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.
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.
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.
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 $L=|\chi\rangle\langle \chi|$.
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 $A\otimes I$: $\mathrm{tr}[(A\otimes I)\rho]=\mathrm{tr}_1(A\,\mathrm{tr}_2\rho)$.
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