Many physicists are inclined to say that there must be a fundamental indeterminism in physics. After all, if we cannot observe an electron's trajectory when it is involved in forming an interference pattern without ruining the interference patter then doesn't that mean that the electron does not have a trajectory in that case? Notice, however that there is a particular philosophy behind this kind of statement — verificationism. Verificationism is the doctrine that, other than the tautologies of logic, the only meaningful truths are those that can be verified empirically (i.e. by an experiment). Given that quantum physics does not give us a clear-cut picture of reality, a retreat into verificationism is a popular position among physicists. If we only focus on those statements that can be directly verified in an experiment, and dismiss anything else as meaningless, then we probably will not get confused about how to use quantum mechanics to make predictions.

Nevertheless, verificationism is not a particularly popular position in the philosophy of science. I do not wish to get too far into philosophical questions here. Suffice to say, there are interpretations of quantum mechanics in which electrons do have well defined trajectories at all times and others in which they do not. The interpretations where there are always definite trajectories are not very popular among physicists, partly because they violate the principle of locality, but the option is still open at this stage and I will not take a strong stance for or against the idea.

However, one thing is for sure: if we cannot observe precise trajectories and interference in the same experiment, then we will generally be uncertain about where an electron will be detected in an interference experiment. We are at best going to be able to predict a probability for where it will land. This could be because the particle does not have a definite trajectory and so the location that it will be detected is not determined until it is measured, or because we simply do not know the trajectory of the particle. Either way, we will only be able to predict a probability.

In general, quantum mechanics only predicts the probability for where a particle will be detected. The rule for predicting these probabilities was proposed by Max Born in 1927, and is now known as the Born rule. For the special case of a particle in one-dimension, the Born rule works as follows.

Suppose $\psi(x,t)$ is the wavefunction of a one-dimensional system. We first ensure that it is normalized such that

\[\int_{-\infty}^{+\infty} |\psi(x,t)|^2\,\mathrm{d}x = 1.\]

Then, at time $t$, the probability for finding the particle at a position $x$ such that $a<x<b$ when you measure its position is

\[p(a<x<b) = \int_a^b |\psi(x,t)|^2 \,\mathrm{d}x.\]

In other words, $|\psi(x,t)|^2$ is the probability density for finding the particle at $x$ when you measure its position.

A few comments are in order here. First, note that the intensity of a wave is proportional to $|\psi(x,t)|^2$. Since interference patterns in quantum mechanics are built out of large numbers of individual particle detections, the proportion of the particles that hit a given location should be proportional to the intensity. Otherwise, we would not reproduce the classical interference patterns in the limit of a large number of particle detections. This is just what the Born rule says.

Second, the reason we work with normalized wavefunctions is that the Born rule tells us that the probability of finding the particle somewhere is \[p(-\infty<x<+\infty) = \int_{-\infty}^{+\infty} |\psi(x,t)|^2 \,\mathrm{d}x\] Since the particle will always be found somewhere if measured, this quantity needs to equal $1$.

You will not always be lucky enough to be given neatly normalized wavefunctions, so what should you do if you are given a wavefunction $\phi(x,t)$ such that $\int_{-\infty}^{+\infty} |\phi(x,t)|^2\,\mathrm{d}x \neq 1$? What I recommend doing is to always normalize the wavefunction as the first step in any calculation. To do this, define $\psi(x,t) = A\phi(x,t)$, where $A$ is a constant that you determine by imposing $\int_{-\infty}^{+\infty} |\psi(x,t)|^2\,\mathrm{d}x = 1$, and then work with $\psi(x,t)$ for the rest of the calculation.

Equivalently, the Born rule can be generalized to unnormalized wavefunctions as follows. \[p(a<x<b) = \frac{\int_a^b |\phi(x,t)|^2 \,\mathrm{d}x}{\int_{-\infty}^{+\infty} |\phi(x,t)|^2\,\mathrm{d}x}.\] Doing things this way makes the formulas look more complicated, so I recommend just working with normalized wavefunctions from the outset. Whenever you are doing any calculation with wavefunctions in quantum mechanics, always ask yourself “Is this wavefunction normalized?” and if not then normalize it straight away.

To practice normalizing wavefunctions and calculating probabilities, please do the in-class activity now.

In 1927, Werner Heisenberg proposed the uncertainty principle relating our uncertainty about the position of a particle to our uncertainty about momentum. For now, $\Delta x$ denotes a quantitative measure of our uncertainty about position and $\Delta p$ denotes a quantitative measure of our uncertainty about momentum.

For a one-dimensional system, the uncertainty principle states \[\boxed{\Delta x \Delta p \geq \frac{\hbar}{2}}.\]

For a three-dimensional system, this applies in all three coordinate directions. \[\boxed{\Delta x\Delta p_x\geq \frac{\hbar}{2},\qquad \Delta y\Delta p_y\geq \frac{\hbar}{2},\qquad \Delta z\Delta p_z\geq \frac{\hbar}{2}}.\]

There are several possible meanings for these equations:

1. Preparation uncertainty: At any given time, $\Delta x$, is our uncertainty about what the outcome of a position measurement would be were we to make it and $\Delta p$ is our uncertainty about what the outcome of a momentum measurement would be were we to make it. (We can only actually make one of these two measurements.)
2. Measurement-disturbance: If we make a measurement of $x$ with accuracy $\Delta x$ then the momentum will be disturbed by an amount $\Delta p$ and vice versa.
3. Joint measurement accuracy: It is not possible to measure both $x$ and $p$ precisely at the same time. If we attempt to make a measurement that tells us both $x$ and $p$ then the accuracy of those measurements is limited by $\Delta x \Delta p \geq \hbar/2.$

All three of these are true in quantum theory. 2 and 3 have to be handled carefully because a quantum particle does not necessarily have a momentum or position that can be disturbed, so how can we tell the accuracy of a measurement of a quantity that does not necessarily exist before the measurement. 2 and 3 are the subject of ongoing research and controversy, but it is possible to come up with definitions of “disturbance” and “accuracy” such that they are true.

Preparation uncertainty is the version that is usually proved in undergraduate quantum mechanics textbooks and we will do so in section 2. However, Heisenberg originally argued for 2.

Heisenberg's argument is a semi-classical argument that does not pass muster in full-blown quantum mechanics, but it will give you an idea of where the principle came from. He imagines observing the position of an electron by scattering light off it. He treats the electron classically, as if it had a well defined position and momentum, and applies quantum principles to the light. This setup is often called the Heisenberg Microscope. I will give a toy version of the argument here that gets the order of magnitude right. Deriving precisely $\hbar/2$ as the limit requires a more detailed argument, which you can find at this link.

To observe an electron's position with accuracy $\Delta x$, we need to use light of wavelength $\lambda \sim \Delta x$ or smaller. By the de Broglie relation, this corresponds to a photon of momentum $p_{\text{light}} = h/\lambda \sim h/\Delta x$.

To see the electron, the light must scatter off the electron, so there will be Compton scattering. The change in momentum of the electron will be of the same order of magnitude as the initial momentum of the photon. Therefore, \[\Delta p \sim p_{\text{light}} \sim \frac{h}{\Delta x} \qquad\qquad \Rightarrow \qquad\qquad \Delta x \Delta p \sim \hbar.\]

The Heisenberg uncertainty relations can be generalized to any pair of complementary observables, i.e., variables that are canonically conjugate in classical mechanics. In particlular, energy and time are complementary, so \[\boxed{\Delta E \Delta t \geq \frac{\hbar}{2}.}\]

The interpretation of the energy-time uncertainty relation is more difficult than that of position and momentum because time is not treated as an observable in quantum mechanics, i.e. it is just a parameter that appears in the equation of motion and not something we usually calculate probabilities for. There are several different versions of energy-time uncertainty that have appeared in the literature. One version says that if your uncertainty about $E$ at time $t$ is $\Delta E$ and $\Delta t$ is the time it takes for the average energy $\bar{E}$ to change by an amount $\Delta E$ then $\Delta E\Delta t \geq \hbar/2$. For a proof of this see the textbook “Quantum Mechanics” by Messiah, Section VIII.13.

1. Consider the wavefunction $\psi(x) = A\sqrt{|x|}e^{-x^2}$, where $A$ is a constant chosen such that $\int_{-\infty}^{+\infty} |\psi(x)|^2 = 1$.

   1. Find $A$.
   2. Determine $p(-1<x<1)$.