Two Qubits and Entanglement

The previous notes proved one thing in the chapter on tensor products: product states are not closed under addition, and the superposition principle forces entanglement into existence. But that was existence only — entanglement got a name, not an opening. These notes build it (two gates), detect it (one line), take it apart (density matrices and the partial trace), measure it (how mixed the parts are), and finally ask whether it is a conjuring trick at all (the CHSH value $2\sqrt2$).

Everything from the first two installments is assumed: the tensor product rule, the projective-measurement recipe $P=\langle\psi|\Pi|\psi\rangle$, the expectation-value sandwich, the Bloch sphere, and the no-cloning theorem. Exactly one new piece of mathematics appears — the trace — introduced in chapter 6, where it is first needed.

1. The read-write system for two qubits

The shape of a quantum circuit is locked down by two constraints: the no-cloning theorem forbids wires from branching (fan-out is identically 1, the conclusion of chapter 3 last time), and a unitary matrix is square, so $k$ wires in means $k$ wires out and they cannot merge either. Only one picture survives — $n$ parallel horizontal lines, gates as boxes sitting on them, the wire count conserved. The world of two lines starts here.

1.1 Components and coefficients: the addresses of four slots

The tensor product rule came last time, $\binom pq\otimes\binom rs=(pr,ps,qr,qs)^T$, along with the four basis vectors it produces, $|00\rangle=(1,0,0,0)^T$ through $|11\rangle=(0,0,0,1)^T$. What needs filling in is the correspondence between components and coefficients. One addition does it:

The 1st component is the coefficient of $|00\rangle$, the 2nd that of $|01\rangle$, and so on; $|a|^2$ is the probability of measuring 00. The basis vectors obey the same rule: $|00\rangle$ is the vector with ”$a=1$, everything else 0”, hence $(1,0,0,0)^T$. (The single-qubit case was already like this: the first component of $\binom\alpha\beta$ is the coefficient of $|0\rangle$, while $|0\rangle$ itself is $\binom10$.)

The labels 00, 01, 10, 11 on the four components are binary counting, with the first qubit as the high bit:

• the first qubit decides whether you are in the first two components or the last two;
• the second qubit decides which of those two you are in.

This correspondence is the key to reading every $4\times4$ matrix in these notes.

1.2 Side-by-side gates $A\otimes B$ and the golden rule

States are now 4-dimensional vectors, so gates must be $4\times4$ matrices. The simplest batch: two single-qubit gates each acting on one qubit, whose matrix is their tensor product (same rule as for vectors — every element on the left multiplies the entire matrix on the right):

The golden rule (the mixed-product property):

Physically: the qubit in $|u\rangle$ passes through gate A while the qubit in $|v\rangle$ passes through gate B, two machines each doing their own job with no communication between them — on a circuit diagram, two parallel lines each carrying a box, which is why they are called “side-by-side”.

Its practical value: when checking a calculation you never have to build the $4\times4$. For instance,

The first equality hands out the assignments, the second has each X consult its table ($X|0\rangle=|1\rangle,\ X|1\rangle=|0\rangle$) — X is consumed at that step and leaves an output behind, just as $f$ “vanishes” in $f(3)=9$.

Scope: this decomposition requires the input itself to be a product state. If it cannot be written as $|u\rangle\otimes|v\rangle$ (such states show up shortly), expand it in the basis, apply the golden rule term by term, and add the results back; linearity makes that legal.

1.3 $X\otimes I$ versus $I\otimes X$: ⊗ does not commute

“Doing nothing” has to be written out as $I$ — the state is a 4-dimensional vector, and a $2\times2$ X simply cannot multiply it.

Multiplying $|00\rangle$, with all four rows written out:

All four basis vectors:

In each line the first bit flips and the second does not — that is what ”$X\otimes I$ = flip only the first qubit” means. On a general vector:

The first two components swap wholesale with the last two, order preserved inside each half. Compare $I\otimes X$:

The halves do not swap; the two entries inside each half do — that is, only the second qubit flips. The two matrices look nothing alike, and the block form makes it obvious at a glance:

The left and right slots of ⊗ are the qubit indices; they cannot be exchanged. When a circuit draws a gate on one line and leaves the others empty, the mathematical content is “that gate ⊗ a string of $I$‘s” (an H on qubit 2 of three qubits is $I\otimes H\otimes I$).

1.4 A trick worth keeping for life

The reason: $M|00\rangle$ is $M$ times $(1,0,0,0)^T$, and the row-by-row dot products pick out exactly the first column. Check: the first column of $X\otimes I$ is $(0,0,1,0)^T=|10\rangle$ ✓, matching the calculation above.

The converse holds too: knowing the truth table lets you write the matrix out one column at a time — which is exactly how the CNOT matrix gets built in the next chapter.

1.5 Unitarity: verify once, exempt forever

$U_1\otimes U_2$ is a freshly built matrix and needs its credentials checked; the worry is concrete, since modules that each pass on their own need not pass once assembled. Distribute the dagger slot by slot and use the matrix version of the golden rule, $(A\otimes B)(C\otimes D)=AC\otimes BD$:

The two certificates are re-examined in their own slots without interfering — ⊗ preserves unitarity, so from now on gates may be placed side by side freely without rechecking.

1.6 The boundary: side-by-side gates can never create entanglement

The golden rule is itself the proof:

Product state in, product state out. Making two qubits genuinely “talk” requires a gate that cannot be decomposed as $A\otimes B$.

2. Controlled gates: installing the action on one half

2.1 CNOT: truth table → matrix → block reading

Two sentences of rule (basis states only, for now): control is 0 → do nothing; control is 1 → apply $U$ to the target. Take $U=X$ (that is CNOT, control = qubit 1, target = qubit 2) and apply the rule case by case:

Using the trick of §1.4, copy the four destinations down as four columns:

That last formula — first two fixed, last two swapped — gets used over and over below.

Block reading: the first two components (control = 0) are preserved as they are, so the top-left block is $I$; the last two (control = 1) swap internally, so the bottom-right block is $X$:

What “controlled” means is plain in the blocks: the action is installed only on the half where the control is 1. Contrast §1.3’s $X\otimes I=\begin{pmatrix}0&I\\I&0\end{pmatrix}$, which is unconditional. Unitarity comes free: $\mathrm{diag}(I,U)^\dagger\mathrm{diag}(I,U)=\mathrm{diag}(I,U^\dagger U)=I$ ✓

Circuit notation: a filled dot on the control line, a vertical stroke down to a box on the target line; the target end of a CNOT is abbreviated ⊕.

2.2 XOR compiled into physics

One line summarizes the truth table: target ^= control. This picks up the thread of reversible computation — classical AND takes two in and gives one out, destroys information, is irreversible, and must be rewritten before it can enter a quantum circuit; XOR, on the other hand, is reversible for free as long as the control bit is kept. CNOT is that reversible XOR, and it is its own inverse: $(a,b,d,c)$ run through again returns $(a,b,c,d)$, so $\mathrm{CNOT}^2=I$.

Last time it wore another hat: the orthogonal-state copier — $|00\rangle\to|00\rangle$, $|10\rangle\to|11\rangle$. On the two orthogonal states 0 and 1 it really is copying, and that is the entire essence of classical copying. What follows shows what it becomes when fed non-orthogonal inputs.

2.3 An “if” that is unitary, not a measurement

A classical if reads the control bit first and then decides. In quantum mechanics “reading” = measuring = collapse. Compare two machines, both fed $|+\rangle\otimes|0\rangle$:

Machine A (measure, then act): measure the control → 50% gives 0 (output $|00\rangle$), 50% gives 1 (output $|11\rangle$). The output is one of two at random, the superposition destroyed, the process irreversible.

Machine B (CNOT): one matrix multiplication. First compute the input:

Apply $(a,b,c,d)\to(a,b,d,c)$:

Definite input, definite output, no probability, no reading, no randomness, and fully reversible. That is the whole content of “this is not a measurement”: nothing looks at the control bit, and the two branches coexist inside the superposition (each carrying its control-bit amplitude) rather than one being chosen. The if-intuition is a nickname that works only on basis states; off them, it has to go.

2.4 Phase kickback: even “only the target changes” is false

Feed in $|+\rangle\otimes|-\rangle$. As a 4-dimensional vector:

Apply the formula:

Try to split it as $\binom pq\otimes\binom rs$ with $p=r=\tfrac{1}{\sqrt2},\ q=s=-\tfrac{1}{\sqrt2}$: $pr=\tfrac12$ ✓, $ps=-\tfrac12$ ✓, $qr=-\tfrac12$ ✓, $qs=\tfrac12$ ✓. So the output is $|-\rangle\otimes|-\rangle$:

The target did not change; the control did. The mechanism is shorter through eigenvectors: $|-\rangle$ is an eigenvector of $X$ with $X|-\rangle=-|-\rangle$. On the $|1\rangle$ branch of the control, applying X leaves the target alone and produces an overall $-1$; on its own that $-1$ is a global phase, but because the control is in superposition it becomes a relative phase inside the control, turning $|+\rangle$ into $|-\rangle$. This is phase kickback, and the oracle tricks of Deutsch and Grover are all built on it.

(It is also another appearance of the principle from last time: a global phase is unobservable when it acts on the whole system, and demotes itself into a measurable relative phase the moment it acts on only one branch of a superposition.)

2.5 The shortest recipe for entanglement

CNOT only does its work with a control in superposition, and the hardware naturally hands you $|00\rangle$, so push the control into superposition with H first:

That is exactly the input from §2.3, so run it through CNOT:

Two gates, one entangled state. This is the first line of code in a whole family of protocols, teleportation and superdense coding among them.

2.6 When entanglement is created

The second row is precisely the “orthogonal-state copier” face from last time; the third is its other face — feed it a non-orthogonal superposition, copying fails, and entanglement comes out instead. Creating entanglement is not an accident of CNOT; it is its day job.

One structural fact follows: CNOT cannot be written as any $A\otimes B$ — if it could, it would only ever output product states, contradicting the Bell output above. The $U_1\otimes U_2$ family is a very small subset of the $4\times4$ unitaries, just as product states are a very small part of the 4-dimensional state space.

3. Separable versus entangled: one line of criterion

3.1 The quantifiers are asymmetric

Separable = product state = not entangled: there exist single-qubit states with $|\psi\rangle=|\psi_1\rangle\otimes|\psi_2\rangle$. Entangled = not separable.

The asymmetry of the quantifiers settles how hard each is to prove: separability is existential, so exhibiting one factorization wins; entanglement is universal, so every possible factorization must be excluded (by contradiction or by a criterion). “I tried a few and none worked” is not a proof.

Factoring is just pulling out a common factor:

It looks like a joint state, but it is two independent stories: qubit 1 is $|1\rangle$, qubit 2 is $|+\rangle$.

3.2 The criterion $ad=bc$

Last time $\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ was shown inseparable by contradiction. Doing that every time is clumsy; for a general state $(a,b,c,d)^T$ there is a mechanical test:

Forward: if it factors, $(a,b,c,d)=(pr,ps,qr,qs)$, so $ad=(pr)(qs)=pqrs$ and $bc=(ps)(qr)=pqrs$ — the same four factors, hence equal. Backward: suppose $ad=bc$ with $a\ne0$ and take $\binom ac\otimes\binom ab=(a^2,ab,ca,cb)^T$; the first three entries differ from $(a,b,c,d)$ by the common factor $a$, and the fourth, $cb=bc=ad$ (by the criterion), differs by the same factor, so the two are proportional and normalization finishes the job (the case $a=0$ goes the same way by cases) ∎

Checking the two protagonists:

• $|1\rangle\otimes|+\rangle=\tfrac{1}{\sqrt2}(0,0,1,1)^T$: $ad=0=bc$ → separable ✓
• $\tfrac{1}{\sqrt2}(1,0,0,1)^T$: $ad=\tfrac12\ne0=bc$ → entangled ✓

A measuring stick comes free: $2|ad-bc|$ has a name, the concurrence, and it measures how strong the entanglement is — 0 for a product state, 1 for the strongest entangled states. The criterion does not merely detect; it quantifies.

3.3 Entanglement comes in degrees

Bell states are the extreme case, and the middle ground genuinely exists:

Same shape as $\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$, but with unbalanced amplitudes. Here $ad=\sqrt{0.24}\approx0.49\ne0=bc$, so it is genuinely entangled, yet the concurrence is $2\sqrt{0.24}\approx0.98<1$ — close, but not maxed out.

Where is the difference? Measure the first qubit: $P(0)=(\sqrt{0.6})^2=\tfrac35\ne\tfrac12$. The correlation is still perfect (read 0 and the other side must be 0), but there is less shared randomness — you can already guess that 0 is the likely answer. Maximal entanglement tops out precisely because it is both perfectly correlated and perfectly random.

As for what should quantify entanglement in general, $2|ad-bc|$ only works for pure two-qubit states. The real answer waits until chapter 7, where the tool generalizes to any system.

4. Bell states and the Bell basis

4.1 The quartet

The state built in §2.5 is not a one-off; it belongs to a family of four:

The naming runs along two dimensions: the Greek letter records whether the two bits agree ($\Phi$ uses 00 and 11, the same; $\Psi$ uses 01 and 10, opposite), and the superscript $\pm$ records the relative phase. As column vectors:

The criterion verifies in one second that all four are entangled: for the $\Phi$ family $ad=\pm\tfrac12,\ bc=0$; for the $\Psi$ family $ad=0,\ bc=\pm\tfrac12$. And every concurrence is maxed out at 1, which is why they are called maximally entangled states.

The correlation signature: take $|\Phi^+\rangle$ and measure only the first qubit. Probability $\tfrac12$ gives 0 and collapses to $|00\rangle$ (the other side is 0 too); probability $\tfrac12$ gives 1 and collapses to $|11\rangle$ (the other side is 1 too). The $\Phi$ family agrees on both sides, the $\Psi$ family disagrees. But either side alone is always a fair 50/50 coin; the correlation appears only when the two records are compared. Chapter 7 builds “entanglement cannot signal” on exactly this fact.

4.2 The Bell basis: a coordinate system made of entangled states

The four Bell states are mutually orthogonal. Verifying two pairs (inner product = conjugate-multiply matching components and sum):

(Between the $\Phi$ and $\Psi$ families the nonzero positions are completely disjoint, so the inner product has to vanish.) Four normalized, mutually orthogonal vectors form a complete orthogonal basis of the 4-dimensional space: the Bell basis.

It is worth weighing what that means. This 4-dimensional space was built out of product states, and yet a basis consisting entirely of entangled states turns out to be equally legitimate. Any two-qubit state can be expanded in $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$ or in $\{|\Phi^\pm\rangle,|\Psi^\pm\rangle\}$ — neither basis is more “real” than the other.

4.3 The H-then-CNOT circuit is a dictionary

The recipe of §2.5 works on $|00\rangle$. Feed the other three basis states to the same circuit $\mathrm{CNOT}\cdot(H\otimes I)$:

Input $|01\rangle$: $(H\otimes I)|01\rangle=|+\rangle\otimes|1\rangle=\tfrac{1}{\sqrt2}(0,1,0,1)^T$; the formula sends $(0,1,0,1)\to(0,1,1,0)$, giving $|\Psi^+\rangle$.

Input $|10\rangle$: $(H\otimes I)|10\rangle=|-\rangle\otimes|0\rangle=\tfrac{1}{\sqrt2}(1,0,-1,0)^T$; the formula gives $\tfrac{1}{\sqrt2}(1,0,0,-1)^T=|\Phi^-\rangle$.

Input $|11\rangle$: $(H\otimes I)|11\rangle=|-\rangle\otimes|1\rangle=\tfrac{1}{\sqrt2}(0,1,0,-1)^T$; the formula gives $\tfrac{1}{\sqrt2}(0,1,-1,0)^T=|\Psi^-\rangle$.

The pattern can be read straight out of the calculations: the first input bit sets the phase $\pm$ (since $H|0\rangle=|+\rangle$ has matching signs and $H|1\rangle=|-\rangle$ opposite ones, and CNOT carries that minus sign into the result untouched); the second input bit picks $\Phi$ or $\Psi$ (it is the target’s initial value, which decides whether the final two bits agree or differ). Two input bits control two output properties without interfering — a one-to-one dictionary.

4.4 Decoding: run the circuit backwards

The reverse problem: someone hands you two qubits, guaranteed to be one of the four Bell states. Which one?

The naive approach fails. $|\Phi^+\rangle$ and $|\Phi^-\rangle$ differ only by a minus sign, and measuring directly returns the squared moduli of the components:

The minus sign is eaten by the modulus squared — the two states give identical statistics. This is the two-qubit reprise of “measuring Z is blind to longitude” from last time.

The fix: since the generating circuit $U=\mathrm{CNOT}\cdot(H\otimes I)$ maps the computational basis one-to-one onto the Bell basis, its inverse sends the Bell basis back to the computational basis. Quantum gates are always invertible, and the inverse circuit is the reverse order with each gate daggered:

(Both gates are their own inverses: $H^2=I$ and $\mathrm{CNOT}^2=I$.) The circuit is CNOT first, then H on the first qubit, then measure both — exactly the reverse of the generating circuit. Verifying by hand with $|\Phi^-\rangle$:

Reading out 10 ✓, matching the dictionary. The full result: $|\Phi^+\rangle\to00$, $|\Psi^+\rangle\to01$, $|\Phi^-\rangle\to10$, $|\Psi^-\rangle\to11$.

The point is the word “definite”: given one of the four Bell states, the output is not random — after decoding, the state lands exactly on a computational basis state, so the measurement necessarily gives a definite result. The identity of an entangled pair is translated losslessly into two classical bits. The operation is called a Bell measurement, and it is the core step of teleportation and superdense coding.

The general principle it embodies is the same one behind “measuring X ≡ padding an H and then measuring Z” from last time: if the information you want is not in the computational basis, rotate it there with a unitary gate first, then measure ordinarily.

4.5 The tap in the laboratory

So far every Bell state has been “built with gates in a circuit”. Optics labs have a physical process that produces entanglement directly: spontaneous parametric down-conversion (SPDC) — an intense laser (the pump, frequency $\omega_p$) strikes a special crystal, and occasionally one pump photon splits into two lower-energy photons, with energy conservation giving $\omega_p=\omega_1+\omega_2$.

Here is where the entanglement comes from. The two “daughter photons” emerge along two cones, photons on one cone vertically polarized (V) and on the other horizontally (H). The two cones intersect along two lines, and a photon coming out along an intersection line belongs to both cones at once, so its polarization has no definite value — it is a superposition of H and V; meanwhile conservation requires the two photons’ polarizations to be opposite. Put the two constraints together and the pair’s joint state is

With the encoding $|0\rangle=|H\rangle$ and $|1\rangle=|V\rangle$, this is precisely $|\Psi^+\rangle$.

Worth a pause: there is no CNOT and no gate here. Entanglement is not “two independent particles correlated by a gate afterwards” — it comes out naturally when one particle splits, with the conservation law itself as the generating mechanism. This source is the hardware behind most quantum key distribution demonstrations and Bell experiments, and an undergraduate teaching lab can build one.

5. Measuring two qubits

5.1 Measuring both: four projectors

The projective-measurement recipe from last time upgrades without a single change. The four projectors are four outer products:

The other three put a 1 in the 2nd, 3rd, or 4th diagonal slot. Hermitian and idempotent (the diagonal entries satisfy $1^2=1$, $0^2=0$) ✓. For $|\psi\rangle=(a_{00},a_{01},a_{10},a_{11})^T$:

and the post-measurement state is $|01\rangle$. Four components, four probabilities, two classical bits out — the Born rule with a wider bus, no new mechanism.

5.2 Measuring one: projector ⊗ identity

To measure qubit 1 without touching qubit 2, the instrument is $\Pi_0\otimes I$:

Branches whose first bit is 1 are zeroed; branches whose first bit is 0 survive untouched, along with their relative magnitudes and phases. Compare §5.1: measuring both qubits leaves one branch, measuring one leaves a whole group of branches.

Dividing by $\sqrt P$ is needed because projection cut away the incompatible branches and the length is no longer 1; renormalizing is the old habit. The recipe is unchanged; only $\Pi$ became $\Pi\otimes I$.

5.3 On a product state, measurement is local

Take $|\psi\rangle=\tfrac12(1,1,1,1)^T$ and measure qubit 1:

Suppose the result is 1; normalizing (with $\sqrt P=\tfrac{1}{\sqrt2}$):

Reading the result: the measured qubit 1 collapsed to $|1\rangle$, while qubit 2 is still in the superposition $|+\rangle$ — coherent, unmeasured, free to pass through more gates (measuring X on it gives + with certainty, not a random 0 or 1). And in the input $\tfrac12(1,1,1,1)^T=|+\rangle\otimes|+\rangle$ qubit 2 was already $|+\rangle$ — not a finger was laid on it, before or after. That is what “the violence of measurement lands only on what is measured” means, for product states.

5.4 On an entangled state, the far half settles

Run $|\Phi^+\rangle=\tfrac{1}{\sqrt2}(1,0,0,1)^T$ through the same procedure, measuring qubit 1:

(A result of 1 gives $|1\rangle\otimes|1\rangle$ the same way.)

Compare §5.3: there qubit 2 was $|+\rangle$ before and after, untouched; here only qubit 1 was measured — even a kilometre away — and qubit 2 becomes a definite $|0\rangle$ on the spot, with the two results perfectly correlated forever. The same formula: gentle and local on a product state, reaching across space to change the other side’s situation on an entangled one.

One preemptive clarification: this cannot send a message faster than light — looking only at qubit 2’s local statistics, whether or not the other side measured and whatever they got, the answer is 50/50. The full proof has to wait for §7.8, because it needs a tool that does not exist yet.

5.5 The open case: what qubit 2 was before the measurement

The claim that qubit 2 has no definite state of its own is not rhetoric but a verifiable assertion. For $|\Phi^+\rangle$:

• measuring Z on qubit 2: $P(0)=|a_{00}|^2+|a_{10}|^2=\tfrac12$;
• expanding in the X basis (substituting $|0\rangle=\tfrac{1}{\sqrt2}(|+\rangle+|-\rangle)$ and $|1\rangle=\tfrac{1}{\sqrt2}(|+\rangle-|-\rangle)$):

measuring X on qubit 2 gives $P(+)=\big|\tfrac12\big|^2+\big|\tfrac12\big|^2=\tfrac12$. The Y basis behaves the same way, again 50/50.

Every pure state has a Bloch direction of its own, and measuring along that direction gives a certain result. Qubit 2 answers 50/50 in every direction — so no pure state can describe it. It is not $|0\rangle$, not $|+\rangle$, not any point on the sphere.

When the previous notes distinguished ”$|+\rangle$ (on the surface, quantum superposition)” from “the centre (classical ignorance)”, the centre was still an abstract role. Its occupant has now been found: the local part of an entangled whole lives at the centre of the sphere. Giving that “part” a legitimate language requires a container bigger than a ket.

6. Density matrices: what a ket cannot hold

6.0 The new tool: the trace

The only new piece of mathematics in these notes. The trace of a matrix is the sum of its diagonal entries:

For example $\mathrm{Tr}\big[|0\rangle\langle0|\big]=1+0=1$ and $\mathrm{Tr}\big[|+\rangle\langle+|\big]=\tfrac12+\tfrac12=1$. The property needed below is cyclicity: $\mathrm{Tr}[AB]=\mathrm{Tr}[BA]$. That is all.

6.1 A puzzle: two machines

Picture two machines, each emitting a stream of qubits:

• Machine P (pure): every qubit is prepared in the same state $|+\rangle$.
• Machine M (mixed): each time it first flips a hidden fair coin — heads prepares $|0\rangle$, tails prepares $|1\rangle$.

Measure both streams in the computational basis. Each of machine P’s $|+\rangle$ gives 0 or 1 with probability $\tfrac12$ by the Born rule; half of machine M’s qubits are $|0\rangle$ (certain to read 0) and half are $|1\rangle$ (certain to read 1), which is still 50/50 statistically. The statistics are identical, yet the two situations are physically nothing alike:

The trouble is that machine M cannot be described by a ket at all. ”$|0\rangle$ or $|1\rangle$, 50% each” is not a ket, and forcing it into $\tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)$ is simply wrong — that is $|+\rangle$, which is machine P.

The open case of §5.5 needs exactly the same container.

6.2 The density operator

Definition: an ensemble that “is in state $|\psi_i\rangle$ with classical probability $p_i$” is described by a density operator (also called a density matrix):

The division of labour is clean: the outer product $|\psi_i\rangle\langle\psi_i|$ holds the quantum superposition (the amplitudes and phases inside each ket), and the real weight $p_i$ holds the classical probability (we do not know which one was drawn). Two kinds of uncertainty are stored in one object, in separate layers, without being confused.

Pure states are the special case: with a single term, $\rho=|\psi\rangle\langle\psi|$. Anything requiring more terms is a mixed state. Note that the $|\psi_i\rangle$ in the definition are not required to be orthogonal.

The two machines’ density matrices:

6.3 The diagonal is probability, the off-diagonal is phase

The two matrices differ in exactly two slots. Diagonal entries = measurement probabilities in the computational basis: $P(0)=\rho_{00}$, $P(1)=\rho_{11}$, and both matrices have diagonal $(\tfrac12,\tfrac12)$ — that is the mathematical origin of “identical statistics”. Off-diagonal entries = coherences, recording the phase relation between $|0\rangle$ and $|1\rangle$: $\rho_P$ has $\tfrac12$ off the diagonal (a genuine coherent superposition), $\rho_M$ has 0 (there never were “two components”, so phase is meaningless).

What a general pure state’s $\rho$ looks like is most convincing when computed yourself. With $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$:

The diagonal holds probabilities (moduli squared, phase eaten), and the off-diagonal holds $\alpha\beta^*$ — exactly the angle between the two arrows, the relative phase the first installment chased for a whole article. Here it finally gets a slot of its own.

This also makes an important phenomenon literally visible. The number one enemy of a real quantum computer is decoherence — the environment keeps “peeking” at the qubit and phase leaks into the surroundings. On the matrix that is precisely the off-diagonal entries decaying to 0, $\rho_P$ slowly turning into $\rho_M$. That is where the name comes from.

6.4 Different recipes, the same $\rho$

Compute one more recipe: flip a coin to decide between preparing $|+\rangle$ and $|-\rangle$. The outer product is $|-\rangle\langle-|=\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$, so

Exactly machine M. Two utterly different preparation procedures give the same $\rho$.

This is not a defect but a feature: ensembles with the same $\rho$ cannot be told apart by any measurement, since every measurement probability is determined by $\rho$. So $\rho$ holds exactly “everything knowable, and nothing more”, discarding the surplus history automatically. It is the same principle that discarded the global phase last time: a difference that cannot be measured is not a physical difference.

6.5 Credentials and purity

Not every matrix qualifies as a $\rho$. Three credentials plus a measuring stick:

① Hermitian: $\rho^\dagger=\rho$, which follows from the definition (each outer product is Hermitian and the weights are real). Consequence: all eigenvalues are real.

② Unit trace: $\mathrm{Tr}[\rho]=1$. The diagonal entries are probabilities of outcomes, and the total must be 1. Check: $\mathrm{Tr}[\rho_P]=\tfrac12+\tfrac12=1$ ✓ and $\mathrm{Tr}[I/2]=1$ ✓

③ Positive semidefinite: all eigenvalues $\ge0$. Hermiticity only guarantees real eigenvalues, but probabilities cannot be negative. Together ② and ③ say that the eigenvalues of $\rho$ themselves form a probability distribution.

④ Purity: $\mathrm{Tr}[\rho^2]$, a measuring stick for how mixed a state is. With eigenvalues $\lambda_i$, $\mathrm{Tr}[\rho^2]=\sum_i\lambda_i^2$; since $\sum\lambda_i=1$ with all terms non-negative, the sum of squares is largest when one $\lambda=1$ and the rest vanish (value 1, a pure state) and smallest when they are equal (for one qubit $\lambda_1=\lambda_2=\tfrac12$, value $\tfrac12$). For Hermitian matrices there is a convenient shortcut: $\mathrm{Tr}[\rho^2]=\sum_{i,j}|\rho_{ij}|^2$, the sum of the squared moduli of all entries. Check:

That is why $I/2$ is called the maximally mixed state.

6.6 Every old formula has a new version

Each pure-state formula has a density-matrix twin:

Why is applying a gate the sandwich $U\rho U^\dagger$ rather than $U\rho$? Derive it: with $\rho=|\psi\rangle\langle\psi|$, the gate turns the ket into $U|\psi\rangle$, and since the bra is the dagger of the ket it becomes $\langle\psi|U^\dagger$; assembling gives $\rho'=U\rho U^\dagger$. The ket and the bra each take a share — that is where the sandwich comes from.

The new formulas strictly contain the old ones, provable in one line with cyclicity. For a pure $\rho=|\psi\rangle\langle\psi|$:

(Last step: $\langle\psi|O|\psi\rangle$ is a number, and the trace of a number is itself.) The upgrade is lossless; nothing you computed correctly before goes wrong under the new formulas.

What can turn a pure state mixed? Not gates — a unitary sandwich leaves the eigenvalues alone, so purity is constant. The real culprits are three: measuring and forgetting (measured but never looked at the result: $|+\rangle$ ends up $|0\rangle$ or $|1\rangle$ with 50% each and you do not know which, which is exactly machine M), decoherence (every interaction with the environment is an uninvited little measurement whose “result” you can never read), and looking at only half of an entangled pair — the subject of chapter 7.

6.7 The solid Bloch ball

The Bloch sphere of the first installment can now be upgraded. Any single-qubit $\rho$ can be written

and inside those brackets are precisely the three average readings measured by tomography last time. Writing $r=\sqrt{x^2+y^2+z^2}$ for the arrow’s length gives $\mathrm{Tr}[\rho^2]=\tfrac{1+r^2}{2}$, and hence a complete map:

Pure states live on the surface, mixed states inside — the Bloch sphere turns from a shell into a solid ball. The centre has a peculiar property: it is equidistant from every pure state, reads 50/50 in every basis, and commits to nothing. The two kinds of dynamics are cleanly separated in this picture: a unitary gate is a rigid rotation (the whole ball turns, every point keeping its $r$ — the geometric version of “gates do not change purity”), and decoherence is a pull inward ($r$ shrinks and the point slides from the surface toward the centre). The picture of a real quantum computer is exactly this: you navigate precisely on the surface with gates while the environment drags you toward the centre.

6.8 Case closed: the two machines can be told apart

Back to the puzzle of §6.1. Measuring in a different basis suffices — measure X, that is, ask “is it $|+\rangle$ or $|-\rangle$”. Using the new formula $P=\mathrm{Tr}[\Pi\rho]$ with $\Pi=|+\rangle\langle+|$:

Every one of machine P’s qubits is $|+\rangle$, so measuring X gives + with certainty; machine M’s $|0\rangle$ and $|1\rangle$ are each 50/50 in the X basis, and the mixture is still 50/50. The practical test: take 100 qubits and measure X — all + means machine P almost surely, half and half means machine M.

So the off-diagonal coherence is not a bookkeeping symbol but a measurable physical quantity — it simply cannot be read in the computational basis. Summarized as a principle these notes will invoke repeatedly:

7. The partial trace: what half of an entangled pair is

7.1 Bob’s problem

Alice and Bob share a pair of entangled qubits, Alice holding A and Bob holding B, possibly a galaxy apart. Bob has no idea what Alice is doing, not even whether she has measured. What should he use to describe the single qubit in his hands?

Not a ket; §5.5 verified that. But chapter 6 has just built the bigger container, so the answer has a shape: a density matrix, and generally a mixed one. What remains is to compute it.

7.2 The intuitive route: one worked example

Use an example that is deliberately not maximally entangled, so the numbers are interesting (convention: the first symbol is Alice’s A, the second is Bob’s B):

Regroup by Alice’s qubit, collecting the A = 0 terms into one pile (pulling out the common factor) and the A = 1 term into another, then normalizing the B state inside each bracket ($|0\rangle+|1\rangle$ has length $\sqrt2$, so pull it out):