Something Certain About Uncertainty

I was motivated by a comment on Jim Pivarski’s recent post to speak about the Heisenberg Uncertainty Principle. Someone asked,

If uncertainty in quantum mechanics comes from (or is inseparable from) quantization, then where does it come from in its mathematical formulation i.e in terms of a space and its Fourier transform?

The Heisenberg Uncertainty Principle is a curious fact: it requires no physical intuition whatsoever and yet has profound physical ramifications. It is also interesting because it is among a small group of facts which are both physically and mathematically interesting. It is an important (to harmonic analysis) and commonly known fact that a function and its Fourier transform cannot both be compactly supported. There are stronger statements than that, though, of the following flavor: if a function is a narrow spike near a point, then its Fourier transform will be more spread out. The Heisenberg Uncertainty Principle is a quantitative statement about this kind of fact.

The Principle follows from several simple but fundamentally powerful aspects of the Fourier transform. First, polynomials in derivatives acting on a function can be pulled outside the Fourier transform into corresponding polynomials in frequency variables. More specifically,

$\displaystyle{p(2\pi i\xi)\hat{f}(\xi) = \left(\widehat{p(\partial)f}\right)(\xi).}$

where $p$ is a polynomial and $\partial$ means derivative with respect to $x.$ Second, translation in space variables leads to modulation in frequency variables:

$\displaystyle{\widehat{f(x-x_0)} = e^{-2\pi ix_0\xi}\hat{f}(\xi).}$

Third, modulation in space variables leads to translation in frequency variables:

$\displaystyle{\widehat{e^{2\pi i\xi_0x}f}(\xi)=\hat{f}(\xi-\xi_0).}$

Now, let’s derive the Uncertainty Principle. Let $L^2$ denote the functions $f$ which are square integrable — that is, $\int |f(x)|^2 dx < \infty.$ We will write

$\displaystyle{||f||_2 = \left(\int |f(x)|^2 dx\right)^{1/2}.}$

On $L^2$ , define the operators $X$ and $D$ by

$\displaystyle{Xf(t) = tf(t)}$ — position.

$\displaystyle{Df(t) = \frac{h}{2\pi i}f'(t) \textrm{ so that }\widehat{Df}(t) = \xi \hat{f}(\xi)}$ — momentum.

Here $h$ is Planck’s constant which is on the order of $10^{-34}$ in the macroscopic units of Joules-seconds. The presence of $h$ is unimportant mathematically, so let’s assume we’re using units so that $h$ is 1. That $X$ corresponds to the position operator of a particle comes from the interpretation of the wave function: if $f$ is a solution to the Schrödinger equation, then it is a probability distribution — or rather, $|f|^2$ is — so that $\int x|f(x)|^2$ gives the expected value for position. That $D$ corresponds to the momentum operator basically comes from the fact that it’s like the velocity of the wave function, and momentum is mass times velocity. More precisely, the Schrödinger equation,

$\displaystyle{-\frac{1}{2m}\left(\frac{h}{2\pi}\right)^2\frac{d^2}{dx^2}\Psi + V(x)\Psi = E\Psi.}$

is nothing more than a statement about conservation of energy. The kinetic energy is $p^2/2m$ which corresponds to the first term, and so our choice for the momentum operator $D$ reflects that.

One should note here that, as indicated above, taking the Fourier transform of momentum gives the position operator in the frequency variable:

$\displaystyle{\widehat{Df}(t) = \xi \hat{f}(\xi).}$

The Fourier transform of position is basically the same thing as momentum, but it’s off by a constant multiple.

Observe that $D$ and $X$ do not commute; specifically,

$\displaystyle{[D,X]f = DXf - XDf = \frac{1}{2\pi i}f.}$

We know that $L^2$ has as an inner product $<f,g> = \int f \bar{g}dx$ . Also, it’s pretty easy to see that both $X$ and $D$ are self-adjoint, meaning that $<Xf,g>=<f,Xg>,$ and likewise for $D$ . Using these two facts, we find that for arbitrary real constants $a$ and $b$ ,

$\displaystyle{\|(aX-ibD)f\|_2^2 =<(aX-ibD)f,(aX-ibD)f>}$

$\displaystyle{= a^2<Xf,Xf> +iab<Xf,Df>}$

$\displaystyle{-iab<Df,Xf>+b^2<Df,Df>}$

$\displaystyle{= a^2\|Xf\|_2^2-iab<(DX-XD)f,f>+b^2\|Df\|_2^2}$

$\displaystyle{= a^2\|Xf\|_2^2 + b^2\|Df\|_2^2 - \frac{ab}{2\pi}\|f\|_2^2.}$

The first quantity is non-negative which means the last quantity is also, and so:

$\displaystyle{a^2\|Xf\|_2^2 + b^2\|Df\|_2^2 \ge \frac{ab}{2\pi}\|f\|_2^2.}$

Picking $a=\|Df\|_2$ and $b=\|Xf\|_2$ gives

$\displaystyle{\|Xf\|_2\|Df\|_2\ge \frac{\|f\|_2^2}{4\pi}.}$

For any fixed pair of numbers $x_0$ and $\xi_0$ , we can apply the above relation to $\displaystyle{g(x) = f(x+x_0)e^{-2\pi i x\xi_0}.}$ and deduce that

$\displaystyle{\left(\int|x-x_0|^2|f(x)|^2 dx\right)^{1/2}\left(\int|\xi-\xi_0|^2|\hat{f}(\xi)|^2 d\xi\right)^{1/2}\ge \frac{\|f\|_2^2}{4\pi}.}$

The second term in the product on the left side of above comes from a fact called Plancherel’s formula:

$\displaystyle{\|f\|_2 = \|\hat{f}\|_2.}$

For any wave function $f$ , we must have that $\|f\|_2=1$ since $|f|^2$ must be a probability distribution and thus have a total mass of 1. By choosing

$\displaystyle{x_0 = \int x|f(x)|^2 dx,}$

the expected position, and

$\displaystyle{\xi_0 = \int \xi|\hat{f}(\xi)|^2 d\xi,}$

the expected momentum, we get that

$\displaystyle{\sqrt{Var x}\sqrt{Var \xi} \ge \frac{1}{4\pi}.}$

where $Var$ denotes variance — the variance is the square of the standard deviation and so for an arbitrary distribution the square root of variance is the standard way to measure deviation from the mean.

More commonly, the above relation is written as

$\displaystyle{\Delta x \Delta p \ge \frac{1}{4\pi}.}$

By carefully retracing how Planck’s constant fits in, one can see that the physically relevant equation is

$\displaystyle{\Delta x \Delta p \ge \frac{h}{4\pi} = \frac{\hbar}{2}.}$

From a physical point of view, this means that our confidence in a measurement of position is at best inversely proportional to our confidence in a measurement of momentum. This gives us the usual qualitative interpretation of the Uncertainty Principle: one cannot simultaneously know with perfect certainty both position and momentum, or, as Heisenberg himself said, “The more precisely the position is determined, the less precisely the momentum is known in this instant, and vice versa.”

Tags: math.CA, math.MP

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13 Responses to “Something Certain About Uncertainty”

This week in the arXivs… « It’s Equal, but It’s Different… Says:
February 27, 2008 at 12:26 am | Reply
[…] Something Certain About Uncertainty […]
Greg Muller Says:
February 27, 2008 at 1:11 am | Reply
Nice post! Reminds me of a couple things.

First, relating the rate at which a function drops off at infinity to how smooth its Fourier transform. Specifically, if f(x) is less than $|x|^{-n}$ for some $n$ and for $x$ far enough for zero, then $\hat{f}(\zeta)$ is in $C^n$ (and vice versa). So, for instance, if we only care about functions who are smooth and have smooth Fourier transform, then they drop off really fast.

Second, heres a thing. So, lets take the Schwartz space $S(\mathbb{R})$ . Then the $L^2$ inner product makes this into an inner-product space, and the Fourier transform becomes an orthogonal operator which squares to -1. So we can think of this as giving $S(\mathbb{R})$ a complex structure, with $L^2$ a unitary inner product.

On consequence of this is that we get an anti-symmetric quadratic form if we take $\{f,g\}:=(f,Ig)=\int_\mathbb{R}f(x)\hat{g}(x)dx$ . With respect to this quadratic form, we can formalize the correspondence between position and momentum (and also modulation and translation), by saying they are adjoint with respect to this quadratic form.

Ok, so that was rather rambly and ill-thought out, but my general questions are along the lines of: is this anti-symmetric form significant? Are adjoints with respect to it interesting? Im hoping that maybe Uncertainty is a more general statement about functions failing to commute with their adjoints.
John Armstrong Says:
February 27, 2008 at 1:26 am | Reply
Okay, a couple things aren’t clear here.

First: how are you getting from the anticommutation relation between $X$ and $D$ ? You seem to be anticommuting copies of $X$ and $D$ acting on two different copies of $f$ .

Second: something is screwey with your next choices. Do you mean $a=\|Df\|_2$ and $b=\|Xf\|_2$ (instead of their squares)? And if so, why does it seem that you’re using $\|Df\|_2\|Xf\|_2\geq1$ in the next inequality?
Aaron Bergman Says:
February 27, 2008 at 1:52 am | Reply
Unless I’m missing a functional analytic subtlety in my on-the-fly attempt to convert physics-math to the real thing (which is always possible), the uncertainty principle follows for any Hermitian operators (I won’t completely vouch that you don’t need self-adjoint, but on first glance it looks like Hermitian suffices) acting on a complex vector space. In particular, if you have Hermitian operators

[A,B] = iC

then

(DA)^2 (DB)^2 >= (1/4)[ (v, (AB + BA)v)^2 + (v,Cv)^2 ]

Here DA^2 means (v, A^2v) – (v,Av)^2, similarly for DB, and v is in the domain of everything in sight.

Specializing to the present case is an easy exercise.
Peter Luthy Says:
February 27, 2008 at 1:59 am | Reply
Hi John,

For your first question, I edited the page to include the computations. You just need to view the $L^2$ norm as induced by the usual inner product and everything falls out, provided you are careful to note that it is conjugate linear in the second variable.

For your second question, you are correct that I was overly zealous and put in some extra squares there where I shouldn’t have. So really $a=||Df||_2$ and $b=||Xf||_2$ . Anyway, once you have those corrections in there, just add everything together and cancel some things out and it all works.
Peter Luthy Says:
February 27, 2008 at 2:38 am | Reply
Aaron,

You are indeed correct that a general uncertainty principle holds for any self-adjoint operators, and indeed a formula very similar to the one you’ve written down gives exactly the result I’ve proven as a special case. I even saw the proof you are referring to in the quantum mechanics courses I took while pursuing my undergraduate physics degree. However, I didn’t find that proof very enlightening. The fact that position and momentum are intertwined by the Fourier transform is not illuminated by the algebraic manipulations involved in that proof. This proof gives glimpses of the relationship between a function and its Fourier transform and explains why the Uncertainty Principle is actually deeper than just a result about self-adjoint operators. In time-frequency analysis, one talks about Heisenberg boxes when proving things completely unrelated to physics. I think this proof brings that more into the foreground.

Also the original question was asked in relation to the Fourier transform, so I thought it most appropriate to meander in that direction.
John Armstrong Says:
February 27, 2008 at 3:12 am | Reply
What happened to the $f$ s from the first to the second line of the new computations? In the first line you’ve got inner products of functions (like $(aX-ibD)f$ ) and in the second line you’ve got inner products of operators. If I assume that it’s just a typo and you meant $Xf$ instead of $X$ , the next step doesn’t follow.

I think what you’re leaving out is the self-adjoint property that lets you move the $D$ and $X$ operators from one side of the inner product to the other. Then you can invoke the commutation relation. The self-adjointness is essential, but you haven’t mentioned it anywhere in the post.

As for the other, there was another factor of $\|Xf\|_2\|Df\|_2$ that you removed in the edit. I get it now. Thanks.
Peter Luthy Says:
February 27, 2008 at 4:11 am | Reply
John, thanks for all the corrections. Originally I intended to completely suppress the computation and just say it’s easy to check using a few properties, but then I decided to write it out. Unfortunately I ended up doing both in a relatively half-assed manner which apparently did not produce a full… err… well I didn’t end up doing a good job with it anyway. Hopefully it’s fixed now.
jonathon lough Says:
February 28, 2008 at 4:23 am | Reply
Thanks for such an excellent response Peter. When I wrote my previous post, I was pondering on a certain line of thought in “Quantum Bears” that is exemplified by sentences like “Quantisation makes fundamental uncertainty possibe” and “Without quantisation, there would be no fundamental uncertainty”.

In light of these comments, the thing I was curious about at the time, was that in the proof for the uncertainty principle (in the context of Fourier transforms), there is no need to refer to quantisation, nor does it seem possible to. It seems that in this context we have no quantisation, yet we do have fundamental uncertainty. What was I missing?

A couple of days on…, and having re-read “Quantum Bears”, I can see that “quantisation” usually refers to quantisation of phase space. In which case, uncertainty really does imply (or is the same thing as) at least one sort of quantisation. When I hear the word “quantisation”, I always think of quantisation of an observable or something like that, but never of the phase space of the system.

So does quantisation always refer to quantisation of phase space? If so, then in the language of “Quantum Bears”, this would imply that all parameters in the relevent parameter space would come in transform pairs (am I thinking straight?). But what of those discrete energy levels of the hydrogen atom, how is this quantisation of phase space?

Interesting stuff. I really like this site by the way so thanks to all involved.
Peter Luthy Says:
February 28, 2008 at 3:29 pm | Reply
Jonathon,

This Uncertainty Principle says nothing about quantization in terms of the Hydrogen atom, at least as far as I can tell. In fact it doesn’t even really say anything physical. It’s a statement about a property of the Fourier transform. It does imply that phase space is blocky and that one can’t hope to get meaningful information about of boxes in the phase plane which are too small.

The term quantization usually refers to the discretization of energy states (or other things). I believe this idea first arose when studying black body radiation — a black body is basically just something which shines only when heated up, like a lightbulb filament. Classical mechanics completely failed to model the situation correctly. There were a number of attempts, but each time the models failed. Then just at the turn of the century, Max Planck tried something he didn’t even fully believe or understand — he assumed that the oscillators (atoms) making up a black body were only allowed to have integer multiples of some base energy. The trick worked, and the function he derived accurately predicted the intensity of light given off by a blackbody at various temperatures and wavelengths.

The reason, I suppose, the uncertainty principle and quantization are inseparable is that the operators $X$ and $D$ only appear physically by way of the Schrodinger equation (otherwise what exactly they both act on is a big question mark). At least that has been my experience — perhaps someone knows better. But this immediately leads to quantization when one attempts to solve the equation — you end up with some free parameters in your solution but are later forced to choose them as integers. Likewise, if you start with the Schrodinger equation, one naturally produces fundamental uncertainty simply by the existence of the momentum and position operators. That is more what Jim meant, I think. Classical mechanics does not have fundamental uncertainty because these operators don’t mean much in classical mechanics; quantum mechanics is intertwined with fundamental uncertainty.

Glad you enjoy the blog. Also, thanks for asking your question!
mousomer Says:
March 5, 2008 at 10:58 am | Reply
Great post.
One small comment, though:
There is a simple physical interpretation of the uncertainty principle. I believe it’s due to Dirac (though I might have confused it).

The idea, in short, is that in order to know anything at all about a particle, one has to interact with it – which means, basically, bombard it with photons. Now photons have finite (nonzero) wavelength. Hence information about the particle’s position can only be exact up to the photon’s wavelength. Use a “long” photon, and you get a big spatial uncertainty. To get a better spatial resolution, you have to use a shorter wavelet photon – which is a higher energy photon. But that means that you have just kicked that particle hard – so you lose information about the particle’s energy.
Of course, it all stems from knowing the wave nature of light, where energy and wavelength are inversely proportional. This is a special case of the theorem you just discussed, but it is a classical one, and fairly easy to grasp.
Peter Luthy Says:
March 9, 2008 at 8:58 pm | Reply
mousomer,

Yes, that is a pretty good way to think of the physical justification for the weaker form of the uncertainty principle, i.e. the no perfect certainty in both momentum and position statement. It explains how the old view of measurement is wrong since measurement biases the experiment.

The quantitative statement, though, is of course stronger. Aside from being extremely nice in its formulation — involving just f, its Fourier transform, and a simple constant — it also expresses, for instance, exactly what scale the fundamental uncertainty becomes relevant or exactly how far away from a neutron the pi mesons (particles involved in strong force interactions) will tend to travel, and so forth.
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