6.5: Incompatible Measurements

Plane Waves

We have spent a lot of time talking about particles associated plane waves, because they are easy to conceive – they have a single, specific wavelength, and therefore a definite momentum.  These are particles free from any forces, moving at a constant, well-defined speed.  Easy, right? Well, let's take a look at this wave function in terms of locating the position of the particle.  Choosing a cosine function to describe this wave function moving in the \(+x\)-direction, we have a function familiar from Physics 9B:

\[\Psi\left(x,t\right)=A\cos\left(\frac{2\pi}{\lambda}x-2\pi ft\right)\]

Let's simplify this discussion by looking at the wave only at time \(t=0\):

\[\psi\left(x\right)=\Psi\left(x,0\right)=A\cos\left(\frac{2\pi}{\lambda}x\right)\]

If we form a probability density from this probability amplitude, we get:

\[\mathcal P\left(x\right)=\left|\psi\left(x\right)\right|^2=A^2\cos^2\left(\frac{2\pi}{\lambda}x\right)\]

Okay, let's normalize our probability density (i.e. find the value of \(A\)):

\[1=\int\limits_{\text{all}~x}\mathcal P\left(x\right)dx=\int\limits_{-\infty}^{+\infty}A^2\cos^2\left(\frac{2\pi}{\lambda}x\right)dx\]

Uh-oh. We have a problem. This integral blows up, making \(A=0\), and \(\mathcal P\equiv 0\). This makes no sense, what is going on here?

Actually, it does make sense – with a wave function that has the same amplitude along the entire \(x\)-axis, the particle must be equally-probable to be found anywhere, so if we look for it in any finite interval, the measure of that interval is zero compared with the measure of the remaining infinite space where it can be found. A simple way to express this is that we have no idea where the particle is. This complete lack of knowledge about the particle's position was the small price we had to pay for knowing the particle's momentum precisely. We will see here that this price cuts both ways.

Localizing Free Particles

We have a hint that the precise knowledge of a particle's momentum goes together with a complete lack of knowledge of its location, so let's see if reducing what we know about momentum (or equivalently, wavelength) has the effect of improving our ability to discern the particle's position. We'll start simple: Let's see what happens when we superpose two plane waves with different wavelengths.

Figure 6.5.1 – Superposition of Two Plane Waves

The top graph depicts a plane wave. The one below it is what happens when another plane wave with a slightly shorter wavelength (the wave number \(k=\frac{2\pi}{\lambda}\) is larger) is superposed with it. The contribution of this second wave to the superposition is slightly less as well (i.e. its amplitude is smaller than that of the original wave):

\(\psi\left(x\right)=A\cos kx\)

\(\psi\left(x\right)=A\cos kx+0.6A\cos\left(k+4\delta k\right)x\)

The first thing that jumps out is the way that the amplitude (not the displacement!) varies when you evaluate it at different places on the \(x\)-axis. In probability terms, this means that the probability of finding the particle in a region near the maximum bulges ("antinodes") is quite high compared to finding it near the narrower regions ("nodes"). While this wave is still infinitely-long, and our knowledge of the position of the particle is still zero, in a relative sense, we have a better sense of where the particle is than when we were dealing with a single harmonic wave function.

If we can do this much just by using two wavelengths, perhaps we can improve things even more by adding some additional harmonic waves. If we choose a third plane wave appropriately, we find that the bulges become more defined:

Figure 6.5.2 – Superposition of Three Plane Waves

All we did here was add a third plane wave with a wave number below that of the original wave by the same amount as the second wave's wave number was larger:

\(\psi\left(x\right)=A\cos kx+0.6A\cos\left(k+4\delta k\right)x+0.6A\cos\left(k-4\delta k\right)x\)

Let's see what happens if we "fill in" a couple of the wave number gaps. That is, to get the above result, we added & subtracted \(4\delta k\) to the wavenumber of the original wave for the added waves. Let's add & subtract \(2\delta k\), and since it is closer to the original frequency, we'll weight its amplitude more as well (\(0.8A\)). The result is:

Figure 6.5.3 – Superposition of Five Plane Waves

The wave function for this fourth case is:

\(\psi\left(x\right)=A\cos kx+0.8A\cos\left(k+2\delta k\right)x+0.8A\cos\left(k-2\delta k\right)x+0.6A\cos\left(k+4\delta k\right)x+0.6A\cos\left(k-4\delta k\right)x\)

It should be clear what happens if we continue this program indefinitely – we are left with a single wave packet, as all the other bulges get pushed out to infinity. This localizes our free particle, giving it a finite probability of being found within a given region.  This localization improved as we added the number of possible wavelengths (momenta) that could be measured. So we can improve our knowledge of the particle's position at the cost of knowing about its momentum, and vice-versa.

Spectral Content

Notice that the bulges separated when we inserted plane waves with wave numbers between the ones we already had in place.  In order to get the bulges separated to infinity we need to fill in all of those in-between wave numbers.  Of course, there is a whole continuum of these available, so we need to use an integral to capture all of the harmonics. Writing a general function as an integral (essentially a sum) of a continuum of harmonic functions is called taking a Fourier transform, which looks like this:

\[\psi\left(x\right)=\int\limits_{-\infty}^{+\infty}\left[A\left(k\right)\cos kx+iA\left(k\right)\sin kx\right]dk\]

The function \(A\left(k\right)\) is the "recipe" for the transform, meaning it tells us how many "tablespoons" of the harmonic wave of wave number \(k=\frac{2\pi}{\lambda}\) to mix into the bowl to get the function \(\psi\). We often refer to this function as the spectral content of the particle's quantum state.  Yes, \(A\left(k\right)\) is the same function multiplying both the cosine and sine functions. Also, yes, the imaginary 'i' makes an appearance here. This gives us a more compact way to express the transform, using the Euler identity:

\[\psi\left(x\right)=\int\limits_{-\infty}^{+\infty}A\left(k\right)e^{ikx}dk\]

If we have the function and want to know the recipe, we can extract it using a mathematical trick (that we won't go into here):

\[A\left(k\right)=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\psi\left(x\right)e^{-ikx}dx\]

This is usually referred to as the inverse Fourier transform.

This "recipe" is a function of wavelength, and thanks to deBroglie, we know that wavelength is related to momentum.  In fact it turns out that this recipe function is itself a wave function.  Just as \(\psi\left(x\right)\) is a probability amplitude for position \(x\), \(A\left(k\right)\) is a probability amplitude for momentum.  It's probably helpful to explicitly show the link between \(k\) (wavenumber) and \(p\) momentum.  They differ only by a constant factor:

\[k=\frac{2\pi}{\lambda},~~~p=\frac{h}{\lambda}~~~\Rightarrow~~~p=\frac{h}{2\pi}k=\hbar k\]

Planck's constant often appears in a fraction with \(2\pi\) in the denominator, so that fraction is given it's own symbol, shown above: \(\hbar\equiv\) "h-bar".

When treating the function \(A\left(k\right)\) as a wave function for momentum, it is common to write it differently, so that it's nature is more evident:

\[A\left(k\right)~~\rightarrow~~\phi\left(p\right)\]

The Heisenberg Uncertainty Principle

Now that it's quite clear that there is an inverse relationship between the uncertainty of measurements of position and momentum, we can state it formally as was first done by Werner Heisenberg. The specific predicament of a particular particle will define what the uncertainty will be in measuring either the position or the momentum, and we can change the conditions of our experiment to improve the certainty of our measurement of either of these quantities, but as we improve one, we necessarily worsen the other. According to Heisenberg's math, we get the following inequality expressing the principle that bears his name:

\[\Delta x\Delta p\ge\frac{\hbar}{2}\]

No matter how we devise our experiment to measure \(x\) and \(p\), when we compute their uncertainties statistically, we will find that the product of these uncertainties will always come out to a number no less than \(\frac{\hbar}{2}\).

This principle is really well demonstrated through the Fourier transform. If we consider a localized (in position) wave packet (by "localized," we mean that the probability amplitude  for measuring various positions drops off very rapidly far away from the center), and then Fourier-transform this function, we get the wave function of the same particle expressed in terms of its spectral content (probability amplitude for measuring various momenta).

Figure 6.5.4 – Particle Wave Function Expressed in Terms of Position and Momentum

The uncertainties in position and momentum can be calculated in the usual way from the probability densities that come from these wave functions, and these are expressed in the diagram above.

Now suppose we change the physical conditions that brought about this quantum state. For example, suppose we change the way we take measurements so that we better-confine our knowledge of the position of the particle. This will serve to "tighten" its wave packet. When we take the Fourier transform of this new position wave packet, the momentum wave packet broadens.

Figure 6.5.5 – Same Particle with Position Measured More Precisely

Heisenberg's principle states that even if one provides the ideal conditions for the particle, this inverse relationship between the position and momentum uncertainties results in a limit to the minimum value that the product of these uncertainties can attain.

One last comment here: Notice that above we used the phrase "change the physical conditions" and "measure differently" interchangeably. This is a very important aspect of quantum theory. The probabilities we measure are dependent upon physical conditions, and the process of making measurements necessarily affects these conditions. We have encountered this once before, when we discussed watching electrons pass through a double slit.