x 0x_0, to within an accuracy σ\sigma. Here, I’ve chosen a Gaussian; but really any acceptance function peaked at x=x 0x=x_0, and dying away sufficiently fast away from x 0x_0 will do (and may more-accurately reflect the properties of your actual detector).

But I only know how to do Gaussian integrals, so this one is a convenient choice.

This is, indeed, a projection operator: P x 0P x 0=P x 0P_{x_0}P_{x_0} = P_{x_0}. But integrating over x 0x_0 doesn’t quite give the completeness relation one would want dx 02σπP x 0𝟙 \int \frac{d x_0}{2\sigma\sqrt{\pi}} P_{x_0} \neq \mathbb{1} Instead we find dx 02σπP x 0:ψ(x)du2σπe u 2/4σ 2ψ(x+u) \int \frac{d x_0}{2\sigma\sqrt{\pi}} P_{x_0}:\quad \psi(x) \mapsto \int \frac{d u}{2\sigma\sqrt{\pi}} e^{-u^2/4\sigma^2} \psi(x+u) Rather than getting ψ(x)\psi(x) back, we get ψ(x)\psi(x) smeared against a Gaussian.

To fix this, we need to consider a more general class of projection operators (here, again, the Gaussian acceptance function proves very convenient):

(6)P x 0,k 0:ψ(x)dyσπexp[((xx 0) 2+(yx 0) 2)/2σ 2+ik 0(xy)]ψ(y)P_{x_0,k_0}:\quad \psi(x)\mapsto \int \frac{d y}{\sigma \sqrt{\pi}}\exp\left[-\left((x-x_0)^2 + (y-x_0)^2\right)/2\sigma^2 +i k_0(x-y)\right] \psi(y)

These are still projection operators, P x 0,k 0P x 0,k 0=P x 0,k 0 P_{x_0,k_0} P_{x_0,k_0} = P_{x_0,k_0} But now they obey the completeness relation

(7)dx 0dk 02πP x 0,k 0=𝟙\int \frac{d x_0 d k_0}{2\pi} P_{x_0,k_0} =\mathbb{1}

so we can now assert that the density matrix after measuring x\mathbf{x} is

(8)ρ^=dx 0dk 02πP x 0,k 0ρP x 0,k 0\hat{\rho} = \int \frac{d x_0 d k_0}{2\pi} P_{x_0,k_0} \rho P_{x_0,k_0}

If ρ\rho is represented by the integral kernel, K(x,y)K(x,y): ρ:ψ(x)dyK(x,y)ψ(y) \rho:\quad \psi(x) \mapsto \int d y\, K(x,y) \psi(y) then the new density matrix, ρ^\hat{\rho} is represented by the integral kernel K^(x,y)=e (xy) 2/2σ 2duσ2πe u 2/2σ 2K(x+u,y+u) \hat{K}(x,y) = e^{-(x-y)^2/2\sigma^2}\int \frac{d u}{\sigma \sqrt{2\pi}} e^{-u^2/2\sigma^2} K(x+u,y+u) Here we see clearly that it has the desired properties:

  1. The off-diagonal terms are suppressed; K^(x,y)0\hat{K}(x,y)\to 0 for |xy|σ|x-y|\gg \sigma.
  2. The near-diagonal terms are smeared by a Gaussian, representing the finite resolution of the detector.

Moreover, the moments of the probability distribution for the measured value of xx are given by taking traces with ρ^\hat{\rho}: x n=Tr(x nρ^) \langle x^n\rangle = Tr(\mathbf{x}^n \hat{\rho}) One easily computes Tr(xρ^) =Tr(xρ) Tr(x 2ρ^) =σ 2+Tr(x 2ρ) \begin{split} Tr(\mathbf{x} \hat{\rho})&=Tr(\mathbf{x} \rho)\\ Tr(\mathbf{x}^2 \hat{\rho})&=\sigma^2+ Tr(\mathbf{x}^2 \rho)\\ \end{split} So the intrinsic quantum-mechanical uncertainty of the position, xx, in the state, ρ\rho, adds in quadrature with the systematic uncertainty of the measuring apparatus to produce the measured uncertainty (Δx) measured 2=(Δx) ρ^ 2=(Δx) ρ 2+σ 2 (\Delta x)^2_{\text{measured}} = (\Delta x)^2_{\hat{\rho}} = (\Delta x)^2_{\rho} + \sigma^2 exactly as we expect.

There’s one feature of this Gaussian measuring apparatus which is a little special. Of course, we expect that measuring xx should change the distribution for values of pp. Here, the effect (at least on the first few moments) is quite simple Tr(pρ^) =Tr(pρ) Tr(p 2ρ^) =1σ 2+Tr(p 2ρ) \begin{split} Tr(\mathbf{p} \hat{\rho})&=Tr(\mathbf{p} \rho)\\ Tr(\mathbf{p}^2 \hat{\rho})&=\frac{1}{\sigma^2}+ Tr(\mathbf{p}^2 \rho)\\ \end{split} If we wanted to compute the effect of measuring pp, using a Gaussian detector with systematic uncertainty σ p=1/σ\sigma_p =1/\sigma, we would use the same projectors (6) and obtain the same density matrix (8) after the measurement. This leads to very simple formulæ for the uncertainties resulting from successive measurements. Say we start with an initial state, ρ\rho, measure xx with a Gaussian detector with systematic uncertainty σ x\sigma_x, and then measure pp with another Gaussian detector with systematic uncertainty σ p\sigma_p. The measured uncertainties are

(9)(Δx) 2 =(Δx) ρ 2+σ x 2 (Δp) 2 =(Δp) ρ^ 2+σ p 2=(Δp) ρ 2+1σ x 2+σ p 2\begin{split} {(\Delta x)}^2&= {(\Delta x)}^2_\rho +\sigma_x^2\\ {(\Delta p)}^2&= {(\Delta p)}^2_{\hat{\rho}} +\sigma_p^2={(\Delta p)}^2_\rho +\frac{1}{\sigma_x^2}+\sigma_p^2 \end{split}

You can play around with other, non-Gaussian, acceptance functions to replace (6). You’re limited only by your ability to find a complete set of projectors, satisfying the analogue of (7) and, of course, by your ability to do the requisite integrals.

What you’ll discover is that the Gaussian acceptance function provides the best tradeoff (when, say, you measure xx) between the systematic uncertainty in xx and the contribution to the quantum-mechanical uncertainty in pp, resulting from the measurement.

Update (9/20/2012):

I looked some more at the Ozawa paper whose “Universally valid reformulation” of the uncertainty principle this PRL proposes to test. Unfortunately, it doesn’t seem nearly as interesting as it did at first glance.

  • For observables, A,BA,B, with pure point spectra, we can assume “ideal” measuring apparati (whose measured uncertainty equals the inherent quantum-mechanical uncertainty of the observable in the quantum state in which the measurement is made). In that case, his uncertainty relation (see (17) of his paper) reduces to the “uninteresting” (5). Of course that’s trivially satisfied. I believe that a stronger bound can be derived, in this case. But doing so requires more sophisticated techniques than Ozawa uses.
  • For unbounded observables, like x,px,p, we can see from what I’ve said above that the actual lower bound is stronger than the one Ozawa derives. Consider a measurement of xx, followed by a measurement of pp. From (9), the product of the measured uncertainties2 satisfies
(10)(Δx) 2(Δp) 2 54+(Δx) ρ 2(1σ x 2+σ p 2)+(Δp) ρ 2σ x 2+σ x 2σ p 2 (12+1+σ x 2σ p 2) 2\begin{split} {(\Delta x)}^2{(\Delta p)}^2 &\geq \frac{5}{4} + {(\Delta x)}^2_\rho \left(\frac{1}{\sigma_x^2}+\sigma_p^2\right) + {(\Delta p)}^2_\rho \sigma_x^2 +\sigma_x^2\sigma_p^2\\ &\geq {\left(\frac{1}{2} +\sqrt{1+\sigma_x^2\sigma_p^2}\right)}^2 \end{split}

where the last inequality is saturated by an initial state, ρ\rho, which is a pure state consisting of a Gaussian wave packet with carefully-chosen width, (Δx) ρ 2=σ x 221+σ x 2σ p 2,(Δp) ρ 2=1+σ x 2σ p 22σ x 2 {(\Delta x)}^2_\rho = \frac{\sigma_x^2}{2\sqrt{1+\sigma_x^2\sigma_p^2}},\qquad {(\Delta p)}^2_\rho = \frac{\sqrt{1+\sigma_x^2\sigma_p^2}}{2\sigma_x^2}

Update (9/28/2012):

Here is, at least, one lower bound (stronger than Ozawa’s stupid bound) for the product of measured uncertainties when A,BA,B have pure point spectra. Let B^= iP i (A)BP i (A) \hat{B} = \sum_i P^{(A)}_i B P^{(A)}_i and M i=P i (A)B(𝟙P i (A)) M_i = P^{(A)}_i B \left(\mathbb{1} - P^{(A)}_i\right) We easily compute (ΔB) ρ^ 2=(ΔB^) ρ 2+ iTr(M iM i ρ) {(\Delta B)}^2_{\hat{\rho}} = {(\Delta \hat{B})}^2_{\rho} + \sum_i Tr(M_i \M_i^\dagger \rho) and hence

(11)(ΔA) ρ 2(ΔB) ρ^ 2=(ΔA) ρ 2(ΔB^) ρ 2+(ΔA) ρ 2 iTr(M iM i ρ){(\Delta A)}^2_\rho {(\Delta B)}^2_{\hat{\rho}} = {(\Delta A)}^2_\rho {(\Delta \hat{B})}^2_{\rho} + {(\Delta A)}^2_\rho \sum_i Tr(M_i \M_i^\dagger \rho)

Since [B^,A]=0[\hat{B},A]=0, the first term is bounded below3 by

(12)(ΔA) ρ 2(ΔB^) ρ 2(12Tr({A,B^}ρ)Tr(Aρ)Tr(B^ρ)) 2{(\Delta A)}^2_\rho {(\Delta \hat{B})}^2_{\rho} \geq {\left(\frac{1}{2} Tr(\{A,\hat{B}\}\rho)-Tr(A\rho)Tr(\hat{B}\rho)\right)}^2

The second term is also positive-semidefinite.

For the classic case of a 2-state system, with A=J xA=J_x and B=J yB=J_y (the system considered by the aforementioned PRL), we see that B^0\hat{B}\equiv 0, and the product of uncertainties is entirely given by the second term of (11).

The most general density matrix for the 2-state system is parametrized by the unit 3-ball ρ=12(𝟙+aσ),aa1 \rho = \frac{1}{2}\left(\mathbb{1}+ \vec{a}\cdot\vec{\sigma}\right), \qquad \vec{a}\cdot\vec{a}\leq 1 The points on the boundary S 2={aa=1}S^2=\{\vec{a}\cdot\vec{a}=1\} correspond to pure states.

Upon measuring A=J x12σ xA=J_x\equiv\tfrac{1}{2}\sigma_x, the density matrix after the measurement is ρ^=12(𝟙+a xσ x) \hat{\rho} = \frac{1}{2}\left(\mathbb{1}+ \a_x\sigma_x\right) and, for a subsequent measurement of J yJ_y, (ΔJ x) ρ 2(ΔJ y) ρ^ 2=116(1a x 2) {(\Delta J_x)}^2_\rho{(\Delta J_y)}^2_{\hat{\rho}} = \frac{1}{16}(1-a_x^2) as “predicted” by (11).

1 Frequently, one wants to ask questions about conditional probabilites: “Given that a measurement of AA yields the value λ 1\lambda_1, what is the probability distribution for a subsequent measurement of …”. To answer such questions, one typically works with a new (“projected”) density matrix, ρ=1ZP 1 (A)ρP 1 (A)\rho' = \frac{1}{Z} P^{(A)}_1 \rho P^{(A)}_1, where the normalization factor Z=Tr(P 1 (A)ρ)Z=Tr(P^{(A)}_1 \rho) is required to make Tr(ρ)=1Tr(\rho')=1. The formalism in the main text of this post is geared, instead, to computing joint probability distributions.

2 Ozawa’s inequality isn’t for the product of the measured uncertainties, (Δx)(Δp)(\Delta x)(\Delta p), but rather for the product, (Δx)(Δp) ρ^(\Delta x){(\Delta p)}_{\hat{\rho}}, of the measured uncertainty in xx with the quantum-mechanical uncertainty in pp in the state ρ^\hat{\rho} which results from the xx-measurement. To obtain this, just mentally set σ p=0\sigma_p=0 in the above formulæ.

3 Let S=(AA ρ𝟙)+e iθα(BB ρ𝟙)S= \left(A-{\langle A\rangle}_\rho\mathbb{1}\right)+e^{i\theta}\alpha \left(B-{\langle B\rangle}_\rho\mathbb{1}\right) for α\alpha\in\mathbb{R}. Consider Q(α)=Tr(SS ρ) Q(\alpha) = Tr(S S^\dagger \rho) This is a quadratic expression in α\alpha, which is positive-semidefinite, Q(α)0Q(\alpha)\geq 0. Thus, the discriminant must be negative-semidefinite, D0D\leq 0. For θ=π/2\theta=\pi/2, this yields the conventional uncertainty relation, (ΔA) ρ 2(ΔB) ρ 214(Tr([A,B]ρ)) 2 {(\Delta A)}^2_\rho{(\Delta B)}^2_\rho\geq\frac{1}{4}{\left(Tr([A,B]\rho)\right)}^2 For θ=0\theta=0, it yields (ΔA) ρ 2(ΔB) ρ 2(12Tr({A,B}ρ)Tr(Aρ)Tr(Bρ)) 2 {(\Delta A)}^2_\rho{(\Delta B)}^2_\rho\geq{\left(\frac{1}{2}Tr(\{A,B\}\rho)-Tr(A\rho)Tr(B\rho)\right)}^2 which is an expression you sometimes see, in the higher-quality textbooks.

Posted by distler at September 16, 2012 12:27 AM

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29 Comments & 0 Trackbacks

Re: Uncertainty

What do you think about this analysis?

Posted by: Albert on September 16, 2012 3:24 PM | Permalink | Reply to this

Re: Uncertainty

It is vaguely possible that there are some correct statements on that page. But you would be hard-pressed to disentangle them from the incorrect ones.

Posted by: Jacques Distler on September 16, 2012 4:01 PM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

I wish Lubos would tell us what he really thinks :)

Posted by: Robert McNees on September 16, 2012 4:39 PM | Permalink | Reply to this

Re: Uncertainty

You should write more often on this blog.

Posted by: Marco Frasca on September 25, 2012 11:20 AM | Permalink | Reply to this

Re: Uncertainty

Hey, doesn’t this paper show that superstring theory is incorrect?

Posted by: Rick on September 27, 2012 6:30 PM | Permalink | Reply to this

Re: Uncertainty

Short answer:


(Slightly) longer answer:

Are you a troll?

Posted by: Jacques Distler on September 27, 2012 10:18 PM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

I found the following video more understandable than the aricle by Rozema et. al:
Yuji Hasegawa: Quantum Physics With Neutrons – Theory and Experiment (zweiter Teil)

Posted by: David Brown on October 19, 2012 7:14 AM | Permalink | Reply to this

Re: Uncertainty

Good to see you’ve put that to print form. Regarding your question, I’d suggest:
1) European Journal of Physics (IOP) - very decent while avoids super high-tech math,
2) or perhaps Foundations of Physics (Springer) - but it does host more speculative stuff.

Posted by: nobodyreally on November 20, 2012 1:32 PM | Permalink | Reply to this

Re: Uncertainty

Am. J. Phys., perhaps?

Posted by: Blake Stacey on November 20, 2012 3:35 PM | Permalink | Reply to this

Re: Uncertainty

European Journal of Physics and Am. J. Phys. are both devoted to papers about physics pedagogy. While it’s my contention that what Sonia and I did should be in all the textbooks, the lamentable fact of the matter is that it isn’t.

Foundations of Physics (at least, based on browsing the recent issues) has a pretty high crank quotient.

Phys Rev A seems to be popular with the Quantum Information people.

That may be the most likely choice …

Posted by: Jacques Distler on November 21, 2012 12:38 AM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

Looking over my notes on a quantum-information-related research project, I see a large proportion of references to Physical Review A papers. So, based on what little evidence I have, that’s probably a good choice. Other journals represented more than once are Journal of Physics A, Journal of Mathematical Physics and Theoretical and Mathematical Physics.

Posted by: Blake Stacey on November 23, 2012 1:58 PM | Permalink | Reply to this

Re: Uncertainty

What do you think of this new paper:

Posted by: rickyboy on November 21, 2012 1:52 PM | Permalink | Reply to this

Re: Uncertainty

What do you think of this paper by Wald and Hollands?

What do you think of this paper by esteemed professors Smolin and Alexander?

Posted by: Steven on December 24, 2012 10:38 AM | Permalink | Reply to this

Re: Uncertainty


What do you think of this interesting paper :

Posted by: Leonard on January 26, 2013 2:25 PM | Permalink | Reply to this

Re: Uncertainty

What do you think of this paper about firewalls? Lubos is a fan.

Posted by: Leonard on February 5, 2013 12:13 PM | Permalink | Reply to this

Re: Uncertainty

Publishing elementary QM on a preprint server is more germaine than in some firewall-protected pay-per-view journal, which is more about politics (publicity, etc.) than simply making info widely available.

Feynman argued (“QED”, Princeton U.P., 1985) that multipath interference in QM (i.e. the Coulomb field quanta of 2nd quantization) provide a simple mechanism to replace the uncertainty principle of non-relativistic 1st quantization.

Can I ask a question: why not replace the usual complex path amplitude exp(iS) (where action S is in h-bar units) with simply its real component, cos S? [Taken from Euler’s equation: exp(iS) = i sin S + cos S.]

When you think about it mathematically, exp(iS) is a vector on the complex plane (Argand diagram), and cos S is a scalar amplitude. All cross-sections and other observables calculated from a path integral (summing exp(iS) contributions) are real numbers, hence the resultant arrow must always be parallel to the real axis, so you get exactly the same result using exp(iS) or cos S. You aren’t losing complex plane directional information. It seems that the only reason to stick to exp(iS) is historical, going back to Dirac’s derivation of exp(iHt) as the amplitude for a single wavefunction from Schroedinger’s equation, where the periodic real solutions produce the quantization. If you’re doing 2nd quantization, multipath interference for large path actions is the mechanism for quantization, so you don’t need Schroedinger’s equation (which is just a non-relativistic approximation). Just curious to see if you have any interest or prefer to delete.

Posted by: Nige on March 1, 2013 6:03 PM | Permalink | Reply to this

Re: Uncertainty

Can I ask a question: why not replace the usual complex path amplitude exp(iS)exp(iS) (where action SS is in h-bar units) with simply its real component, cosScos S?

Because that would not lead to unitary time evolution. (In other words, it would violate the conservation of probability.)

I think you would find it instructive to go through the derivation which shows that Feynman’s prescription leads to unitary time evolution. It would then be completely obvious to you that your prescription does not (for the impatient: if UU is unitary, 12(U+U )\tfrac{1}{2}(U+U^\dagger) is not).

You don’t need to consult Feynman and Hibbs’s book. Any textbook, which treats the path-integral approach to quantum mechanics, will explain this.

Posted by: Jacques Distler on March 1, 2013 10:19 PM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

Thanks for your reply! Your reaction is to come up with a formal no-go theorem to reject it. I do intent to write a paper, and tackling this kind of objection is important. Dirac’s “interaction picture” wavefunction amplitude of exp(-iHt) obeys unitary time evolution.

But there’s no time-evolution for cross-sections and probabilities from the path integral. Experiments are not concerned with unitary time evolution, since you can’t determine a photon’s wavefunction amplitude at a series of intervals in time. There’s multipath interference due to many wavefunction amplitudes interfering, which maximises eigenvalues for the paths of least action. So the path integral’s exp(iS) is redundant and you lose nothing by the substitution exp(iS) -> cos (S). The lagrangian is integrated over time to give the action for a particular path, but there’s lots of different virtual paths, so unitary time evolution isn’t a checkable fact. Integrating cos (S) produces exactly the same real eigenvalues as exp(iS) for calculating cross-sections. Agreeing with the textbooks, sure, you need exp(-iHt) to produce eigenvalues in 1st quantization, where there is no mechanism for eigenvalues to arise by multipath interference. But with the path integral, it’s superfluous, and an amplitude of cos S will do. Feynman in “QED” (1985) plots amplitudes as unit-length arrows of varying direction, plotted nose-to-tail on Argand diagrams, with the path integral equal to a resultant arrow (the overall sum of all contributions). It’s obvious that the resultant length of an arrow is a scalar equal to the sum of cos S contributions.

Posted by: Nige on April 18, 2013 10:13 AM | Permalink | Reply to this

Re: Uncertainty

I do intent to write a paper, and tackling this kind of objection is important.

Where, by “tackling”, you apparently mean “bury under a mountain of irrelevant verbiage.”

Since your prescription for the path integral does not yield a unitary S-matrix, scattering cross-sections (your only example of something “physical” to calculate) — which are extracted from the matrix elements of the S-matrix — will be garbage, too.

Consider (as one of an infinite number of constraints that your prescription will fail to reproduce), for instance, the Optical Theorem σ total=4πpImf(0) \sigma_{\text{total}} = \frac{4\pi}{p}Im f(0) where f(0)f(0) is the amplitude for forward-scattering (which you want to take to be real).

Posted by: Jacques Distler on April 18, 2013 11:37 AM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

But consider Haag’s theorem against demonstrating the self-consistency of the renormalization procedure for complex (Hilbert) space. There’s your no-go theorem! Paths of large action always cancel out in the path integral, so they don’t have any significant contribution to probabilities or cross-sections whether you sum in complex or real space. So physically, there’s no problem. In Bethe’s Optical Theorem, Maxwell’s equations for a light wave give a classical wavefunction amplitude (i.e. electric field strength) solution, exp(ikz). That’s a perfectly valid classical approximation theorem.

Posted by: Nige Cook on April 19, 2013 3:45 AM | Permalink | Reply to this

Re: Uncertainty

That paragraph is a bravura combination of word salad

In Bethe’s Optical Theorem, Maxwell’s equations for a light wave give a classical wavefunction amplitude (i.e. electric field strength) solution …

and non-sequiturs

So physically, there’s no problem.

I begin to suspect that “Nigel” is a pseudonym and that you are actually the genius behind the snarXiv.

Posted by: Jacques Distler on April 19, 2013 8:03 AM | Permalink | PGP Sig | Reply to this

Re: Uncertainty


What do you think of this interesting paper:

Posted by: steve on March 15, 2013 8:25 PM | Permalink | Reply to this

Re: Uncertainty

You must be using the word “interesting” in some sense with which I am unfamiliar.

Posted by: Jacques Distler on March 15, 2013 8:42 PM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

Any new interesting developments in string theory?

Posted by: teve on March 18, 2013 12:03 AM | Permalink | Reply to this

Re: Uncertainty

What do you think of Lee Smolin’s latest book? Have you read it?

Posted by: Ricky on May 1, 2013 8:15 AM | Permalink | Reply to this

Re: Uncertainty

What do you think of Witten’s new paper?

What are the implications to string theory?

Posted by: Ricky on May 1, 2013 9:26 AM | Permalink | Reply to this

Re: Uncertainty

I have already discussed this at great length.

The only thing to be changed is that the statement, “Supermoduli space is probably non-split.” can be replaced with, “Supermoduli space is non-split.”

One of the puzzling features of their proof, though, is that increasing the number of (NS) punctures, at fixed genus, removes the obstruction to splitness (at least, the lowest obstruction, which they calculate). This is puzzling, as one would not expect that adding punctures would “help.”

Presumably, demanding that the splitting extend to the Deligne-Knudsen-Mumford compactification of the moduli space would remove this lacuna.

(Reading the comment threads attached to those old posts is something of a hoot, as we get to see Luboš at his inimitably-obtuse best.)

Posted by: Jacques Distler on May 1, 2013 1:22 PM | Permalink | PGP Sig | Reply to this

Re: Uncertainty

What do you think of this paper?

Posted by: Ed on May 29, 2013 1:40 AM | Permalink | Reply to this

That is now the tenth occurrence on this page of the words, “What do you think…”

Jacques, why is the sky blue?

But why?

But why?

Posted by: Ask Jacques Anything on May 29, 2013 5:30 PM | Permalink | Reply to this

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