No-Nonsense Quantum Mechanics: Part 1 Bonus Material

Although elementary particles are discrete things (we observe "click click click" in our detectors), we need waves to describe them.

Surprisingly, we observe for elementary particles like electrons an interference pattern, which we usually only expect for waves like water waves. For classical discrete objects like balls or objects, we don't get an interference pattern. Most of them end up around in the middle and progressively less as we move farther away from the middle. However, for waves, we get a sequence of maxima and minima.

While usually, we expect that any value is possible for a physical quantity, this is no longer the case when it is quantized. For a quantized quantity, only a discrete set of values is possible and no longer any value.

We can understand this by recalling the fundamental lesson of the double slit experiment: we need waves to describe particles. Wave naturally quantize quantities. Imagine a rope that is fixed at one end and with the other end in your hand. If you start swinging it, you can only produce a discrete set of wave modes. The allowed modes are quantized in discrete steps.

Similarly, in Quantum Mechanics the fact that we need waves to describe particles leads us naturally to the result that, for example, the energy levels of a particle in a box are quantized.

• Electrical charge is always quantized.
• Energy is quantized for some quantum systems, but not in general.
• Angular momentum is also quantized for some systems.

Even when quantities like the energy of electrical charge are quantized, the discrete steps between allowed values are so tiny that it's impossible to see them in everyday life.

• Using the Fourier transform we can understand any localized wave package as a superposition of plane waves.
• Only plane waves have well-defined wavelengths, but only wave packages are localized. A plane wave spreads out all over space.

"Heisenberg sometimes explained the uncertainty principle as a problem of making measurements. His most well-known thought experiment involved photographing an electron. To take the picture, a scientist might bounce a light particle off the electron's surface. That would reveal its position, but it would also impart energy to the electron, causing it to move. Learning about the electron's position would create uncertainty in its velocity; and the act of measurement would produce the uncertainty needed to satisfy the principle." - Scientific American

"Bohr, for his part, explained uncertainty by pointing out that answering certain questions necessitates not answering others. To measure position, we need a stationary measuring object, like a fixed photographic plate. This plate defines a fixed frame of reference. To measure velocity, by contrast, we need an apparatus that allows for some recoil, and hence moveable parts. This experiment requires a movable frame. Testing one therefore means not testing the other." - NYTimes

The expectation value is \[ < x> = (2+4+1+3+3+6+3+1+4+5) \cdot \frac{1}{10} = 3.2. \] An alternative way of computing this is collecting equal results and weighing them by their empirical probability: \[ < x> = \frac{2}{10} \cdot 1 + \frac{1}{10} \cdot 2 + \frac{3}{10} \cdot 3 + \frac{2}{10} \cdot 4 + \frac{1}{10} \cdot 5 + \frac{1}{10} \cdot 6 = 3.2. \]

First, we calculate the projections along the axis defined by the new basis vectors: \begin{align} \vec{e}_1 \cdot \vec{v} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\0 \end{pmatrix} \cdot \begin{pmatrix} 9\\ 1 \\ -2 \end{pmatrix} = \frac{10}{\sqrt{2}} \\ \vec{e}_2 \cdot \vec{v} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\0 \end{pmatrix} \cdot \begin{pmatrix} 9\\ 1 \\ -2 \end{pmatrix} = \frac{8}{\sqrt{2}} \\ \vec{e}_3 \cdot \vec{v} = \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\ 0 \\1 \end{pmatrix} \cdot \begin{pmatrix} 9\\ 1 \\ -2 \end{pmatrix} = -\sqrt{2} \\ \end{align}

This yields the components in the new basis and the vector $ \vec{v}$, therefore, reads in the new basis:

\begin{align} \vec{v} = \begin{pmatrix} \frac{10}{\sqrt{2}} \\ \frac{8}{\sqrt{2}} \\-\sqrt{2} \end{pmatrix} _{\text{new basis}} = \frac{10}{\sqrt{2}} \vec{e}_1 + \frac{8}{\sqrt{2}}\vec{e}_2 -\sqrt{2} \vec{e}_3 \end{align}

The momentum operator is given by $\hat{p}_i = -i \hbar \partial_i$, where $\partial_i \equiv \frac{ \partial}{\partial i}$ and $i \in \{x,y,z\}$.

It makes a difference whether we first measure the momentum and then the location or first the location and then the momentum.

The Schrödinger equation: $$ i \hbar \partial_t |{\Psi} \rangle = -\frac{ \hbar^2 \partial_i^2}{2m} |{\Psi}\rangle + V( \hat x) | {\Psi}\rangle \, .$$

We have orbital angular momentum and spin. While orbital angular momentum describes the rotation of one object around another (think: electron around the proton), spin describes internal angular momentum (think: a spinning ball).

We calculate: $$ \langle p_7 |\Psi\rangle = \frac{1}{\sqrt{30}} \left( \langle p_7 |p_1 \rangle -2 \langle p_7 |p_2 \rangle + 5 \langle p_7 | p_7 \rangle\right) = \frac{5}{\sqrt{30}} \, . $$ This is the corresponding probability amplitude and the probability is therefore: $$ P(p_7) = |\frac{5}{\sqrt{30}} |^2 = \frac{25}{30} = 0.8333 = 83.33 \% $$