# Quantum mechanics - a basic introduction

by Michael Brunnbauer, 2021-03-07

Writing this really helped me understand the subject matter. Understanding even the basics of quantum mechanics is hard and everyone has to find his/her own way into it. I don't claim that this text will suit anyone but with a bit of luck, maybe it will suit you. The general idea is to look at things in the energy basis and stick to it - while avoiding abstract vector space notation as much as possible. At least for me, things are more perceptible this way. I also lay out some fundamentals that might be presupposed in other introductions.

For a professional introduction, check out The Feynman Lectures on Physics, Volume III. In case you lack the basics for Volume III, start with Volume I. Feynman and the other authors really start from zero - a remarkable style I have not seen anywhere else.

I assume that the reader is familiar with trigonometric functions, rules for obtaining derivatives, complex numbers, complex exponentials and has some basic knowledge about vector spaces, integrals and special relativity.

I might occasionally improve this text or add sections. If you spot any error, please send a note to brunni@netestate.de

### Signals in time

Let's look at some audio signals where we have the time t on the x-axis and the amplitude A (e.g. displacement of the diaphragm in a loudspeaker or your eardrum) on the y-axis. Our unit of time will be milliseconds.

$$A(t) = cos(t)$$

What is the period T of the signal? As expected from $$cos(t)$$, it repeats after $$2 \pi$$ milliseconds. The frequency f of the signal is $$\frac{1}{T} = \frac{1}{2 \cdot \pi \cdot 0.001} \approx 160 Hz$$. The angular frequency $$\omega = 2 \cdot \pi \cdot f = \frac{2 \cdot \pi}{T}$$ is 1 radian per millisecond.

What is the time of the signal? When does it happen? Kind of all the time or maybe not at all. The time is not definite. Let's say our uncertainty about the time is infinite.

Here is a another signal:

$$A(t) = 1$$ for $$t=0$$.
$$A(t) = 0$$ else.

The signal clearly "happens" at $$t = 0$$. But what is its frequency? Again we are at a loss as it does not repeat.

Watch what happens when we add 21 cosine waves from -1000 to 1000 Hz:

$$A(t) = \sum\limits_{n=-10}^{10} { cos( 2 \cdot \pi \cdot \frac{n}{10} \cdot t ) }$$ br>
Now we can say something "happens" in the signal ca. every 10 milliseconds and although there are permanent ups and downs, we can clearly see a main period of 10 milliseconds - which corresponds to a frequency of 100 Hz.

While this is mostly about the underlying math, one might ask whether real sound waves really add like this? They do - the acoustic wave equation is linear, so the Superposition principle holds. The acoustic wave equation is only an approximation though and does not hold under extreme conditions. See Nonlinear acoustics.

What would happen when we add $$N + 1$$ cosine waves from -f to +f? The peaks would happen at intervals of $$\frac{N}{2f}$$. They will also be more thin with higher f. It is of course possible to move the time of the peaks by changing the phase P of all cosines, e.g. $$cos( 2 \cdot \pi \cdot \frac{n}{10} \cdot (t+P) )$$ instead of $$cos( 2 \cdot \pi \cdot \frac{n}{10} \cdot t )$$. Let's try this while dividing the result through $$N + 1$$ to keep the peaks at constant height:

$$A(t) = \frac{1}{N+1} \cdot \sum\limits_{n=-N/2}^{N/2} { cos( 2 \cdot \pi \cdot \frac{2 \cdot f \cdot n}{N} \cdot (t+P) ) }$$

The more we increase the number of cosine waves while increasing the frequency range, the more our signal will look like in the previous example - going towards an infinitesimally thin peak at 0. We can say that if we add all possible frequencies, both our uncertainty about the frequency of the signal and our certainty about when it happens are - in a way - infinite.

This uncertainty between the frequency domain and the time domain of a signal can be quantified. If we know f within $$\delta f$$ and t within $$\delta t$$, then $$\delta t \cdot \delta f \ge \frac{1}{ 4 \cdot \pi }$$. This is known as the Gabor limit of signal processing.

It is possible to transform any arbitrary signal in the time domain to the frequency domain: Information about the amplitudes of each sine frequency and its phase. This operation is called the Fourier transform. The process of adding those sine waves together to recover the original signal is called the inverse Fourier transform. As a signal is basically a function, this also applies to functions but they must be well behaved. See Fourier inversion theorem.

### Signals in time and space

The x-axis will now represent space and we add a slider t to see if and how the wave travels. Think of waves travelling in water. The most simple example is: $$A(x,t) = cos(\omega \cdot t + k \cdot x)$$ Does this obey the general Wave equation? $$\frac{d^2 A(x,t)}{d t^2} = c^2 \cdot \frac{d^2 A(x,t)}{d x^2}$$ The second time-derivative needs to be the second space-derivative times a constant. The constant is squared in the equation - so it might have some significance. $$\frac{d A(x,t)}{d t} = - \omega \cdot sin(\omega \cdot t + k \cdot x)$$ $$\frac{d^2 A(x,t)}{d t^2} = - \omega^2 cos(\omega \cdot t + k \cdot x)$$ $$\frac{d A(x,t)}{d x} = -k \cdot sin(\omega \cdot t + k \cdot x)$$ $$\frac{d^2 A(x,t)}{d x^2} = -k^2 cos(\omega \cdot t + k \cdot x)$$ We see that the wave equation holds and that that c must be $$\frac{\omega}{k}$$. So let's play a bit:

$$A(x,t) = cos(\omega \cdot t + k \cdot x)$$

k is called the wavenumber and is related to the wavelength $$\lambda$$ by $$k = \frac{2 \cdot \pi}{\lambda}$$ - as can be verified from the graph above.

It seems the velocity of the peaks is directly proportional to $$\omega$$ and indirectly proportional to k (whose sign determines which direction the wave travels). So the velocity of the peaks - the phase velocity - seems to be proportional to $$\frac{\omega}{k}$$ - our constant c from the wave equation, which turns out to be the phase velocity. It should be 1 units of space per unit of time here with the default settings - try it out!.

Adding two waves with opposite k (travelling direction) results in a standing wave (again we divide the result by the number of waves added):

$$A(x,t) = \frac{1}{2} \cdot ( cos(\omega \cdot t + k \cdot x) + cos(\omega \cdot t - k \cdot x) )$$

That's what happens when waves are being reflected back by some obstruction. When the wave is confined by both sides, we have a resonator which will only generate standing waves that fit exactly between both obstructions (the amplitude has to be zero where the obstructions is).

Adding many waves travelling in the same direction can create a travelling localized disturbance - a wavepacket. The group velocity $$v_g$$ of a wavepacket does not have to be equal to the phase velocity $$v_p$$. It is determined by the derivative of the angular frequency with respect to the wavenumber: $$v_g = \frac{d \omega}{d k}$$ Some examples:

$$\text{ 1) } \omega = a \implies v_p = \frac{a}{k} \text{ and } v_g = 0$$
$$\text{ 2) } \omega = k \cdot a \implies v_p = a \text{ and } v_g = a$$
$$\text{ 3) } \omega = k + a \implies v_p = 1 + \frac{a}{k} \text{ and } v_g = 1$$
$$\text{ 4) } \omega = k^2 \cdot a \implies v_p = k \cdot a \text{ and } v_g = 2 \cdot k \cdot a$$

The first example shows that just varying the wavenumber is not sufficient to create a travelling wavepacket. $$\omega$$ must be a function of k - or the wavepacket will stay. The relation between k and $$\omega$$ is called the dispersion relation. We will see why in a moment.

The second example corresponds to electromagnetic waves. Their frequency is directly proportional to their wavenumber so they all travel at the same speed - the speed of light. And so do wavepackets of light.

The third example results in a group velocity that is lower than the phase velocity.

The fourth example has a non-linear dispersion relation which results in a group velocity that is a function of k. The wavepacket constructed of several k cannot hold together - it will disperse.

Here you can test all four examples with $$a = 1$$.

$$\text{Mode 1: } A(x,t) = \frac{1}{N} \cdot \sum\limits_{n=1}^{N} { cos( t + \frac{k}{N} \cdot x ) }$$
$$\text{Mode 2: } A(x,t) = \frac{1}{N} \cdot \sum\limits_{n=1}^{N} { cos( \frac{k}{N} \cdot t + \frac{k}{N} \cdot x ) }$$
$$\text{Mode 3: } A(x,t) = \frac{1}{N} \cdot \sum\limits_{n=1}^{N} { cos( ( \frac{k}{N} + 1 ) \cdot t + \frac{k}{N} \cdot x ) }$$
$$\text{Mode 4: } A(x,t) = \frac{1}{N} \cdot \sum\limits_{n=1}^{N} { cos( \frac{k^2}{N^2} \cdot t + \frac{k}{N} \cdot x ) }$$

### The Schrödinger equation

Quantum mechanics is about waves of complex amplitude travelling through space. Not real space but configuration space. The complex amplitude is a probability amplitude whose absolute value squared is the probability (or probability density) of finding the system in a certain configuration - or definite state - when measuring. Those waves obey the Schrödinger equation: $$i \cdot \hbar \cdot \frac{d}{dt} \left| \psi(t) \right> = \widehat{H} \left| \psi(t) \right>$$ t is time, i is the imaginary unit and $$\hbar$$ is the reduced Planck constant. The equations says that if you take the derivative of the wave function vector $$\left| \psi(t) \right>$$ with respect to time and multiply with i times $$\hbar$$, you get the same result as when applying the hamiltonian operator $$\widehat{H}$$ to the wave function vector.

The components of those vectors are complex numbers. Now for the start it is easier not to consider the vectors as a whole but only their components and too look at everything in the energy basis of the vector space. For simplicity, you may consider $$\psi(x,t)$$ the $$x_{th}$$ component of the wave vector $$\left| \psi(t) \right>$$. The complex function $$\psi(x,t)$$ can be seen as a vector by plugging in all possible x.

The solution of the Schrödinger equation for a definite energy state looks like this: $$\psi(x,t) = \psi_{E_n}(x) \cdot e^{ -i \cdot E_n \cdot t / \hbar }$$ The time dependent wave function $$\psi(x,t)$$ is expressed in terms of a time independent wave function $$\psi_{E_n}(x)$$.

• $$E_n$$ is the energy of definite energy state n
• The possible values of x represent all the definite (measurably different) states of the system. They are taken from the set called configuration space. This set can be finite (e.g. only two possible outcomes like spin up / spin down), countably infinite or uncountably infinite (e.g. position). x can be a tuple (e.g. x,y,z coordinate for one or several particles, maybe combined with additional parameters like spin). When x is a tuple like x1,y1,z1,x2,y2,z2,..., the wave function is usually written as $$\psi(x1,y1,z1,x2,y2,z2,...,t)$$.
• The value of the wave functions $$\psi(x,t)$$ and $$\psi_{E_n}(x)$$ is a complex number. Its absolute value squared $$|\psi(x,t)|^2$$ and $$|\psi_{E_n}(x)|^2$$ is the probability to find the system in definite state x (at time t for $$\psi(x,t)$$). Note that $$|A|^2 = A \cdot A^*$$ where $$A^*$$ is the complex conjugate of A.
• The vector $$\left| \psi_{E_n} \right>$$ spanned from $$\psi_{E_n}(x)$$ by letting x range over configuration space is a quantum state vector (or short, quantum state) of the system. In this case, it is a definite energy state vector.
• The vector $$\left| \psi(t) \right>$$ spanned from $$\psi(x,t)$$ by holding t fixed and letting x range over configuration space is the quantum state vector of the system at time t.
• As the probability of measuring any possible definite state must be 1: $$\sum_{x} |\psi(x,t)|^2 = 1$$ $$\sum_{x} |\psi_{E_n}(x)|^2 = 1$$ "The quantum state vector has norm 1".
• The n definite energy state vectors form a complete basis of the underlying vector space so they must be orthogonal: $$E_a \neq E_b \implies \sum_{x} \psi_{E_a}(x) \cdot \psi_{E_b}(x)^* = 0$$
• In the case of uncountably infinite configuration space, $$|\psi(x,t)|^2$$ and $$|\psi_{E_n}(x)|^2$$ give a probability density instead of a probability. A probability can be determined by integrating over a region of configuration space. Also the sum for the norm has to be replaced with an integral.
How does the probability of measuring some x change over time? Not at all! $$e^{ -i \cdot E_n \cdot t / \hbar }$$ is a complex number of absolute value 1 whose phase changes with t. Multiplying it with $$\psi_{E_n}(x)$$ will only change the phase of $$\psi_{E_n}(x)$$, not the absolute value. But the absolute value of $$\psi(x,t)$$ squared determines the probability of measuring x at t.

Because of this, two quantum states (or state vectors) are considered equal if they differ only by multiplication with a complex number p of absolute value 1 (all components of the vector multiplied with p yield the other vector).

We can already see that dynamics in time requires us to add several definite energy solutions (a superposition of energies) and that the time vs. frequency uncertainty from signal processing carriers over to time vs. energy here.

In fact, the angular frequency of the phase rotation in our definite energy state is $$\frac{E_n}{\hbar}$$, so the frequency is $$\frac{E_n}{\hbar \cdot 2 \cdot \pi }$$ and $$E_n = f \cdot \hbar \cdot 2 \cdot \pi$$. The time vs. frequency uncertainty is $$\delta t \cdot \delta f \ge \frac{1}{ 4 \cdot \pi }$$ so the time vs. energy uncertainty is $$\delta t \cdot \delta E \ge \frac{\hbar}{2}$$

Does our equation for definite energy satisfy the Schrödinger equation? It does! $$i \cdot \hbar \cdot \frac{d}{dt} \psi(x,t) = i \cdot \hbar \cdot ( -i \cdot E_n \cdot \frac{1}{\hbar} \cdot \psi(x,t) ) = E_n \cdot \psi(x,t)$$ The hamiltonian for this simple case is $$E_n$$.

Because the Schrödinger equation is a linear differential equation, every linear combination $$A_1 \cdot \psi1(x,t) + A_2 \cdot \psi2(x,t)$$ of solutions is again a solution. As the definite energy states form a complete basis we can represent any quantum system by adding together solutions for the definite energy states.

So the general solution of the Schrödinger equation in the basis of definite energy states looks like this: $$\psi(x,t) = \sum_{n} A_n \cdot \psi_{E_n}(x) \cdot e^{ -i \cdot E_n \cdot t / \hbar }$$ $$A_n$$ is the amplitude with which the definite energy solution n contributes. The probability to measure energy n will be $$|A_n|^2$$, so: $$\sum_{x} |A_n|^2 = 1$$ We can also see that what we define as zero energy does not really matter. Suppose we add some constant amount of energy $$\delta E$$ to every $$E_n$$. This amounts to a muliplication with $$e^{ -i \cdot \delta E \cdot t / \hbar }$$ everywhere - the probabilities will not change.

### Systems with two definite states

We will now look at a system with only two definite states and will call them $$\uparrow$$ and $$\downarrow$$. So x can take the value $$\uparrow$$ or $$\downarrow$$.

In quantum physics, the number of possible definite states is equal to the number of energy levels as both form a complete basis for the underlying vector space. Usually, those two bases are different. Our equations are: $$\psi(\uparrow,t) = A_1 \cdot \psi_{E_1}(\uparrow) \cdot e^{ -i \cdot E_1 \cdot t / \hbar } + A_2 \cdot \psi_{E_2}(\uparrow) \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ $$\psi(\downarrow,t) = A_1 \cdot \psi_{E_1}(\downarrow) \cdot e^{ -i \cdot E_1 \cdot t / \hbar } + A_2 \cdot \psi_{E_2}(\downarrow) \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ Can $$\psi_{E_n}(x)$$ be zero? Yes, but the $$\psi_{E_n}(x)$$ vectors for fixed n need to stay orthogonal and have norm 1. The only way to assure it is like this (we ignore the symmetrical solution with 0 and 1 swapped): $$|\psi_{E_1}(\uparrow)| = 1, \psi_{E_1}(\downarrow) = 0$$ $$\psi_{E_2}(\uparrow) = 0, |\psi_{E_2}(\downarrow)| = 1$$ Solution 1: $$|A_1| = 1, |A_2| = 0$$ $$\psi(\uparrow,t) = e^{ -i \cdot E_1 \cdot t / \hbar }$$ $$\psi(\downarrow,t) = 0$$ Solution 2: $$|A_1| = 0, |A_2| = 1$$ $$\psi(\uparrow,t) = 0$$ $$\psi(\downarrow,t) = e^{ -i \cdot E_2 \cdot t / \hbar }$$ The definite energy levels correspond to the definite states. Measuring the state is the same as measuring energy. There is no way to put the system into a superposition of states or energies by measuring something. However - these superpositions are still useful, as we will see later.

The interesting case is $$\psi_{E_n}(x) \ne 0$$. This means: In a definite energy state, the probability for measuring any definite state is never zero. The definite energy states must be a superposition of both definite states. We suspect that by symmetry, each definite state must be a superposition of both definite energy states. Our analysis will show that this is true.

Let's look at $$\psi(\uparrow,t)$$ and its absolute value squared: $$a = \frac{ A_1 \cdot \psi_{E_1}(\uparrow) }{ e^{i \cdot \alpha}}$$ $$b = \frac{ A_2 \cdot \psi_{E_2}(\uparrow) }{ e^{i \cdot \beta}}$$ Where $$\alpha$$ is the overall phase of $$A_1 \cdot \psi_{E_1}(\uparrow)$$ and $$\beta$$ is the overall phase of $$A_2 \cdot \psi_{E_2}(\uparrow)$$ such that a and b are real. $$\psi(\uparrow,t) = A_1 \cdot \psi_{E_1}(\uparrow) \cdot e^{ -i \cdot E_1 \cdot t / \hbar } + A_2 \cdot \psi_{E_2}(\uparrow) \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ $$= a \cdot e^{i \cdot \alpha} \cdot e^{-i \cdot E_1 \cdot t / \hbar } + b \cdot e^{i \cdot \beta} \cdot e^{-i \cdot E_2 \cdot t / \hbar }$$ $$= a \cdot e^{-i \cdot ( E_1 \cdot t / \hbar - \alpha ) } + b \cdot e^{-i \cdot ( E_2 \cdot t / \hbar - \beta ) }$$ $$\psi(\uparrow,t) \cdot \psi(\uparrow,t)^* = ( a \cdot e^{-i \cdot ( E_1 \cdot t / \hbar - \alpha ) } + b \cdot e^{-i \cdot ( E_2 \cdot t / \hbar - \beta ) } ) * ( a \cdot e^{i \cdot ( E_1 \cdot t / \hbar - \alpha ) } + b \cdot e^{i \cdot ( E_2 \cdot t / \hbar - \beta ) } ) =$$ $$a^2 + b^2 + a \cdot b \cdot e^{i \cdot ( t \cdot (E2-E1) / \hbar - \beta + \alpha ) } + a \cdot b \cdot e^{i \cdot ( t \cdot (E1-E2) / \hbar -\alpha + \beta ) } =$$ $$a^2 + b^2 + a \cdot b \cdot ( e^{i \cdot ( t \cdot (E2-E1) / \hbar - \beta + \alpha ) } + e^{-i \cdot ( t \cdot (E2-E1) / \hbar - \beta + \alpha ) } ) =$$ $$a^2 + b^2 + 2 \cdot a \cdot b \cdot \cos (t \cdot ( E2 - E1 ) / \hbar - \beta + \alpha )$$ We can see that the probability of measuring $$\uparrow$$ oscillates with time. Where does the probability go? Of course into and out of $$\psi(\downarrow,t)$$ to preserve the norm of 1. $$c = \frac{ A_1 \cdot \psi_{E_1}(\downarrow) }{ e^{i \cdot \gamma}}$$ $$d = \frac{ A_2 \cdot \psi_{E_2}(\downarrow) }{ e^{i \cdot \delta}}$$ $$\psi(\downarrow,t) = c^2 + d^2 + 2 \cdot c \cdot d \cdot \cos (t \cdot ( E2 - E1 ) / \hbar - \delta + \gamma )$$ So to preserve the norm of 1: $$- \beta + \alpha = - \delta + \gamma$$ $$a \cdot b = - c \cdot d$$ $$a^2 + b^2 + c^2 + d^2 = 1$$ As the phase of $$A_1, A_2, \psi_{E_1}(\uparrow), \psi_{E_1}(\downarrow), \psi_{E_2}(\uparrow), \psi_{E_2}(\downarrow)$$ will only cause a phase shift in the resulting oscillation, we will regard them as real from now on ($$\beta = \alpha = \delta = \gamma = 0$$). Our third requirement then already follows from the general assumptions that $$\psi_{E_1}(\uparrow)^2 + \psi_{E_1}(\downarrow)^2 = 1$$, $$\psi_{E_2}(\uparrow)^2 + \psi_{E_2}(\downarrow)^2 = 1$$ and $$A_1^2 + A_2^2 = 1$$: $$a^2 + b^2 + c^2 + d^2 = A_1^2 \cdot \psi_{E_1}(\uparrow)^2 + A_2^2 \cdot \psi_{E_2}(\uparrow)^2 + A_1^2 \cdot \psi_{E_1}(\downarrow)^2 + A_2^2 \cdot \psi_{E_2}(\downarrow)^2 =$$ $$A_1^2 \cdot ( \psi_{E_1}(\uparrow)^2 + \psi_{E_1}(\downarrow)^2 ) + A_2^2 \cdot ( \psi_{E_2}(\uparrow)^2 + \psi_{E_2}(\downarrow)^2 ) =$$ $$A_1^2 + A_2^2 = 1$$ From $$a \cdot b = - c \cdot d$$, $$A_1 \neq 0$$, $$A_2 \neq 0$$ we get: $$A_1 \cdot \psi_{E_1}(\uparrow) \cdot A_2 \cdot \psi_{E_2}(\uparrow) = - A_1 \cdot \psi_{E_1}(\downarrow) \cdot A_2 \cdot \psi_{E_2}(\downarrow)$$ $$\psi_{E_1}(\uparrow) \cdot \psi_{E_2}(\uparrow) = - \psi_{E_1}(\downarrow) \cdot \psi_{E_2}(\downarrow)$$ If this is valid for $$A_1 \neq 0$$, $$A_2 \neq 0$$, it must also be valid for the special case where one of the A is 0. Taking the square on both sides gives: $$\psi_{E_1}(\uparrow)^2 \cdot \psi_{E_2}(\uparrow)^2 = \psi_{E_1}(\downarrow)^2 \cdot \psi_{E_2}(\downarrow)^2$$ Now because $$\psi_{E_1}(\uparrow)^2 + \psi_{E_1}(\downarrow)^2 = 1$$ and $$\psi_{E_2}(\uparrow)^2 + \psi_{E_2}(\downarrow)^2 = 1$$, we get: $$\psi_{E_1}(\uparrow)^2 \cdot ( 1 - \psi_{E_2}(\downarrow)^2 ) = ( 1 - \psi_{E_1}(\uparrow)^2 ) \cdot \psi_{E_2}(\downarrow)^2$$ $$\psi_{E_1}(\uparrow)^2 - \psi_{E_1}(\uparrow)^2 \cdot \psi_{E_2}(\downarrow)^2 = \psi_{E_2}(\downarrow)^2 - \psi_{E_2}(\downarrow)^2 \cdot \psi_{E_1}(\uparrow)^2$$ $$\psi_{E_1}(\uparrow)^2 = \psi_{E_2}(\downarrow)^2$$ Furthermore: $$\psi_{E_2}(\uparrow)^2 = 1 - \psi_{E_2}(\downarrow)^2 = 1 - \psi_{E_1}(\uparrow)^2 = 1 - ( 1 - \psi_{E_1}(\downarrow)^2 ) = \psi_{E_1}(\downarrow)^2$$ To summarize: $$\psi_{E_1}(\uparrow) \cdot \psi_{E_2}(\uparrow) = - \psi_{E_1}(\downarrow) \cdot \psi_{E_2}(\downarrow)$$ $$\psi_{E_1}(\uparrow)^2 = \psi_{E_2}(\downarrow)^2$$ $$\psi_{E_2}(\uparrow)^2 = \psi_{E_1}(\downarrow)^2$$ We just pick one solution - it does not seem to matter: $$\psi_{E_1}(\uparrow) = -\psi_{E_2}(\downarrow)$$ $$\psi_{E_2}(\uparrow) = \psi_{E_1}(\downarrow)$$ This corresponds with our requirement that the definite energy state vectors have to be orthogonal. So we have our final description of the system: $$A_1^2 + A_2^2 = 1$$ $$\psi_{E_1}(\uparrow)^2 + \psi_{E_1}(\downarrow)^2 = 1$$ $$\psi(\uparrow,t) = A_1 \cdot \psi_{E_1}(\uparrow) \cdot e^{ -i \cdot E_1 \cdot t / \hbar } + A_2 \cdot \psi_{E_1}(\downarrow) \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ $$\psi(\downarrow,t) = A_1 \cdot \psi_{E_1}(\downarrow) \cdot e^{ -i \cdot E_1 \cdot t / \hbar } - A_2 \cdot \psi_{E_1}(\uparrow) \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ $$a := A_1 \cdot \psi_{E_1}(\uparrow)$$ $$b := A_2 \cdot \psi_{E_1}(\downarrow)$$ $$|\psi(\uparrow,t)|^2 = a^2 + b^2 + 2 \cdot a \cdot b \cdot \cos (t \cdot ( E2 - E1 ) / \hbar )$$ $$|\psi(\downarrow,t)|^2 = 1 - ( a^2 + b^2) - 2 \cdot a \cdot b \cdot \cos (t \cdot ( E2 - E1 ) / \hbar )$$ Here is a graph of $$|\psi(\uparrow,t)|^2$$ (red) and $$|\psi(\downarrow,t)|^2$$ (blue) vs time. For simplicity, $$\hbar = 1$$ and $$E_1 = 6$$. You can adjust $$A_1$$, $$E_2$$ and $$\psi_{E_1}(\uparrow)$$ with a slider ($$A_2$$ and $$\psi_{E_1}(\downarrow)$$ will be adjusted accordingly). The default for $$A_1$$ and $$\psi_{E_1}(\uparrow)$$ is $$\frac{1}{\sqrt{2}}$$.

### Ammonia

Let's take a closer look at the equations for the default setting of $$\psi_{E_1}(\uparrow) = \psi_{E_1}(\downarrow) = \frac{1}{\sqrt{2}}$$. This describes a lot of physical systems. One of them is an aspect of the Ammonia molecule, whose nitrogen atom can be above (state $$\uparrow$$) or below (state $$\downarrow$$) the plane formed by the hydrogen atoms. If you measure the nitrogen to be above the plane, you will find it to be below the plane a little time later, which is exactly what the equations describe: $$\psi(\uparrow,t) = A_1 \cdot \frac{1}{\sqrt{2}} \cdot e^{ -i \cdot E_1 \cdot t / \hbar } + A_2 \cdot \frac{1}{\sqrt{2}} \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ $$\psi(\downarrow,t) = A_1 \cdot \frac{1}{\sqrt{2}} \cdot e^{ -i \cdot E_1 \cdot t / \hbar } - A_2 \cdot \frac{1}{\sqrt{2}} \cdot e^{ -i \cdot E_2 \cdot t / \hbar }$$ $$|\psi(\uparrow,t)|^2 = \frac{1}{2} + A_1 \cdot A_2 \cdot \cos (t \cdot ( E_2 - E_1 ) / \hbar )$$ $$|\psi(\downarrow,t)|^2 = \frac{1}{2} - A_1 \cdot A_2 \cdot \cos (t \cdot ( E_2 - E_1 ) / \hbar )$$ For the nitrogen being above the plane is symmetric and not otherwise different from being below the plane, which is reflected in $$\psi_{E_1}(\uparrow) = \psi_{E_1}(\downarrow) = \frac{1}{\sqrt{2}}$$.

Here are some states of this system written as state vector $$(\psi(\uparrow,t),\psi(\downarrow,t))$$:

$$A_2 = 0$$ / definite energy $$E_1: (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$$

$$A_1 = 0$$ / definite energy $$E_2: (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$$
$$A_1 = A_2 = \frac{1}{\sqrt{2}}, t = 0$$ / definite state $$\uparrow$$: $$(1,0)$$
$$A_1 = A_2 = \frac{1}{\sqrt{2}}, t = \frac{\pi \cdot \hbar}{E_1-E_2}$$ / definite state $$\downarrow$$: $$(0,-1)$$

The system has at least four discernible states but if we measure a definite energy, the state will be maximally unclear and the energy will stay definite. If we measure the state, the energy will be maximally unclear and the state will not stay definite - the probabilities for the state oscillate.

Note that adding the $$E_1$$ and $$E_2$$ state and multiplying with $$\frac{1}{\sqrt{2}}$$ yields the $$\uparrow$$ state. Subtracting them and multiplying with $$\frac{1}{\sqrt{2}}$$ yields the $$\downarrow$$ state (if you get flipped signs by subtracting in the "wrong" order remember this is the same state - it just has been multiplied with a phase factor of -1).

Adding the $$\uparrow$$ and $$\downarrow$$ state and multiplying with $$\frac{1}{\sqrt{2}}$$ yields the $$E_2$$ state. Subtracting them and multiplying with $$\frac{1}{\sqrt{2}}$$ yields the $$E_1$$ state.

What about the cases with other values for $$A_1$$ and $$A_2$$? Those are superpositions of the definite energy states with unequal probabilities. It seems such states cannot be prepared by measuring anything. Suppose we had such a state. Once we have measured a definite state, $$A_1$$ and $$A_2$$ are reset to $$\frac{1}{\sqrt{2}}$$. For the ammonia molecule, such "non-prepareable" states do not seem to be relevant but they definitely are for other systems.

When ammonia is subjected to an electric field, the nitrogen above and nitrogen below states are no longer symmetrical because the molecule has a dipole moment perpendicular to the plane of the hydrogen atoms. This reflects in $$\psi_{E_1}(\uparrow) \neq \psi_{E_1}(\downarrow)$$. For example set $$\psi_{E_1}(\uparrow)$$ to 0.90. You will find that in the two definite energy states $$A_1 = 1$$ or $$A_1 = 0$$, the nitrogen is more likely to be found up or down. A beam of ammonia shot into an electric field will split into two beams corresponding to the two definite energies. The electric field measures the energy.

Two-state systems can switch between the definite energy levels by emitting or absorbing a photon that has the same energy as the difference between the energy levels. The lower energy state plus photon has higher entropy, which is why those systems tend to be in the lower energy state. This explains chemical bonding (the electron neigher belongs to atom A nor to atom B, but forms the lower energy superposition of both), nuclear force and many other phenomena.

### Electron spin

We will now learn how those "unprepareable" states with nonzero but unequal amplitudes can be useful. Both the definite energies and the definite states form a complete basis for the vector space of a two-state system. In the most simple case, they are identical. But there are of course many other possible bases. In fact every possible base corresponds to two opposite points on a unit sphere - the so called Bloch sphere. Now there are quantum systems - like electron spin - which can be characterized by two states plus a direction and it turns out they can be described by a simple two-state system with changes of basis (which correspond to a three-dimensional direction).

Our definite energy basis is given by the state vectors $$\left| \psi_{E_1} \right>$$ and $$\left| \psi_{E_2} \right>$$. Let us look at a general superposition of those states: $$|A_1|^2 + |A_2|^2 = 1$$ $$\left| \psi \right> = A_1 \cdot \left| \psi_{E_1} \right> + A_2 \cdot \left| \psi_{E_2} \right>$$ $$A_1$$ and $$A_2$$ are assumed to be complex for full generality. $$A_1 = a_1 \cdot e^{i \cdot \alpha_1}$$ $$A_2 = a_2 \cdot e^{i \cdot \alpha_2}$$ $$\left| \psi \right> = a_1 \cdot e^{i \cdot \alpha_1} \cdot \left| \psi_{E_1} \right> + a_2 \cdot e^{i \cdot \alpha_2} \cdot \left| \psi_{E_2} \right>$$ We can multiply with $$e^{-i \cdot \alpha_1}$$ and still have the same state: $$\left| \psi \right> = a_1 \cdot \left| \psi_{E_1} \right> + a_2 \cdot e^{i \cdot ( \alpha_2 - \alpha_1 ) } \cdot \left| \psi_{E_2} \right>$$ The splitting of the real amplitudes can be represented with an angle $$\theta$$ - or some multiple of $$\theta$$. For reasons which will become clear soon, we use $$\frac{\theta}{2}$$. The angle $$\alpha_2 - \alpha_1$$ is renamed to $$\varphi$$. $$a_1^2 + a_2^2 = 1$$ $$cos( \frac{\theta}{2} )^2 + sin( \frac{\theta}{2} )^2 = 1$$ $$\left| \psi \right> = cos( \frac{\theta}{2} ) \cdot \left| \psi_{E_1} \right> + sin( \frac{\theta}{2} ) \cdot e^{i \cdot \varphi } \cdot \left| \psi_{E_2} \right>$$ So all possible states can be represented by choosing a basis and supplying two angles. These angles can be seen as polar coordinates on a unit sphere:

Source: Wikimedia

$$\theta$$ ranges from 0 to $$\pi$$ and $$\varphi$$ ranges from 0 to $$2 \cdot \pi$$. If we set $$\theta = 0$$, we get $$\left| \psi_{E_1} \right>$$ - named $$\left| 0 \right>$$ in the picture. If we set $$\theta = \pi$$, we get $$\left| \psi_{E_2} \right>$$ on the opposite end - named $$\left| 1 \right>$$ in the picture.

We can always reach the opposite end of a point on the sphere by adding $$\pi$$ to $$\varphi$$ and subtracting $$\theta$$ from $$\pi$$: $$\left| \psi_1 \right> = cos( \frac{\pi - \theta}{2} ) \cdot \left| \psi_{E_1} \right> + sin( \frac{\pi - \theta}{2} ) \cdot e^{i \cdot (\varphi + \pi ) } \cdot \left| \psi_{E_2} \right> =$$ $$= sin( \frac{\theta}{2} ) \cdot \left| \psi_{E_1} \right> - cos( \frac{\theta}{2} ) \cdot e^{i \cdot \varphi } \cdot \left| \psi_{E_2} \right> =$$ $$= a_2 \cdot \left| \psi_{E_1} \right> - a_1 \cdot e^{i \cdot (\alpha_2 - \alpha_1)} \cdot \left| \psi_{E_2} \right>$$ We can multiply with $$e^{-i \cdot \alpha2}$$ without changing the state: $$\left| \psi_1 \right> = a_2 \cdot e^{-i \cdot \alpha_2} \cdot \left| \psi_{E_1} \right> - a_1 \cdot e^{-i \cdot \alpha_1} \cdot \left| \psi_{E_2} \right> =$$ $$A_2^* \cdot \left| \psi_{E_1} \right> - A_1^* \cdot \left| \psi_{E_2} \right>$$ This state $$\left| \psi_1 \right>$$ will alway be orthogonal to $$\left| \psi \right>$$. $$\left| \psi \right> = A_1 \cdot \left| \psi_{E_1} \right> + A_2 \cdot \left| \psi_{E_2} \right>$$ $$\left| \psi_1 \right> = A_2^* \cdot \left| \psi_{E_1} \right> - A_1^* \cdot \left| \psi_{E_2} \right>$$ $$\sum_{x} \psi(x) \cdot \psi_1(x)^* =$$ $$\sum_{x} ( A_1 \cdot \psi_{E_1}(x) + A_2 \cdot \psi_{E_2}(x) ) \cdot ( A_2^* \cdot \psi_{E_1}(x) - A_1^* \cdot \psi_{E_2}(x) )^* =$$ $$\sum_{x} ( A_1 \cdot \psi_{E_1}(x) + A_2 \cdot \psi_{E_2}(x) ) \cdot ( A_2 \cdot \psi_{E_1}(x)^* - A_1 \cdot \psi_{E_2}(x)^* ) =$$ $$A_1 A_2 \sum_{x} \psi_{E_1}(x) \psi_{E_1}(x)^* - A_1^2 \sum_{x} \psi_{E_1}(x) \psi_{E_2}(x)^* + A_2^2 \sum_{x} \psi_{E_2}(x) \psi_{E_1}(x)^* - A_2 A_1 \sum_{x} \psi_{E_2}(x) \psi_{E_2}(x)^* =$$ $$A_1 \cdot A_2 - A_2 \cdot A_1 = 0$$ The absolute values of the probability amplitudes of $$\left| \psi \right>$$ in the energy basis are $$cos( \frac{\theta}{2} )$$ and $$sin( \frac{\theta}{2} )$$. Imagine the line connecting the two opposite points of the basis. Now imagine a line perpendicular to that one leading to the point $$\left| \psi \right>$$. The line connecting the two opposite points of the basis is split into to sections and the length of those sections determines the ratio of the probabilities for the two states of definite energy. $$cos( \frac{\theta}{2} )^2 = \frac{1}{2} \cdot ( 1 + cos( \theta ) )$$ $$sin( \frac{\theta}{2} )^2 = \frac{1}{2} \cdot ( 1 - cos( \theta ) )$$ Of course, this works in any basis. Let's try the state basis of spin up and spin down for the electron. If there is no magnetic field, it falls together with the energy basis and we get that very boring simple two-state system where everything always stays the same. But spin also has a direction. Suppose we know the system to be in the spin up or $$\left| 0 \right>$$ state along the direction indicated by $$\left| \psi \right>$$ in our picture. Then the probabilities of measuring spin up or down in the z direction are given by the equations above. If spin is prepared along any of the three axes x, y or z, then measuring along one of the other two axes will give equal probabilities for spin up or spin down.

### Position and momentum

x now becomes a continuous variable for the position - e.g. of a particle. The state vectors can then be thought of having uncountably infinite dimension - one dimension for each possible particle position.

Suppose we knew the exact rest mass $$m_0$$ and momentum p of a particle. That means we can calculate its exact energy - we must be in a stationary state of definite energy. We are justified to talk about a time-independent wave function $$\psi_p(x)$$ instead of $$\psi_{E_n}(x)$$. Now if we were able to somehow locate the particle with $$\psi_p(x)$$, it would have to move with $$p > 0$$. But $$\psi_p(x)$$ is independent of time, so the position of the particle must be completely indeterminate and we might as well set $$\psi_0(x)$$ to a constant. Now the state vector has to have norm 1: $$\int\limits_{-\infty}^{+\infty} |\psi_0(x)|^2 \, d x = 1$$ We cannot integrate over a constant from $$-\infty$$ to $$+\infty$$ and get 1. To circumvent this problem, we will for now just consider space as continuous but finite - with x ranging from -0.5 to 0.5: $$\int\limits_{-0.5}^{0.5} |\psi_0(x)|^2 \, d x = 1$$ $$\psi_0(x) = 1$$ So for a particle of mass $$m_0$$ at rest with energy $$E_0 = m_0 \cdot c^2$$, we get: $$\psi(x,t) = e^{-i \cdot E_0 \cdot t / \hbar }$$ Let's see how this looks from a reference frame with constant velocity v. The particle should then have a different energy (rest mass + kinetic energy). With a Galilean transformation, $$t = t^\prime$$. And x does not occur on the right side, so we get the same equation again indicating the same energy. That's not what we wanted. With a Lorentz transformation, everything comes out right: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$t = \gamma \cdot (t^\prime - x^\prime \cdot v / c^2 )$$ $$\psi(x^\prime,t^\prime) = e^{-i \cdot E_0 \cdot \gamma \cdot (t^\prime - x^\prime \cdot v / c^2 ) / \hbar } = e^{ ( -i / \hbar ) \cdot (E_0 \cdot \gamma \cdot t^\prime - E_0 \cdot \gamma \cdot v \cdot x^\prime / c^2 ) }$$ Now $$E_0 \cdot \gamma$$ is the correct relativistic energy $$E_p$$ of the particle (containing the kinetic energy) and $$E_0 \cdot \gamma / c^2$$ is its relativistic mass m. Multiplying it with v yields the relativistic momentum p. $$\psi(x^\prime,t^\prime) = e^{ ( -i / \hbar ) \cdot (E_p \cdot t^\prime - p \cdot x^\prime) } = e^{i \cdot p \cdot x^\prime / \hbar} \cdot e^{-i \cdot E_p \cdot t^\prime / \hbar }$$ So we consider: $$\psi_p(x) = e^{i \cdot p \cdot x / \hbar}$$ $$\psi_p(x,t) = \psi_p(x) \cdot e^{-i \cdot E_p \cdot t / \hbar } = e^{ ( -i / \hbar ) \cdot (E_p \cdot t - p \cdot x) }$$ This is a complex wave in time and space. Its absolute value is 1 everywhere but the complex phase changes with time and space. We call it a matter wave or de Broglie wave after physicist Louis de Broglie.

The definite energy state vectors formed by $$\psi_p(x)$$ need to be orthogonal: $$p_1 \neq p_2 \implies \int\limits_{-0.5}^{0.5} \psi_{p_1}(x) \cdot \psi_{p_2}(x)^* \, d x = 0$$ It turns out this is not the case: $$\int\limits_{-0.5}^{0.5} \psi_{p_1}(x) \cdot \psi_{p_2}(x)^* \, d x = \int\limits_{-0.5}^{0.5} e^{i \cdot p_1 \cdot x / \hbar} \cdot e^{-i \cdot p_2 \cdot x / \hbar} \, d x$$ $$\int\limits_{-0.5}^{0.5} e^{i \cdot x \cdot (p_1 - p_2)/ \hbar } \, dx = 0 \text{ if and only if } e^{i \cdot 0.5 \cdot ( p_1 - p_2 ) / \hbar } - e^{-i \cdot 0.5 \cdot ( p_1 - p_2 ) / \hbar } = 0$$ We can save ourselves with a boundary condition: The waves at $$t = 0$$ have to fit exactly into our universe from -0.5 to +0.5 such that each wave starts and ends at the boundaries with a phase of 0. Note that this is different from the particle in a box boundary condition where the amplitude is required to be 0 at the boundaries. So our possible p will be integer multiples of $$2 \cdot \pi \cdot \hbar$$. $$p_1 = n_1 \cdot 2 \cdot \pi \cdot \hbar$$ $$p_2 = n_2 \cdot 2 \cdot \pi \cdot \hbar$$ $$p_1 - p_2 = (n_1 - n_2) \cdot 2 \cdot \pi \cdot \hbar = n_{12} \cdot 2 \cdot \pi \cdot \hbar$$ $$e^{i \cdot 0.5 \cdot ( p_1 - p_2 ) / \hbar } - e^{-i \cdot 0.5 \cdot ( p_1 - p_2 ) / \hbar } = e^{i \cdot \pi \cdot n_{12} } - e^{-i \cdot \pi \cdot n_{12} } = 0$$ In real quantum mechanics, momentum is continuous and space goes from $$-\infty$$ to $$+\infty$$ - in which case the integral turns out fine without any boundary condition. However, I don't have the knowledge to present the fine points of the math involved. The matter wave is an idealization anyway. Real wave functions are well behaved in the sense that they drop off towards infinity.

What's the phase velocity of the matter wave? $$v_p = \frac{\omega}{k} = \frac{E_p \cdot \hbar}{p \cdot \hbar} = \frac{m \cdot c^2}{m \cdot v} = \frac{c^2}{v}$$ The phase velocity is faster than light for particles with rest mass. As the phase of a wave cannot transmit information, this does not contradict relativity. The phase velocity approaches the speed of light as v goes towards c. For photons, the phase velocity is the speed of light.

It is possible to create localized particles with a wavepacket consisting of a superposition of several matter waves. At $$t = 0$$, we would have a superposition of several $$e^{i \cdot p \cdot x / \hbar }$$. The spacial frequency is $$\frac{p}{\hbar \cdot 2 \cdot \pi}$$. So analogous to the time vs. frequency uncertainty $$\delta t \cdot \delta f \ge \frac{1}{4 \cdot \pi}$$, we conclude that the position vs. momentum uncertainty is $$\delta x \cdot \delta p \ge \frac{h}{2}$$.

The relativistic energy $$E_p$$ can be written as $$\sqrt{ p^2 \cdot c^2 + m_0^2 \cdot c^4 }$$ so the dispersion relation is: $$\omega = E_p / \hbar = \sqrt{ p^2 \cdot c^2 + m_0^2 \cdot c^4 } / \hbar = \sqrt{ \hbar^2 \cdot k^2 \cdot c^2 + m_0^2 \cdot c^4} / \hbar$$ The group velocity of a wavepacket is: $$v_g = \frac{d \omega}{d k} = \frac{\hbar \cdot k \cdot c^2}{\sqrt{ p^2 \cdot c^2 + m_0^2 \cdot c^4 }} = \frac{p \cdot c^2}{E_p} = \frac{m \cdot v \cdot c^2}{m \cdot c^2} = v$$ As v is a function of the momentum p, the group velocity is a function of the momentum. This means that a wavepacket of matter waves cannot hold itself together (unless v stays constant with varying momentum as is the case with photons). It will disperse like our real-valued example from the section "Signals in time and space" and the particle it represents will become less localized over time.

When creating wavepackets by adding matter waves of definite momentum, the norm of 1 for the states is automatically kept if the amplitudes reflect the norm: $$A_1^2 + A_2^2 = 1$$ $$p_1 \neq p_2$$ $$\psi(x,t) = A_1 \cdot \psi_{p_1}(x,t) + A_2 \cdot \psi_{p_2}(x,t)$$ $$\int\limits_{-0.5}^{0.5} |\psi(x,t)|^2 \, d x = \int\limits_{-0.5}^{0.5} \psi(x,t) \cdot \psi(x,t)^* \, d x =$$ $$\int\limits_{-0.5}^{0.5} ( A_1 \cdot \psi_{p_1}(x) \cdot e^{-i \cdot E_{p_1} \cdot t / \hbar } + A_2 \cdot \psi_{p_2}(x) \cdot e^{-i \cdot E_{p_2} \cdot t / \hbar } ) \cdot ( A_1 \cdot \psi_{p_1}(x)^* \cdot e^{i \cdot E_{p_1} \cdot t / \hbar } + A_2 \cdot \psi_{p_2}(x)^* \cdot e^{i \cdot E_{p_2} \cdot t / \hbar } ) \, d x =$$ $$\int\limits_{-0.5}^{0.5} A_1^2 \cdot |\psi_{p_1}(x)|^2 \, d x + \int\limits_{-0.5}^{0.5} A_2^2 \cdot |\psi_{p_2}(x)|^2 \, d x = A_1^2 + A_2^2 = 1$$

In the following visualization you can add N electron matter waves with definite velocities v from -500 to 500 meters per second. Each wave contributes with amplitude $$\frac{1}{\sqrt{N}}$$ to keep the norm of 1. The unit of space on the x-axis is micrometers and the unit of time is nanoseconds. The y-axis is the absolute value of the probability amplitude $$|\psi(x,t)|$$ (the probability density is this value squared). The boundary condition of $$p = n \cdot 2 \cdot \pi \cdot \hbar$$ with integer n is respected.