Measuring a single qubit is straightforward — you learned that in W1. But what happens when you measure one qubit of a two-qubit system? Especially when the qubits are entangled?
This is where quantum mechanics gets genuinely counterintuitive. Measuring one qubit doesn't just give you a result — it changes the state of the other qubit, even though you never touched it. The tool for computing exactly what happens is the generalized Born rule.
To measure one qubit of a multi-qubit state in some basis \(\{|v\rangle, |v^\perp\rangle\}\): rewrite the state as \(\alpha|v\rangle \otimes |\phi_0\rangle + \beta|v^\perp\rangle \otimes |\phi_1\rangle\). Then the probabilities are \(|\alpha|^2\) and \(|\beta|^2\), and the post-measurement states are \(|v\rangle \otimes |\phi_0\rangle\) and \(|v^\perp\rangle \otimes |\phi_1\rangle\).
If you measure all qubits in the computational basis, the rule is the same as single-qubit measurement, just extended: each basis state \(\ket{x}\) has probability \(|\alpha_x|^2\), and after measurement the state collapses to \(\ket{x}\).
State: \(\ket{\phi} = \frac{1}{\sqrt{5}}\ket{00} + \frac{2}{\sqrt{5}}\ket{11}\)
$$P(\ket{00}) = \left|\frac{1}{\sqrt{5}}\right|^2 = \frac{1}{5} \qquad\qquad P(\ket{11}) = \left|\frac{2}{\sqrt{5}}\right|^2 = \frac{4}{5}$$
$$P(\ket{01}) = 0 \qquad\qquad P(\ket{10}) = 0$$
After measurement, the state is whichever basis state you got. No surprises here — this is just the Born rule applied to a 4D vector instead of a 2D one.
For a state \(\alpha_{00}\ket{00} + \alpha_{01}\ket{01} + \alpha_{10}\ket{10} + \alpha_{11}\ket{11}\):
Normalization guarantees: \(|\alpha_{00}|^2 + |\alpha_{01}|^2 + |\alpha_{10}|^2 + |\alpha_{11}|^2 = 1\).
Now the interesting question: what if you only measure the first qubit and leave the second alone? The second qubit's state depends on what you measured. This is where partial measurement gets subtle.
For a general two-qubit state \(\alpha_{00}\ket{00} + \alpha_{01}\ket{01} + \alpha_{10}\ket{10} + \alpha_{11}\ket{11}\), measuring the first qubit in the Z-basis \(\{\ket{0}, \ket{1}\}\):
Group terms by the first qubit's value:
$$\ket{\phi} = \underbrace{\ket{0} \otimes (\alpha_{00}\ket{0} + \alpha_{01}\ket{1})}_{\text{first qubit} = \ket{0}} + \underbrace{\ket{1} \otimes (\alpha_{10}\ket{0} + \alpha_{11}\ket{1})}_{\text{first qubit} = \ket{1}}$$
The \(\frac{1}{\sqrt{p}}\) factor renormalizes the remaining qubit's state so it has norm 1.
The key idea: group the terms by the measured qubit's value, then read off the probabilities and leftover states. The leftover state must be renormalized.
State: \(\ket{\phi} = \frac{1}{\sqrt{5}}\ket{00} + \frac{2}{\sqrt{5}}\ket{11}\)
Group by the first qubit:
$$\ket{\phi} = \frac{1}{\sqrt{5}}\ket{0} \otimes \ket{0} + \frac{2}{\sqrt{5}}\ket{1} \otimes \ket{1}$$
This is already grouped! The first qubit is \(\ket{0}\) in the first term and \(\ket{1}\) in the second.
Outcome 0: \(\quad p_0 = \left|\frac{1}{\sqrt{5}}\right|^2 = \frac{1}{5}\). Post-measurement state: \(\ket{0} \otimes \ket{0} = \ket{00}\)
Outcome 1: \(\quad p_1 = \left|\frac{2}{\sqrt{5}}\right|^2 = \frac{4}{5}\). Post-measurement state: \(\ket{1} \otimes \ket{1} = \ket{11}\)
Notice: measuring the first qubit completely determines the second. Get 0? Second qubit is \(\ket{0}\). Get 1? Second qubit is \(\ket{1}\). That's entanglement in action — you already saw this on the previous page.
What if you want to measure in a different basis — not the Z-basis, but the X-basis \(\{\ket{+}, \ket{-}\}\), or any orthonormal basis \(\{|v\rangle, |v^\perp\rangle\}\)? This is where the generalized Born rule comes in.
To measure the first qubit of an \(n\)-qubit state \(\ket{\phi}\) in basis \(\{|v\rangle, |v^\perp\rangle\}\):
Step 1: Rewrite the state so the first qubit is expressed in the measurement basis:
$$\ket{\phi} = \alpha\,|v\rangle \otimes |\phi_0\rangle + \beta\,|v^\perp\rangle \otimes |\phi_1\rangle$$
where \(|\phi_0\rangle\) and \(|\phi_1\rangle\) are \((n-1)\)-qubit states, and \(|\alpha|^2 + |\beta|^2 = 1\).
Step 2: Read off the measurement outcomes:
The Z-basis case above is just a special case where \(|v\rangle = \ket{0}\) and \(|v^\perp\rangle = \ket{1}\). The general rule works for any orthonormal basis.
The hard part is Step 1 — rewriting the state. For the Z-basis it's trivial (just group terms). For other bases, you need to change basis on the first qubit by substituting \(\ket{0}\) and \(\ket{1}\) in terms of \(|v\rangle\) and \(|v^\perp\rangle\), then collecting terms.
Same state: \(\ket{\phi} = \frac{1}{\sqrt{5}}\ket{00} + \frac{2}{\sqrt{5}}\ket{11}\). Now measure the first qubit in the X-basis \(\{\ket{+}, \ket{-}\}\).
The Born rule only works when the state is written in the basis you're measuring in. Our state uses \(\ket{0}\) and \(\ket{1}\), but we're measuring in \(\{\ket{+}, \ket{-}\}\). So we need to express \(\ket{0}\) and \(\ket{1}\) in terms of \(\ket{+}\) and \(\ket{-}\).
You already know: \(\ket{+} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1})\) and \(\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})\). Just invert these with algebra — add them:
$$\ket{+} + \ket{-} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1}) + \frac{1}{\sqrt{2}}(\ket{0} - \ket{1}) = \frac{2}{\sqrt{2}}\ket{0} = \sqrt{2}\,\ket{0}$$
$$\Rightarrow\quad \ket{0} = \frac{1}{\sqrt{2}}(\ket{+} + \ket{-})$$
Subtract them:
$$\ket{+} - \ket{-} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1}) - \frac{1}{\sqrt{2}}(\ket{0} - \ket{1}) = \frac{2}{\sqrt{2}}\ket{1} = \sqrt{2}\,\ket{1}$$
$$\Rightarrow\quad \ket{1} = \frac{1}{\sqrt{2}}(\ket{+} - \ket{-})$$
Now substitute these into the first qubit of each term:
$$\ket{\phi} = \frac{1}{\sqrt{5}} \cdot \frac{\ket{+} + \ket{-}}{\sqrt{2}} \otimes \ket{0} \;+\; \frac{2}{\sqrt{5}} \cdot \frac{\ket{+} - \ket{-}}{\sqrt{2}} \otimes \ket{1}$$
Expand and group by \(\ket{+}\) and \(\ket{-}\):
$$= \frac{1}{\sqrt{10}}\ket{+}\ket{0} + \frac{1}{\sqrt{10}}\ket{-}\ket{0} + \frac{2}{\sqrt{10}}\ket{+}\ket{1} - \frac{2}{\sqrt{10}}\ket{-}\ket{1}$$
$$= \ket{+} \otimes \left(\frac{1}{\sqrt{10}}\ket{0} + \frac{2}{\sqrt{10}}\ket{1}\right) + \ket{-} \otimes \left(\frac{1}{\sqrt{10}}\ket{0} - \frac{2}{\sqrt{10}}\ket{1}\right)$$
Now factor out the norms to get the \(\alpha\) and \(\beta\) coefficients. Both inner states have the same norm:
$$\left\|\frac{1}{\sqrt{10}}\ket{0} + \frac{2}{\sqrt{10}}\ket{1}\right\| = \sqrt{\frac{1}{10} + \frac{4}{10}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$
So we can write:
$$\ket{\phi} = \frac{1}{\sqrt{2}}\ket{+} \otimes \underbrace{\left(\frac{1}{\sqrt{5}}\ket{0} + \frac{2}{\sqrt{5}}\ket{1}\right)}_{|\phi_0\rangle} + \frac{1}{\sqrt{2}}\ket{-} \otimes \underbrace{\left(\frac{1}{\sqrt{5}}\ket{0} - \frac{2}{\sqrt{5}}\ket{1}\right)}_{|\phi_1\rangle}$$
Now we have the form \(\alpha\ket{+} \otimes |\phi_0\rangle + \beta\ket{-} \otimes |\phi_1\rangle\) with \(\alpha = \beta = \frac{1}{\sqrt{2}}\).
Outcome \(\ket{+}\): \(\quad p_+ = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\). Post-measurement state: \(\ket{+} \otimes \left(\frac{1}{\sqrt{5}}\ket{0} + \frac{2}{\sqrt{5}}\ket{1}\right)\)
Outcome \(\ket{-}\): \(\quad p_- = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\). Post-measurement state: \(\ket{-} \otimes \left(\frac{1}{\sqrt{5}}\ket{0} - \frac{2}{\sqrt{5}}\ket{1}\right)\)
Z-basis measurement: probabilities \(\frac{1}{5}\) and \(\frac{4}{5}\). Second qubit collapses to \(\ket{0}\) or \(\ket{1}\) — a clean basis state.
X-basis measurement: probabilities \(\frac{1}{2}\) and \(\frac{1}{2}\). Second qubit collapses to \(\frac{1}{\sqrt{5}}\ket{0} + \frac{2}{\sqrt{5}}\ket{1}\) or \(\frac{1}{\sqrt{5}}\ket{0} - \frac{2}{\sqrt{5}}\ket{1}\) — a superposition.
The choice of measurement basis changes both the probabilities AND what state the unmeasured qubit ends up in. This is a purely quantum phenomenon.
The algebra in Step 1 is where most mistakes happen on exams. Here's the systematic procedure:
Let's see how partial measurement works on a product state to build intuition.
State: \(\ket{\phi} = \ket{+} \otimes \ket{0} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{10})\)
Measure the first qubit in the Z-basis:
$$\ket{\phi} = \frac{1}{\sqrt{2}}\ket{0} \otimes \ket{0} + \frac{1}{\sqrt{2}}\ket{1} \otimes \ket{0}$$
Outcome 0: \(\quad p_0 = \frac{1}{2}\). Post-measurement state: \(\ket{0} \otimes \ket{0} = \ket{00}\)
Outcome 1: \(\quad p_1 = \frac{1}{2}\). Post-measurement state: \(\ket{1} \otimes \ket{0} = \ket{10}\)
The second qubit is \(\ket{0}\) regardless of the outcome. Measuring the first qubit told you nothing about the second. This is exactly what "not entangled" means — the qubits are independent.
Product state: measuring one qubit leaves the other unchanged. The \(|\phi_0\rangle\) and \(|\phi_1\rangle\) parts are the same.
Entangled state: measuring one qubit changes the other. The \(|\phi_0\rangle\) and \(|\phi_1\rangle\) parts are different — the measurement outcome tells you something about the unmeasured qubit.
The generalized Born rule works for any number of qubits. Let's measure the first qubit of a 3-qubit state.
State: \(\ket{\phi} = \frac{1}{\sqrt{2}}\ket{000} + \frac{1}{\sqrt{2}}\ket{111}\) (the GHZ state)
Group by the first qubit:
$$\ket{\phi} = \frac{1}{\sqrt{2}}\ket{0} \otimes \ket{00} + \frac{1}{\sqrt{2}}\ket{1} \otimes \ket{11}$$
Outcome 0: \(\quad p_0 = \frac{1}{2}\). Post-measurement state: \(\ket{0} \otimes \ket{00} = \ket{000}\)
Outcome 1: \(\quad p_1 = \frac{1}{2}\). Post-measurement state: \(\ket{1} \otimes \ket{11} = \ket{111}\)
Measuring just the first qubit collapses all three qubits. This is the power (and the strangeness) of multi-qubit entanglement: one measurement can determine the state of an entire register.