2
$\begingroup$

I am struggling to interpret the probabilistic nature of quantum mechanics and am wondering if uncertainty is an emergent property of nature or if this is really a statement about our measurement tools. I know similar questions have been answered here and I have read the responses and still don't feel I have a good grasp on this. One way I have been thinking about this is with a simple classical example of a wave that can be applied to think about quantum mechanical waves:

Consider a pulse forming a wave, at a short time after we start the pulses it has a small delta uncertainty in position but a large uncertainty in wavelength. Now as we add more pulses one after another we can measure more information about the wave and we get less uncertainty in wavelength but it increases uncertainty in the position. If this uncertainty was an emergent property of the waves the uncertainty wouldn't change as we measured longer duration of pulses and get more certainty in the wavelength but less in the position. So it begs the question from a QM standpoint, is this simply because we don't have enough information/tools to measure properties accurately or are the waves properties truly not well defined. I feel like the answer is yes, since when measuring this wave the wave itself has not changed, only how we are measuring it.

I find this to be an important question because we make a lot of assumptions about the nature of reality based on the idea that states of particles are probabilistic until measured.

$\endgroup$
3
1
$\begingroup$

The quantum Heisenberg uncertainty relations are completely analogous to the Fourier uncertainty relations which are often used in time series analysis and signal processing. Both statements are about the width of wave packets. I think the term uncertainty is a misnomer. As @GuyInchbald stated, it is a historical artifact. The modern understanding is that the Heisenberg uncertainty relations are fundamentally about wavefunctions, not measurements.

You describe an experiment where you start with no wave, and then begin to continuously generate waves that propagate away. If this were a quantum mechanical wavefunction, you would be continuously changing the quantum state. The wavefunction uniquely determines the state. When you add another cycle to the wavefunction, you've made a different quantum state.

... when measuring this wave the wave itself has not changed, only how we are measuring it.

This statement is not true.

What if you make a localized wave packet and send it away? The classical wave equation would say that the wave packet would preserve its shape, so its width would not change. For a wavefunction, the Schrödinger equation would say that the wave would spread out as it propagates away. As time passes the state evolves, its width increases. The $\Delta x$ and $\Delta p$ for a propagating wavepacket would change.

I have gained a lot of insight by thinking about the connection between Fourier and Heisenberg uncertainty and by applying uncertainty reasoning to situations where no measurements occur.

Fourier uncertainty

A sine wave $\psi(t) = \sin(\omega_0 t)$ has a single frequency $\omega_0$, but it is spread out over all time. We can define the width of any signal in time, by its RMS width $$\Delta t = \sqrt{\left< (t-\left<t\right>)^2 \right>},$$ where the brackets $\left<\right>$ denote the average. For the sine wave this would give $\Delta t \rightarrow \infty$.

We can think of the time representation of the wave as specifying a basis to describe the wave. Each instant of time gets a basis vector, and the amplitude of the wave at that time is one element of the vector.

If we take the Fourier transform of the sine wave we would get a frequency representation: $$\tilde\psi(\omega) = \delta(\omega-\omega_0),$$ where $\delta$ is the Dirac delta function. In the frequency domain the sine wave is zero everywhere except $\omega_0$. It is infinitely thin. $$\Delta \omega = \sqrt{\left< (\omega-\left<\omega\right>)^2 \right>} = 0.$$

This frequency representation is a new basis for the wave. There is a basis vector for each frequency value. In this basis it takes just one basis vector to define the wave, the one for $\omega_0$.

If we have a finite wave packet, like a Gaussian enveloped sine wave $$\psi(t) = e^{-t^2/(2\sigma^2)}\sin(\omega_0 t),$$ the wave will have a finite width in both the time and frequency representation. This wave has some spread over multiple $\omega$s.

Using the RMS width it can be shown that $$\Delta t\,\Delta\omega \ge \frac{1}{2}.$$ This is a general property of Fourier transforms that tells us about the RMS width of waves in the two representations.

Heisenberg uncertainty

Now what if our wave were a quantum mechanical wavefunction? The same thing would be true. If we multiply the Fourier uncertainty statement by $\hbar$, we get $$\Delta t\,(\hbar\Delta\omega) = \Delta t\, \Delta E \ge \frac{\hbar}{2}.$$

This says the width of a wavefunction in its energy representation is related to its width in its time representation. An energy eigenstate has $\Delta E=0$, so $\Delta t \rightarrow \infty$. The wavefunction for an energy eigenstate extends infinitely forward through time. This is equivalent to the statement that energy eigenstates are stationary states.

There's no reference to measurement in the previous statement. It is a statement about the wavefunction when written in two different bases: $\psi(t)$ and $\tilde\psi(E)$.

If the wavefunction describes a mixture of energy states, it has finite width in time. It is not stationary. $$\psi(E) = A\delta(E-E_1) + B\delta(E-E_2) \implies \Delta E > 0 \implies \Delta t\,\,\mathrm{finite}$$

There's an analogous Fourier relation for position $x$ and wavenumber $k = \frac{2\pi}{\lambda}$, which leads to the quantum position momentum uncertainty relation since $p = \hbar k$: $$\Delta x\, \Delta k \ge \frac{1}{2}\quad\implies\quad \Delta x\, \Delta p \ge \frac{\hbar}{2}.$$

This uncertainty relation tells us about the width of of the wavefunction in two different basis representations: $\psi(x)$ and $\tilde\psi(p)$.

$\endgroup$
1
  • $\begingroup$ Wow, this is an extremely insightful way of looking at it. It's been a long time since I have done any QM but this was a good refresher and it makes a lot of sense when thinking about no measurement situations. Thanks $\endgroup$ Apr 27 at 20:16
1
$\begingroup$

It used to be thought that uncertainty was essentially a measurement problem - that measuring one property meant disturbing another. Heisenberg himself, who discovered the uncertainty principle, long thought that this was the case.

But eventually it became clear that uncertainty is an intrinsic property of quantum reality and, though Heisenberg and his colleagues fought a tough rearguard action, they eventually had to concede that the measurement explanation is just plain wrong. Reality really is uncertain. For example if you pin down a particle's position with extreme accuracy, its momentum really does become uncertain and, vice versa, if you give it a precise momentum then it really has no particular location.

One example of this can be seen if you shine a beam of say electrons through a small hole whose size you can vary. The smaller the hole, the tighter you pin down their position, which means that their momentum becomes more uncertain and they will spread out wider and make a bigger blob on your detection screen. The measurement argument cannot explain this.

$\endgroup$
1
  • $\begingroup$ Thank you, that makes a lot of sense $\endgroup$ Apr 27 at 20:14

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.