<![CDATA[Quantum Superposition]]>

<![CDATA[Quantum Superposition]]>guest post by Piotr Migdał

In this blog post I will introduce some basics of quantum mechanics, with the emphasis on why a particle being in a few places at once behaves measurably differently from a particle whose position we just don’t know. It’s a kind of continuation of the “Quantum Network Theory” series (Part 1, Part 2) by Tomi Johnson about our work in Jake Biamonte’s group at the ISI Foundation in Turin. My goal is to explain quantum community detection. Before that, I need to introduce the relevant basics of quantum mechanics, and of the classical community detection.

But before I start, let me introduce myself, as it’s my first post to Azimuth.

I just finished my quantum optics theory Ph.D in Maciej Lewenstein’s group at The Institute of Photonic Sciences in Castelldefels, a beach near Barcelona. My scientific interests range from quantum physics, through complex networks, to data-driven approach…. to pretty much anything—and now I work as a data science freelancer. I enjoy doing data visualizations (for example of relations between topics in mathematics), I am a big fan of Rényi entropy (largely thanks to Azimuth), and I’m a believer in open science. If you think that there are too many off-topic side projects here, you are absolutely right!

In my opinion, quantum mechanics is easy. Based on my gifted education experience it takes roughly 9 intense hours to introduce entanglement to students having only a very basic linear algebra background. Even more, I believe that it is possible to get familiar with quantum mechanics by just playing with it—so I am developing a Quantum Game!

Quantum weirdness

In quantum mechanics a particle can be in a few places at once. It sounds strange. So strange, that some pioneers of quantum mechanics (including, famously, Albert Einstein) didn’t want to believe in it: not because of any disagreement with experiment, not because of any lack of mathematical beauty, just because it didn’t fit their philosophical view of physics.

It went further: in the Soviet Union the idea that electron can be in many places (resonance bonds) was considered to oppose materialism. Later, in California, hippies investigated quantum mechanics as a basis for parapsychology—which, arguably, gave birth to the field of quantum information.

As Griffiths put it in his Introduction to Quantum Mechanics (Chapter 4.4.1):

To the layman, the philosopher, or the classical physicist, a statement of the form “this particle doesn’t have a well-defined position” [...] sounds vague, incompetent, or (worst of all) profound. It is none of these.

In this guest blog post I will try to show that not only can a particle be in many places at once, but also that if it were not in many places at once then it would cause problems. That is, as fundamental phenomena as atoms forming chemical bonds, or particle moving in the vacuum, require it.

As in many other cases, the simplest non-trivial case is perfect for explaining idea, as it covers the most important phenomena, while being easy to analyze, visualize and comprehend. Quantum mechanics is not an exception—let us start with a system of two states.

A two state system

Let us study a simplified model of the hydrogen molecular ion $\mathrm{H}_2^+$ , that is, a system of two protons and one electron (see Feynman Lectures on Physics, Vol. III, Chapter 10.1). Since the protons are heavy and slow, we treat them as fixed. We focus on the electron moving in the electric field created by protons.

In quantum mechanics we describe the state of a system using a complex vector. In simple terms, this is a list of complex numbers called ‘probability amplitudes’. For an electron that can be near one proton or another, we use a list of two numbers:

$|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$

In this state the electron is near the first proton with probability $|\alpha|^2,$ and near the second one with probability $|\beta|^2.$

Note that

$|\psi \rangle = \alpha \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \beta \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

So, we say the electron is in a ‘linear combination’ or ‘superposition’ of the two states

$|1\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

(where it’s near the first proton) and the state

$|2\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

(where it’s near the second proton).

Why do we denote unit vectors in strange brackets looking like

$| \mathrm{something} \rangle$ ?

Well, this is called Dirac notation (or bra-ket notation) and it is immensely useful in quantum mechanics. We won’t go into it in detail here; merely note that $| \cdot \rangle$ stands for a column vector and $\langle \cdot |$ stands for a row vector, while $\psi$ is a traditional symbol for a quantum state.).

Amplitudes can be thought as ‘square roots’ of probabilities. We can force an electron to localize by performing a classical measurement, for example by moving protons away and measuring which of them has neutral charge (for being coupled with the electron). Then, we get probability $| \alpha|^2$ of finding it near the first proton and $|\beta|^2$ of finding it near the second. So, we require that

$|\alpha|^2 + |\beta|^2 = 1$

Note that as amplitudes are complex, for a given probability there are many possible amplitudes. For example

$1 = |1|^2 = |-1|^2 = |i|^2 = \left| \tfrac{1+i}{\sqrt{2}} \right|^2 = \cdots$

where $i$ is the imaginary unit, with $i^2 = -1.$

We will now show that the electron ‘wants’ to be spread out. Electrons don’t really have desires, so this is physics slang for saying that the electron will have less energy if its probability of being near the first proton is equal to its probability of being near the second proton: namely, 50%.

In quantum mechanics, a Hamiltonian is a matrix that describes the relation between the energy and evolution (i.e. how the state changes in time). The expected value of the energy of any state $| \psi \rangle$ is

$E = \langle \psi | H | \psi \rangle$

Here the row vector $\langle \psi |$ is the column vector $| \psi\rangle$ after transposition and complex conjugation (i.e. changing $i$ to $-i$ ), and

$\langle \psi | H | \psi \rangle$

means we are doing matrix multiplication on $\langle \psi |,$ $H$ and $| \psi \rangle$ to get a number.

For the electron in the $\mathrm{H}_2^+$ molecule the Hamiltonian can be written as the following $2 \times 2$ matrix with real, positive entries:

$H = \begin{bmatrix} E_0 & \Delta \\ \Delta & E_0 \end{bmatrix},$

where $E_0$ is the energy of the electron being either in state $|1\rangle$ or state $|2\rangle$ , and $\Delta$ is the ‘tunneling amplitude’, which describes how easy it is for the electron to move from neighborhood of one proton to that of the other.

The expected value—physicists call it the ‘expectation value’—of the energy of a given state $|\psi\rangle$ is:

$E = \langle \psi | H | \psi \rangle \equiv \begin{bmatrix} \alpha^* & \beta^* \end{bmatrix} \begin{bmatrix} E_0 & \Delta \\ \Delta & E_0 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix}.$

The star symbol denotes the complex conjugation. If you are unfamiliar with complex numbers, just work with real numbers on which this operation does nothing.

Exercise 1. Find $\alpha$ and $\beta$ with

$|\alpha|^2 + |\beta|^2 = 1$

that minimize or maximize the expectation value of energy $\langle \psi | H | \psi \rangle$ for

$|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$

Exercise 2. What’s the expectation value value of the energy for the states $| 1 \rangle$ and $| 2 \rangle$ ?

Or if you are lazy, just read the answer! It is straightforward to check that

$E = (\alpha^* \alpha + \beta^* \beta) E_0 + (\alpha^* \beta + \beta^* \alpha) \Delta$

The coefficient of $E_0$ is 1, so the minimal energy is $E_0 - \Delta$ and the maximal energy is $E_0 + \Delta$ . The states achieving these energies are spread out:

$| \psi_- \rangle = \begin{bmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix}, \quad \text{with} \quad \quad E = E_0 - \Delta$

and

$| \psi_+ \rangle = \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}, \quad \text{with} \quad \quad E = E_0 + \Delta$

The energies of these states are below and above the energy $E_0,$ and $\Delta$ says how much.

So, the electron is ‘happier’ (electrons don’t have moods either) to be in the state $|\psi_-\rangle$ than to be localized near only one of the protons. In other words—and this is Chemistry 101—atoms like to share electrons and it bonds them. Also, they like to share electrons in a particular and symmetric way.

For reference, $|\psi_+ \rangle$ is called ‘antibonding state’. If the electron is in this state, the atoms will get repelled from each other—and so much for the molecule!

How to classically add quantum things

How can we tell a difference between an electron being in a superposition between two states, and just not knowing its ‘real’ position? Well, first we need to devise a way to describe probabilistic mixtures.

It looks simple—if we have an electron in the state $|1\rangle$ or $|2\rangle$ with probabilities $1/2$ , we may be tempted to write

$|\psi\rangle = \tfrac{1}{\sqrt{2}} |1\rangle + \tfrac{1}{\sqrt{2}} |2\rangle$

We’re getting the right probabilities, so it looks legit. But there is something strange about the energy. We have obtained the state $|\psi_+\rangle$ with energy $E_0+\Delta$ by mixing two states with the energy $E_0$ !

Moreover, we could have used different amplitudes such that $|\alpha|^2=|\beta|^2=1/2$ and gotten different energies. So, we need to devise a way to avoid guessing amplitudes. All in all, we used quotation marks for ‘square roots’ for a reason!

It turns out that to describe statistical mixtures we can use density matrices.

The states we’d been looking at before are described by vectors like this:

$| \psi \rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$

These are called ‘pure states’. For a pure state, here is how we create a density matrix:

$\rho = | \psi \rangle \langle \psi | \equiv \begin{bmatrix} \alpha \alpha^* & \alpha \beta^*\\ \beta \alpha^* & \beta \beta^* \end{bmatrix}$

$\rho = \begin{bmatrix} 1/2 & -1/2\\ -1/2 & 1/2 \end{bmatrix}$

For an electron in the state $|1\rangle$ we get

$\rho = \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}.$

To calculate the energy, the recipe is the following:

$E = \mathrm{tr}[H \rho]$

where $\mathrm{tr}$ is the ‘trace‘: the sum of the diagonal entries. For a $n \times n$ square matrix with entries $A_{ij}$ its trace is

$\mathrm{tr}(A) = A_{11} + A_{22} + \ldots + A_{nn}$

Exercise 3. Show that this formula for energy, and the previous one, give the same result on pure states.

I advertised that density matrices allow us to mix quantum states. How do they do that? Very simple: just by adding density matrices, multiplied by the respective probabilities:

$\rho = p_1 \rho_1 + p_2 \rho_2 + \cdots + p_n \rho_n$

It is exactly how we would mix probability vectors. Indeed, the diagonals are probability vectors!

So, let’s say that our co-worker was drunk and we are not sure if (s)he said that the state is $|\psi_-\rangle$ or $|1\rangle$ . However, we think that the probabilities are $1/3$ and $2/3.$ We get the density matrix:

$\rho = \begin{bmatrix} 5/6 & -1/6\\ -1/6 & 1/6 \end{bmatrix}$

So, how about its energy?

Exercise 4. Show that calculating energy using density matrix gives the same result as averaging energy over component pure states.

I may have given the impression that density matrix is an artificial thing, at best—a practical trick, and what we ‘really’ have are pure states (vectors), each with a given probability. If so, the next exercise is for you:

Exercise 5. Show that a 50%-50% mixture of $|1\rangle$ and $|2\rangle$ is the same as a 50%-50% mixture of $|\psi_+\rangle$ and $|\psi_-\rangle$ .

This is different than statistical mechanics, or statistics, where we can always think about probability distributions as uniquely defined statistical mixtures of possible states. Here, as we see, it can be a bit more tricky.

As we said, for the diagonals things work as for classical probabilities. But there is more—at the same time as adding probabilities we also add the off-diagonal terms, which can add up to cancel, depending on their signs. It’s why it’s mixing quantum states may make them losing their quantum properties.

The value of the off-diagonal term is related to so-called ‘coherence’ between the states $|1\rangle$ and $|2\rangle$ . Its value is bounded by the respective probabilities:

$\left| \rho_{12} \right| \leq \sqrt{\rho_{11}\rho_{22}} = \sqrt{p_1 p_2}$

where for pure states we get equality.

If the value is zero, there are no quantum effects between two positions: this means that the electron is sure to be at one place or the other, though we might be uncertain at which place. This is fundamentally different from a superposition (non-zero $\rho_{12}$ ), where we are uncertain at which site a particle is, but it can no longer be thought to be at one site or the other: it must be in some way associated with both simultaneously.

Exercise 6. For each $c \in [-1,1]$ propose how to obtain a mixed state described by density matrix

$\rho = \begin{bmatrix} 1/2 & c/2\\ c/2 & 1/2 \end{bmatrix}$

by mixing pure states of your choice.

A spatial wavefunction

A similar thing works for position. Instead of a two-level system let’s take a particle in one dimension. The analogue of a state vector is a wavefunction, a complex-valued function on a line:

$\psi(x)$

In this continuous variant, $p(x) = |\psi(x)|^2$ is the probability density of finding particle in one place.

We construct the density matrix (or rather: ‘density operator’) in an way that is analogous to what we did for the two-level system:

$\rho(x, x') = \psi(x) \psi^*(x')$

Instead of a 2×2 matrix matrix, it is a complex function of two real variables. The probability density can be described by its diagonal values, i.e.

$p(x) = \rho(x,x)$

Again, we may wonder if the particle energetically favors being in many places at once. Well, it does.

Density matrices for a classical and quantum state. They yield the same probability distributions (for positions). However, their off-diagonal values (i.e. $x\neq x’$) are different. The classical state is just a probabilistic mixture of a particle being in a particular place.

What would happen if we had a mixture of perfectly localized particles? Due to Heisenberg’s uncertainly principle we have

$\Delta x \Delta p \geq \frac{\hbar}{2},$

that is, that the product of standard deviations of position and momentum is at least some value.

If we exactly know the position, then the uncertainty of momentum goes to infinity. (The same thing holds if we don’t know position, but it can be known, even in principle. Quantum mechanics couldn’t care less if the particle’s position is known by us, by our friend, by our detector or by a dust particle.)

The Hamiltonian represents energy, the energy of a free particle in continuous system is

$H=p^2/(2m)$

where $m$ is its mass, and $p$ is its momentum: that is, mass times velocity. So, if the particle is completely localized:

• its energy is infinite,
• its velocity are infinite, so in no time its wavefunction will spread everywhere.

Infinite energies sometimes happen if physics. But if we get infinite velocities we see that there is something wrong. So a particle needs to be spread out, or ‘delocalized’, to some degree, to have finite energy.

As a side note, to consider high energies we would need to employ special relativity. In fact, one cannot localize a massive particle that much, as it will create a soup of particles and antiparticles, once its energy related to momentum uncertainty is as much as the energy related to its mass; see the Darwin term in the fine structure.

Moreover, depending on the degree of its delocalization its behavior is different. For example, a statistical mixture of highly localized particles would spread a lot faster than the same $p(x)$ but derived from a single wavefunction. The density matrix of the former would be in between of that of pure state (a ‘circular’ Gaussian function) and the classical state (a ‘linear’ Gaussian). That is, it would be an ‘oval’ Gaussian, with off-diagonal values being smaller than for the pure state.

Let us look at two Gaussian wavefunctions, with varying level of coherent superposition between them. That is, each Gaussian is already a superposition, but when we combine two we let ourselves use a superposition, or a mixture, or something in between. For a perfect superposition of Gaussian, we would have the density matrix

$\rho(x,x') = \frac{1}{2} \left( \phi(x+\tfrac{d}{2}) + \phi(x-\tfrac{d}{2}) \right) \left( \phi(x'+\tfrac{d}{2}) + \phi(x'-\tfrac{d}{2}) \right)$

where $\phi(x)$ is a normalized Gaussian function. For a statistical mixture between these Gaussians split by a distance of $d,$ we would have:

$\rho(x,x') = \frac{1}{2} \phi(x+\tfrac{d}{2}) \phi(x'+\tfrac{d}{2}) + \frac{1}{2} \phi(x-\tfrac{d}{2}) \phi(x'-\tfrac{d}{2})$

And in general,

$\begin{array}{ccl} \rho(x,x') &=& \frac{1}{2} \left( \phi(x+\tfrac{d}{2}) \phi(x'+\tfrac{d}{2}) + \phi(x-\tfrac{d}{2}) \phi(x'-\tfrac{d}{2})\right) + \\ \\ && \frac{c}{2} \left( \phi(x+\tfrac{d}{2}) \phi(x'-\tfrac{d}{2}) + \phi(x-\tfrac{d}{2}) \phi(x'+\tfrac{d}{2}) \right) \end{array}$

for some $|c| \leq 1.$

PICTURE: Two Gaussian wavefunctions (centered at -2 and +2) in a coherent superposition with each other (the first and the last plot) and a statistical mixture (the middle plot); the 2nd and 4th plot show intermediary states. Superposition can be with different phase, much like the hydrogen example. Color represents absolute value and hue phase; here red is for positive numbers and teal is for negative.

(Click to enlarge.)

Conclusion

We have seen learnt the difference between the quantum superposition and the statistical mixture of states. In particular, while both of these descriptions may give the same probabilities, their predictions on the physical properties of states differ. For example, we need an electron to be delocalized in a specific way to describe chemical bonds; and we need delocalization of any particle to predict its movement.

We used density matrices to express both quantum superposition and (classical) lack of knowledge on the same ground. We have identified its off-diagonal terms as ones related to the quantum coherence.

But what if there were not only two states, but many? So, instead of $\mathrm{H}_2^+$ (we were not even considering the full hydrogen atom, but only its ionized version), how about electric excitation on something bigger? Not even $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ or some sugar, but a protein complex!

So, this will be your homework (cf. this homework on topology). Just joking, there will be another blog post.

]]>