From polarisation to quantum mechanics: states, observables, Born's law

Like most texts on the theory, I will motivate the mathematics of quantum mechanics from the example of polarisation -- mostly because it's a very accessible example of stuff being wavelike. From this example, we will be able to motivate: the state vector (generalising the polarisation), state vector collapse (the event of polarisation), observables and their eigenvalues (stuff like energy, number of photons, etc.), eigenstates and their orthogonality (polarisation basis), noncommuting operators and uncertainty (the noncommuting of lenses).

The key feature of quantum mechanics -- the fundamentally probabilistic nature -- comes from the following two facts, confirmed by experiments (the famous experiments here are the double-slit experiment and photoelectric effect respectively):

• Everything is a wave -- objects behave as waves, following the superposition principle and the waves represent densities of observations at large scales.
• Everything is a particle -- which manifests itself in the form of some stuff, like energy and momentum, coming in little quanta.

This is the principle of wave-particle duality. You may realise how this implies a probabilistic description, but the following example should make it quite clear: consider a wave of light, with energy $hf$ (so it's a single photon) polarised at angle $\theta$ to the horizontal -- and it passes through a horizontal polarising filter. Well, then the wave that passes through would be a horizontally polarised wave with energy $hf\cos^2\theta$, right?

But this is impossible, since energy levels in quantum mechanics are quantised -- you can't have $\cos^2\theta$ of a photon, you can only have integer multiples of a photon. But the fact that energy drops as $E\cos^2\theta$ is something that you can verify at your home, using sunglasses -- what the heck?

The key point is that the empirical verification of the $\cos^2\theta$ business that you can do at home is on a macroscopic level, when you have a large number of photons $E=Nhf$. So something occurs with the photons on a microscopic level such that when you try it with a large number of photons, $\cos^2\theta$ of the photons pass through.

Well, this is essentially the "definition" of probability! A single photon passes through the filter with a probability of $\cos^2\theta$ so that for a large number of photons, $\cos^2\theta$ of the photons pass through. This is a non-trivial result -- wave-particle duality makes no mentions of probability as such, it just tells us that stuff is both a particle and a wave, but this simple condition in itself implies a probabilistic, non-deterministic reality.

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Similar thought experiments can illustrate the probabilistic nature of other things (the "things" in question here will soon be called "eigenstates"): position is easy -- consider a standing wave photon in a box (this can easily be constructed). This is uniformly distributed throughout the box -- so how much of the energy is in some chunk of the box?

Momentum is trickier, but shouldn't be too hard if you're familiar with Fourier transforms -- what's the analog of a "box" in momentum-space? Well, consider a concentrated pulse of light -- this can be written, via a Fourier transform, as the sum of several light waves of different momenta (i.e. frequencies), each wave with some lower energy. Taking "some chunk" of this "box" amounts to filtering some specific frequencies of the light. This can be done easily, e.g. with a colour filter -- so how much of the energy is contained in the waves with these specific momenta?

In both cases, the key point is that you can't have a fraction of the energy of the photon at these positions/momenta, so you must have a probability of measuring the photon to be in a specific range of positions or a specific range of positions -- to be in a specific region or in a specific region of momentum-space.

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The fundamental point here can be made for any quantity $X$: if you can filter out the "part" of a collection of particles that has $X$ in a certain subset of its range, then on a microscopic level, is probabilistic. The act of "filtering out the parts with a certain $X$", applied to a single particle, is just the act of checking if a particle is in a certain $X$-interval, and is called measurement. Any quantity that you can measure is called an observable. 

Something like polarisation is really a form of measurement -- you're finding out whether or not the photon is in a certain polarisation $|\phi_{\parallel}\rangle$. You may have another observable, corresponding to a different polarisation -- even one that is orthogonal to the first polarisation -- $|\phi_{\perp}\rangle$ and still get that the photon is in $|\phi_\perp\rangle$. There is nothing wrong with this, as we just know beforehand that the photon is in $|\phi_\parallel\rangle$ or $|\phi_\perp\rangle$. If you perform the polarisation with $|\phi_\perp\rangle$ after the polarisation with $|\phi_\parallel\rangle$, you will find that the photon doesn't pass through, as you know for sure that the photon is not in both $|\phi_\parallel\rangle$ and $|\phi_{\perp}\rangle$.

Now, you may have certain psychological issues with this, as have many in history -- however, you might want to note that the aim of quantum mechanics is not to fix your psychological problems but to explain nature. You need to accept logical positivism and learn to shut up and calculate to be comfortable with quantum mechanics.

So whatever calculus we invent to describe these probabilistic phenomena, it is going to apply to all observables.

In our first example, the polarisation of the photon can be represented by a unit vector which we will denote as $|\psi\rangle$. The polarising filter has two special axes, represented by unit vectors $|\phi_{\parallel}\rangle$ and $|\phi_\perp\rangle$ -- these are special in the sense that an incoming photon polarised as $|\phi_{\parallel}\rangle$ or $|\phi_\perp\rangle$ will simply be scaled, by factors of 1 and 0 respectively -- so these form an eigenbasis for a certain operator.

Well, we said that the photon passes through (with polarisation $|\phi_{\parallel}\rangle$) with probability $\cos^2\theta$ -- this arises simply from considering the amplitude of $|\psi\rangle$ in the direction of $|\phi_\parallel\rangle$. So we can write the probability that the photon ends up in a state $|\phi\rangle$ as $|\langle\psi|\phi\rangle|^2$ where $\langle\psi|\phi\rangle$ is called the corresponding "probability amplitude".

This expression, $P(x=\lambda)=|\langle\psi|\phi_\lambda\rangle|^2$ is called Born's rule.

Let's get back to the eigenbasis -- what exactly is this an eigenbasis of? We said that the corresponding eigenvalues are 1 and 0, so this gives us a complete description of the operator. Note that this operator depends only on the observable (namely "number of photons in the $|\phi_{\parallel}\rangle$ direction), not on the state or any other feature of the observation. So we decide to call this operator/matrix the "observable", and its eigenvalues are the values of the observable that can be measured.

To find properties of these observables, the natural way is to note that the only feature we've really required of them is Born's rule, i.e. the probabilistic interpretation -- so we can apply the axioms of probability and see what they apply in the context of these observables.

• $P(E)\ge 0$ -- imply that the observables are over either the reals or complexes, so that $|\langle\psi|\phi\rangle|^2\in \mathbb{R}$ in the first place. The nonnegativity then follows.
• $P(\Omega)=1$ and $P\left(\bigcup_i E_i\right) = \sum_i P(E_i)$ for disjoint $E_i$ -- this, along with the second axiom, implies that $\sum |\langle\phi|\psi\rangle|^2 = 1=|\langle\psi|\psi\rangle|^2$ where the sum is taken over all eigenstates $|\phi\rangle$ of the operator. As this must be true for all states $|\psi\rangle$, the thing on the left must be a Pythagorean sum, so the $|\phi\rangle$s must form an orthogonal basis. This implies that all observables are normal operators.

The latter fact is very important, and can also be seen in the following way -- if you a system is in one eigenstate, it cannot possibly collapse onto another eigenstate (the probabilistic interpretation is: if you know for sure the value of the symbol is a thing, it's that thing) -- so we must have $|\langle \phi_1|\phi_2\rangle|^2=0$ for all eigenstates $|\phi_1\rangle$ and $|\phi_2\rangle$.

Another restriction we add is that the observables be not only normal, but Hermitian operators in particular, so they have real eigenvalues. This may seem an odd choice, but it makes sense, as any normal operator may be uniquely written as $X_H+iX_{AH}$ where $X_H$ and $X_{AH}$ are Hermitian, and $X_H$ and $X_{AH}$ commute, so any complex observation can be done unambiguously as two real observations. So we stick to real eigenvalues.

This also makes it essential that we allow complex operators rather than just real ones (the two choices were given to us from the first probability axiom), so that this decomposition is possible. Later, we will see concrete examples of this with commutators $[X,Y]$, which must be multiplied by $i$ to turn Hermitian. We will also see more fundamental reasons to choose complex numbers in QM.

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Exercise: Show that the expected value of an observable $X$ given a state $\psi$ can be given as $\langle \psi|X|\psi \rangle$ (i.e. $\psi^*X\psi$ in conventional notation).

Exercise: Explain Born's rule with other observables, like position and momentum. Explain why it holds in general.

SVD, polar decomposition, normal matrices; a re-look at transposes and FTLA

Back in Null, row spaces, transpose, fundamental theorem of algebra, we first introduced some hand-wavy intuition for the transpose and the orthogonality of the row space and the null space (and the following fundamental theorem of linear algebra). Here, we solidify this intuition a bit more clearly.

Consider the "symmetric collapse" discussed in the above article. Our study of the transformation relied specifically on looking at it in a specific basis -- an orthogonal basis -- comprised of the column space and the null space. In this basis, the transformation is a scaling on both axes. In the more general case of an asymmetric collapse -- in which we rotated our space before collapsing, we looked at a basis formed by the row space and the null space -- the basis got rotated and scaled into the new basis, that was the column space and an arbitrary other vector (that could be perpendicular to the column space).

A sensible question to ask is if any transformation can be written in this fashion -- as a transformation of an orthogonal basis into another orthogonal basis. Analogous to how an eigenvalue decomposition of a matrix writes it as scalings in some basis, we're looking to represent the matrix as a spiral (i.e. a scaling combined with a rotation) in some basis. But let's stick with the first formulation of our question -- for any linear transformation $A: \mathbb{R}^n \to \mathbb{R}^m$, can we find an orthogonal basis on $\mathbb{R}^n$ that is mapped to an orthogonal basis $\mathbb{R}^m$?

One could, e.g. consider the images under $A$ of the angle of each orthonormal basis in $\mathbb{R}^2$ (i.e. look at the function $AU(\theta)\vec{e}_1 \cdot AU(\theta)\vec {e}_2$ for varying $\theta$ where $U(\theta)$ is the rotation matrix by angle $\theta$) and apply the intermediate value theorem, etc. And such a proof could in principle be extended to $\mathbb{R}^n$.

(See here for a thorough explanation.)

Here's another, more insightful way you might come to prove this -- we've been visualising linear transformations so far by looking at the image of the basis vectors, but another way to do visualise these transformations is by looking at the image of the unit circle under the transformation.

Why does this make sense? Well, one can find an ellipse passing through any two vectors centered at the origin. Elaborate on this argument. Is the resulting ellipse unique? (Hint: no, unless you mark points on the circumference)

Specifically, consider the axes of the image ellipse $\sigma_1 u_1$, $\sigma_2 u_2$ where $u_1$, $u_2$ are unit vectors. In the original unit circle, any pair of orthogonal vectors on the circle can be axes, so consider the pre-image of the axes of the image ellipse $v_1$, $v_2$.  So we have:
$$ A v_1 = \sigma_1 u_1\\
A v_2 = \sigma_2 u_2 $$
Or in general:
$$AV = U\Sigma\\
A = U\Sigma V^*$$
Where $\Sigma$ is diagonal and positive-definite, while $U$ and $V$ are orthogonal/unitary. This is called the Singular-Value Decomposition (SVD) of $A$.

In a sense, one can view this as an alternative to the eigen-decomposition. In the eigendecomposition, one looks for a single basis in which the transformation is a scaling. In the singular value decomposition, one looks at scaling and then "re-interpreting" in another basis, but requires that the bases be orthogonal, and that the diagonal matrix be positive and real-valued.

The entries $\Sigma$ are called the singular values of $A$, the columns of $U$ and $V$ respectively are the left-singular vectors and the right-singular vectors of $A$ respectively.

(Exercise: You know that $\Sigma$ is the scaling of the orthogonal basis, i.e. of the right-singular basis. Convince yourself that the rotation of the basis is given by $UV^*$.)

This gives us a much better intuition for the transpose. The SVD of $A^*$ is clearly:

$$A^* = V\Sigma U^*$$
I.e. the transpose has precisely the opposite rotational effect as $A$ and the same scaling. This is as opposed to the inverse matrix, which has both the opposite rotational and scaling effect as the matrix. For a rotation matrix (or generally an orthogonal matrix $A^*A=I$), the transpose equals the inverse, analogous to how the conjugate equals the inverse for a unit complex number $\bar{z}z=1$. A Hermitian matrix, $A=A^*$, by contrast, is one for which is irrotational, $UV^*=1$, i.e. for which the SVD equals the eigendecomposition.

Use the SVD to get some intuition for transpose identities like $(AB)^*=B^*A^*$

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It's instructive to consider the SVD in the case of our original motivating example -- an asymmetric matrix representing a collapse of $\mathbb{R}^2$ into a line. What are the singular bases of this transformation? Well, it maps the orthogonal basis formed by row space and the null space into the orthogonal basis formed by the column space and the left-nullspace.

Think about its transpose.

Arranging the singular values from largest to smallest, we then have the following relation between the SVD and the items in the fundamental theorem of linear algebra, where $n$ is the dimension of the domain, $m$ is the dimension of the codomain, and $r$ is the dimension of the image/column space:

• The last $n - r$ singular values are zeroes. 
• The first $r$ singular values are positive.
• The last $n - r$ right-singular vectors span the null space.
• The first $r$ right-singular vectors span the row space.
• The last $m-r$ left-singular vectors span the left-null space.
• The first $r$ left-singular vectors span the column space.

Note that the terms kernel, coimage, cokernel and image are also used for the null space, row space, left-null space and column space, sometimes in a more general setting.

This is the full form of the Fundamental Theorem of Linear Algebra, which generalizes the rank-nullity theorem, row rank equals column rank and row space being perpendicular to null space.

Spend some time thinking about the SVD of non-square matrices, relating them to square collapse matrices. Think about their transposes.

Show that the right-singular vectors $V$ of a matrix $A$ are given by the eigenvectors of $A^*A$ (hint: start by considering the two-dimensional case, relating the right-singular vectors to a maximisation/minimisation problem, and extend the idea to more dimensions).

From this, it is clear that the left-singular eigenvectors $U$ (which are the right-singular eigenvectors of $A^*$) are given by the eigenvectors of $AA^*$ and the singular values are the square roots of the singular values of $A^*A$. Well, some of them are (which ones?).

We observed earlier that the rotational effect of the matrix $A = U \Sigma V^*$ can be given by $UV^*$. The scaling effect is given by $\Sigma$ on the basis of $V$. Hence we can write:

$$A = (UV^*)(V\Sigma V^*)$$
Letting $W=UV^*$ and $R = V\Sigma V^*$, this gives us a representation of $A$ as:

$$A=WR$$
Where $W$ is orthogonal and $R$ is positive-semidefinite. This is known as the right-polar decomposition of $A$. Analogously, one may consider the left-polar decomposition:

$$A = (U\Sigma U^*)(UV^*) = R'W$$

Interpret the above decomposition like we did the right decomposition. Note how $R' = WRW^*$, and how a right-polar decomposition leads to a left-polar decomposition of the transpose, and vice versa.

When is the polar decomposition unique? Compare the situation to the polar decomposition of complex numbers.

Here's a question: when is $R=R'$? One way of putting it is that $WRW^* = R$, i.e. $R$ commutes with $W$. This is a bit difficult to work with. Instead, one may show that the matrices $R$ and $R'$ can be given by $(A^*A)^{1/2}$ and $(AA^*)^{1/2}$ respectively (prove it!) -- noting that a principal square root can uniquely be defined for positive semidefinite matrices. So $R=R'\Leftrightarrow A^*A=AA^*$. These are known as normal matrices. Below is an exercise that provides more intuition into the behaviour of normal matrices.

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What does it mean for a matrix to commute with its transpose? Earlier, when discussing commuting matrices, we referred to it as "matrices that do not disturb each other" -- specifically, they preserve each other's (generalised) eigenspaces. In the case of commuting with its transpose, it's easy to show that this means having the same eigenvectors (prove this!).

Here's a fact about the eigenvectors of the Hermitian transpose: the eigenvector of $A$ corresponding to the eigenvalue $\lambda$ is orthogonal to all eigenvectors of $A^*$ corresponding to any eigenvalue other than $\lambda^*$ (prove this!).

From these two facts, it follows that the following are equivalent (write down the proofs clearly!):

• $A$ is normal.
• $A$ commutes with $A^*$.
• $R$ commutes with $W$.
• $A$ is unitarily diagonalisable.

This is known as the spectral theorem.

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(Anyone ever thought about how weird the word "normal" is? Sometimes, it means "perpendicular", sometimes -- as in "orthonormal" and "normalisation" -- it means "unit length", because "norm". What does it mean in this context? It's probably just referring to its eigenvectors being normal/orthogonal, but I like to think it's referring to the fact that the two alternative Hermitian-valued "norms" of the matrix, $A^*A$ and $AA^*$, are equal, so the matrix has a single "norm".)

Fundamentally, normal matrices are "analogous" to complex numbers, or they "generalise" complex numbers, in the sense that each of its eigenvalues acts as a complex number, transforming, acting as spirals on each of the orthogonal eigenvectors within its own copy of $\mathbb{C}$. One may construct the following table of analogies between normal matrices and complex numbers:

Complex numbers	Normal matrices
Zero (sorta)	Singular
Non-zero	Invertible
Real	Hermitian
Positive real	Positive-definite
Nonnegative real	Positive-semidefinite
Imaginary	Anti-Hermitian
Unit	Unitary
Conjugate	Hermitian transpose
Norm-squared	Gram matrix $A^*A$
Magnitude	$(A^*A)^{1/2}=R=V\Sigma V^*$
Argument	$AR^{-1}=W=UV^*$
Real Part	$\frac12(A+A^*)$
Imaginary Part times $i$	$\frac12(A-A^*)$

Exercise: an EP matrix or range-Hermitian matrix is a weakened version of a Hermitian matrix -- the row space of the matrix equals the column space. Although this was a bit hard to understand in the first article and was only briefly mentioned towards the end, we now have the intuition to comprehend them. Explain why a matrix is range-Hermitian if and only if it is unitarily similar to a matrix of the form of a block matrix:

$$\left[ {\begin{array}{*{20}{c}}C&0\\0&0\end{array}} \right]$$
Where $C$ is a non-singular square matrix and the zeroes are zero block matrices. This decomposition is called the core-nilpotent decomposition. Hence, show that being range-Hermitian is a weakened form of being normal.