Mixed states I: density matrix, partial trace, the most general Born rule

In the last article, we saw that sub-systems entangled with other sub-systems did not have well-defined pure states themselves -- just like correlated random variables don't have their own probability distributions. Since pretty much everything you see in the real world is entangled with something -- has correlations with some other thing -- this is a problem. One can't just consider the "state of the entire universe" when you just want to study a single electron or something.

Wait -- why can't we just consider the marginal distributions, like we do in statistics? OK, suppose we start with the system -- with $|\phi\rangle$ and $|\varphi\rangle$ an orthonormal basis:

$$|\psi\rangle = \frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$$
Naively, you may think that the state of the first sub-system $|\psi_1\rangle$ may be given by $|\psi'_1\rangle=\frac1{\sqrt2}|\phi\rangle+\frac1{\sqrt2}|\varphi\rangle$. Certainly, if we're measuring the subsystem with an operator with eigenvalues $|\phi\rangle$ and $|\varphi\rangle$, you have 50% probabilities of each. But to say that two things are in the same state requires that they produce the same outcome for any measurement, not just that one. Does our sub-system behave exactly like $|\psi'_1\rangle$ for all observables? Recall that in the last article, we showed that collapsing the first sub-system onto $|\chi\rangle$ collapses the entire system into the state:

$$|\chi\rangle\otimes\left(
\langle\chi|\varphi\rangle|\phi\rangle+\langle\chi|\phi\rangle|\varphi\rangle\right)$$
To calculate the probability amplitude of this collapse, we may take the inner product of this with the original state -- you can compute this, and see the answer comes down to $1/\sqrt2$, i.e. there's a probability of $1/2$ of the first subsystem collapsing to any such eigenstate $|\chi\rangle$. You can use any observable in this two-dimensional state space, and the sub-system would collapse into either eigenstate with probability exactly $1/2$.

This is a completely different situation from if the state of the first subsystem were simply a pure state like $|\psi'_1\rangle$.

The situation we're dealing with is called a mixed state -- an example of a mixed state, in line with the motivating examples we had at the beginning of the course -- is unpolarised light. In fact, the state we described above models precisely unpolarised light involving two photons (is it obvious why?).

The basic idea behind mixed states is that we have some uncertainty as to what the state of a particle is -- we don't know if the particle has state $|\phi\rangle$ or $|\varphi\rangle$ -- it has a 50% chance of either. This is a classical probability, rather than a quantum one, and different from the state being a superposition of these states, as we just saw above.

Does this sort of thing occur with multivariate distributions in statistics? Suppose we have a multivariate distribution $\psi(x,y)$ and extract the marginal $x$-distribution $\phi(x) = \int_y \psi(x,y) dy$. Certainly this $\phi(x)$ gives us the right probability densities of each $x$-value. But the analog of considering general states like $|\chi\rangle$ is to make a transformation of the domain -- like a Fourier transform -- and consider probability densities in the transformed domain.

As an exercise, write down a multivariate Fourier transform expression for $\hat{\psi}(\omega_1,\omega_2)$ and use it to compute the $\omega_1$-marginal probabilities $\hat{\phi}(\omega_1)$ -- compare this to what you would get if you were to Fourier-transform the $x$-marginal $\phi(x)$ directly.

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But saying "it has 1/2 probability of being in $|\phi\rangle$ and 1/2 probability of being in $|\varphi\rangle$" is clearly an overdetermination. As we saw above, this resulting state has a 1/2 probability of collapsing onto any state -- this is a statement that doesn't depend on $|\phi\rangle$ and $|\varphi\rangle$, the behaviour of the state is the same if you describe it instead as having "1/2 probability of being in $\frac1{\sqrt2}(|\phi\rangle+|\varphi\rangle)$ and 1/2 probability of being in $\frac1{\sqrt2}(|\phi\rangle-|\varphi\rangle)$". These two are different statistical ensembles but in the same mixed state.

You can see similarly that "50% left-polarised + 50% right-polarised" is the same mixed state as "50% left-circular + 50% right-circular" -- they're both just unpolarised light.

What's the general condition for two statistical ensembles to produce the same observations?

Given statistical ensemble $\left(\left(p_i,|\psi_i\rangle\right)\right)$ and $\left(\left(p_i,|\psi'_i\rangle\right)\right)$, they are the same mixed state if for all $|\chi\rangle$, the probabilities of collapsing onto $|\chi\rangle$ is the same, i.e.

$$\sum_i p_i|\langle\psi_i|\chi\rangle|^2=\sum_i p_i|\langle\psi'_i|\chi\rangle|^2$$
Well, each side of this equation is just the evaluation of a quadratic form for the vector $|\chi\rangle$ -- and two quadratic forms are identically equal on all vectors if and only if their matrix representations are the same. Well, what's the matrix representation? In the basis of $|\psi_i\rangle$, it's just the matrix of probabilities $p_i$. The way to write this in Bra-ket notation is to factor out the $|\chi\rangle$s:

$$\left\langle \chi  \right|\left( {\sum\limits_i {{p_i}\left| {{\psi _i}} \right\rangle \left\langle {{\psi _i}} \right|} } \right)\left| \chi  \right\rangle  = \left\langle \chi  \right|\left( {\sum\limits_i {{p_i}\left| {{{\psi '}_i}} \right\rangle \left\langle {{{\psi '}_i}} \right|} } \right)\left| \chi  \right\rangle
$$
This quadratic form in between, representing a mixed state, is called the density matrix and can be used to completely specify mixed states. In this sense, it is a generalisation of the state vector, which can only be used to represent pure states.

$$\rho={\sum\limits_i {{p_i}\left| {{\psi _i}} \right\rangle \left\langle {{\psi _i}} \right|} }
$$
You may confirm that indeed:

$$\frac12|\phi\rangle\langle\phi|+\frac12|\varphi\rangle\langle\varphi|=\frac12\left(\frac{|\phi\rangle+|\varphi\rangle}{\sqrt2}\frac{\langle\phi|+\langle\varphi|}{\sqrt2}\right)+\frac12\left(\frac{|\phi\rangle-|\varphi\rangle}{\sqrt2}\frac{\langle\phi|-\langle\varphi|}{\sqrt2}\right)$$
In fact, there is a simpler way to see that those two ensembles are the same: the density matrix is simply the Gram matrix of the ensemble -- you take the states in the ensemble, weighted by $\sqrt{p_i}$ in a matrix $Y$, and $\rho=Y^*Y$. Well, $Y^*Y=Y'^*Y' \iff Y'=UY$  for some unitary $Y$, i.e. the ensembles are rotations of each other.

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Properties of the density matrix, generalised Born's rule, etc.

Here's something that's obvious: the density matrix is nonnegative-definite ("positive-semidefinite") Hermitian and unit-trace -- and all such matrices can represent density matrices.

Well, so it's a Hermitian operator -- does it represent any interesting observable? Not really. It's an observable, sure, but not an interesting one (you might say it measures something's being in one of the ensemble states -- written in an orthonormal basis -- and whose eigenvalues are the mixing ratios, etc. -- but what if two mixing ratios are the same? Its behaviour is just bizarre and useless, really).

We saw earlier that the probability of a density matrix collapsing into a state $|\chi\rangle$ is given by $\langle\chi|\rho|\chi\rangle$.

This is completely different from the generalised Born's rule we saw earlier which took the form $\langle\psi|L|\psi\rangle$! There, the state was the vector and the information on the projection space was the quadratic form in between. Here, the state is the quadratic form in between while the state being projected onto is the vector. This is just a generalisation of the simple Born rule $\langle\chi|\psi\rangle\langle\psi|\chi\rangle$, as far as I can see. If anyone comes up with a connection between it and the generalised Born rule for pure states, tell me.

This brings the question, though -- what's the most generalised Born's rule we can come up with? What is the probability of a mixed state collapsing into some eigenspace of a Hermitian projection operator?

Well, given the ensemble $((p_i,|\psi_i\rangle))$ (you can start writing ensembles with their density matrices now if you like, like $\sum p_i|\psi_i\rangle\langle\psi_i|$ -- but I just want to reaffirm that our result will indeed be in terms of the density matrix), the probability is:

$$\sum_i p_i\langle\psi_i|L|\psi_i\rangle$$
This is hardly useful -- it's not in terms of the density matrix at all. But look at each term -- what's $p_i\langle\psi_i|L|\psi_i\rangle$? $L|\psi_i\rangle$ is the $i$th column of $L$ in the $(|\psi_i\rangle)$-basis -- the inner product $\langle\psi_i|L|\psi_i\rangle$ is the $i$th entry of this column. Multiplying this by $p_i$ gives us the dot product of the $i$th row of $\rho$ with the $i$th column of $L$. The sum of these for all $i$ gives us the trace of $\rho L$:

$$\ldots = \mathrm{tr}(L\rho)$$
This is the most general form of Born's rule. Note that our derivation could have also applied to finding the expectation value of a general operator $A$ under the density matrix $\rho$ (recall that Hermitian projection operators are basically "indicator variables" whose expectation values represent probabilities), indeed generally:

$$\langle A\rangle_\rho=\mathrm{tr}(L\rho)$$
(note that $\mathrm{tr}(V)=\sum_{i} \langle i | V|i \rangle $ for any basis $(|i\rangle)$, which you should show.)

It is also trivial to show that upon collapse given by Hermitian projection operator $L$, the density matrix collapses to:

$$\rho'=\frac1{\mathrm{tr}\left(L\rho L\right)}{L\rho L}=\frac1{\mathrm{tr}\left(L\rho\right)}{L\rho L}$$
Generalising the pure state collapse to $|\psi'\rangle=\frac{1}{\langle \psi | L | \psi\rangle}L|\psi\rangle$. One may check that the above expression reduces to $|\chi\rangle\langle\chi|$ in the case where $L=|\chi\rangle\langle \chi|$.

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Partial trace, trace

We started our discussion considering the pure state $\frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$ and asking for the mixed state of the first sub-system. We computed the inner product of this state with its projection under the operator $|\chi\rangle\langle\chi|\otimes1$ -- this tells us the evaluation of the quadratic form $\langle\chi|\rho_A|\chi\rangle$ at all vectors $|\chi\rangle$, which determines the quadratic form $\rho_A$ of the first state.

So what exactly did we do -- in general? Starting with a density matrix $\rho$ on $H_1\otimes H_2$, we compute the probability of the first sub-system appearing in state $|\chi\rangle$: it's $\mathrm{tr}((|\chi\rangle\langle\chi|\otimes 1)\rho)$. So we try to find a density matrix $\rho_1$ satisfying, for all states $|\chi\rangle$:

$$\mathrm{tr}((|\chi\rangle\langle\chi|\otimes 1)\rho)=\mathrm{tr}(|\chi\rangle\langle\chi|\rho_1)$$

Exercise: Let $V$ be an operator on $H_1\otimes H_2$. We define its partial trace on $H_2$ as $\mathrm{tr}_2(V)=\sum_{j}\langle j|V|j\rangle $ for basis $(|j\rangle)$ of $H_2$ (where the inner product is done by extending operators by tensoring them with the identity). Show that the density matrix $\rho_1$ is given by:

$$\rho_1=\mathrm{tr}_2(\rho)$$
I.e. show that for operators of the form $A\otimes I$: $\mathrm{tr}[(A\otimes I)\rho]=\mathrm{tr}_1(A\,\mathrm{tr}_2\rho)$.

Projection operators, generalised Born's rule, position basis, wavefunction

At the end of the last article, I asked you to investigate Born's rule for continuous variables like position and momentum.

Well, the problem is that if $x$ is continuously distributed (i.e. we have an operator $X$ whose eigenvalues form a continuous spectrum $\Sigma_X$), typically $P(x=\lambda)=0$ -- and this gives us very little information about the actual probability distribution. What we're really interested in is $P(x\in B)$ for $B$ some subset of $\Sigma_X$.

Technically, we need $B$ to be a "Borel subset", or "measurable subset". We will be omitting several such technicalities in the article, such as the need for the spectral theorem to define a "projection-valued measure" or "spectral measure" on an operator with a continuous spectrum -- this is something that will be covered in the MAO1103: Linear Algebra course.

First, let's think about $P(x\in B)$ in the countable case. One can write $B=\{\lambda_1,\ldots\lambda_n\}$, and then simply say that

$$P(x\in B)=\sum |\langle\psi|\phi_k\rangle|^2$$
But the term on the right is a Pythagorean sum -- specifically, it is the length-squared of the vector formed by summing all the projections of $|\psi\rangle$ onto the eigenstates $|\phi_1\rangle\ldots|\phi_k\rangle$. But this is the same as the length of the projection of $|\psi\rangle$ onto the span of these eigenstates.

(Note on notations: From here onwards, we will use the notation $|\lambda\rangle$ to refer to the eigenvector corresponding to the eigenvalue $\lambda$ (if the eigenspace has dimension more than 1, we'll figure something out). We will use the notation $\{|B\rangle\}$ to refer to the span of the eigenvectors corresponding to the eigenvalues in $B$.)

So we could just define a Hermitian projection operator $L_X(B)$ for any subset $B$ of the spectrum of $X$ -- it is an easy exercise to write down an explicit form for $L_X(B)$ in terms of the eigenvectors of $X$.

Then the probability $P(x\in B)$ is simply $|L_X(B)|\psi\rangle|^2$. Recalling that a Hermitian projection operator satisfies $L^*=L=L^2$, we can write the generalised Born's rule as:

$$\begin{align}P(x\in B) &= |L_X(B)|\psi\rangle|^2\\ &= \langle\psi|L_X(B)|\psi\rangle\end{align}$$
Well, this is interesting! In the last article, you proved that the expected value of an observable $X$ given a state $|\psi\rangle$ is given by $\langle\psi|X|\psi\rangle$. But here we have a probability given by the same expression. So we want to interpret our projection operators as some sort of "observable" -- we can omit the "Hermitian", since all observables are Hermitian.

There's another place you might've seen something like this, and that is with indicator variables in probability and statistics -- the expected value of an indicator variable for an event is the probability that the event occurs.

Try to interpret these projection operators as observables that are analogous to "indicators" in some sense. If you think a little about it, you might see exactly what these observables represent: the eigenvalues of $L_X(B)$ are all 1 and 0 -- if the value "1" is realised, the state has been projected into the $\{|B\rangle\}$ -- and if the value "0" is realised, it hasn't.

So projection operators are a special type of observable, measuring the answer to "Yes/No questions" -- if the answer to "is the system in one of the states $\{|B\rangle\}$?" is yes, the observable $L_X(B)$ takes the value 1 -- if the answer is no, then it takes the value 0. So it is precisely an "indicator variable" for $\{|B\rangle\}$.

We have seen such projection operators, of course, in the context of polarisation -- where the operator represented whether or not the photon has passed through. Indeed, one may formulate quantum mechanics entirely in terms of projection operators, as any question can be formulated with some number of Yes/No questions (the key reason why this can be done, as we will see -- is that these "yes/no questions" all commute, i.e. the corresponding projection operators share an eigenbasis). Let's not.

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Well, this can be generalised in the straightforward way to an operator with a continuous spectrum, resulting in the same expression. We can also calculate probability densities using this result. Let $X$ be an operator with continuous spectrum $\Sigma_X$ -- then we can write the state $|\psi\rangle$ in the eigenbasis of $X$:

$$ |\psi\rangle = \int_{\Sigma_X} |x\rangle\, \Psi(x)\, dx $$
Where $\Psi(x)\, dx=\langle\psi|x\rangle$ are the coefficients of the state in the eigenbasis, i.e. the probability amplitudes -- we call $\Psi(x)$ the wavefunction, and it represents probability amplitude densities. Then for some set $M\subseteq \Sigma_X$ of eigenvalues $L_X(B)|\psi\rangle$ is the projection:

$$ L(M)|\psi\rangle = \int_B |x\rangle\, \Psi(x)\, dx $$
And one may calculate the dot product, noting that complex dot products require taking the complex conjugate:

$$ \langle \psi | L_X(B) | \psi \rangle = \int_B \Psi^*(x)\, \Psi(x)\, dx $$
Which gives us an expression for the probability density function on $\Sigma_X$ as:

$$\begin{align}\rho(x) &=\Psi^*(x)\,\Psi(x) \\
&=|\Psi(x)|^2\end{align}$$
And this applies to any operator with a continuous spectrum, like position and momentum.

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Some texts define the eigenvectors $|x\rangle$ of a continuous-spectrum observable differently from us -- it is often conventional to let $|x\rangle$ be infinitely large so that $\langle x_1|x_2\rangle = \delta(x_1-x_2)$. This is so that the amplitudes $\langle\psi|x\rangle$ are not infinitesimal, but instead $\langle\psi|x\rangle=\Psi(x)$ (without multiplication by $dx$). For consistency with discrete spectra, we do not use this convention.

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.