From Entanglement as a Formula to Entanglement as Behavior
For a general pure two-qubit state
\[|\Psi\rangle=r|00\rangle+s|01\rangle+t|10\rangle+u|11\rangle,\]
we learned that
\[ru=st\]
is the factorability condition for a product state, while
\[ru\neq st\]
means the pure joint state is entangled.
That test answers one question:
Is the joint state separable into an individual pure state for Alice tensor an individual pure state for Bob?
But once we know the state is entangled, another question becomes more interesting:
What do Alice and Bob actually observe when they measure their qubits?
Start with the Bell State
Consider the Bell state
\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}.\]
The first qubit belongs to Alice and the second to Bob.
The state contains only two computational-basis possibilities:
\[|00\rangle\]
and
\[|11\rangle.\]
Each has amplitude \(1/\sqrt{2}\), so each has probability \(1/2\).
If Alice Measures in the 0/1 Basis
Suppose Alice measures only her qubit in the computational basis.
If Alice obtains \(0\), the only term compatible with that result is
\[|00\rangle.\]
Conditioned on Alice’s result, the joint state becomes
\[|\Phi^+\rangle\longrightarrow|00\rangle.\]
Therefore, if Bob now measures in the same basis, he obtains \(0\) with probability 1.
If Alice instead obtains \(1\), the compatible term is
\[|11\rangle,\]
and Bob will obtain \(1\) with probability 1 if he measures in the same basis.
Alice cannot choose which outcome she gets. Her own result is random:
\[P(A=0)=\frac12,\qquad P(A=1)=\frac12.\]
The key pattern is therefore:
Each individual result is random, but the two results are perfectly correlated when measured in the same computational basis.
What Does Bob See by Himself?
Suppose Alice and Bob are very far apart and Alice measures her qubit but does not tell Bob the result.
Bob still sees
\[P(B=0)=\frac12,\qquad P(B=1)=\frac12.\]
His local results look completely random.
He cannot determine from his own measurements whether Alice measured her qubit, what result she obtained, or whether she did anything at all.
The correlation becomes visible only when Alice and Bob later compare their results through ordinary classical communication.
This is why entanglement cannot by itself be used as a controllable faster-than-light communication channel.
Changing the Measurement Basis
Now consider another orthonormal basis:
\[|+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}\]
and
\[|-\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2}.\]
The inverse relations are
\[|0\rangle=\frac{|+\rangle+|-\rangle}{\sqrt2}\]
and
\[|1\rangle=\frac{|+\rangle-|-\rangle}{\sqrt2}.\]
Changing basis does not physically change the Bell state. It only expresses the same state using a different set of coordinates and corresponds to asking a different measurement question.
Rewriting the Bell State in the +/- Basis
Start with
\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}.\]
Expanding \(|00\rangle\) in the \(+/-\) basis gives
\[|00\rangle=\frac12\left(|++\rangle+|+-\rangle+|-+\rangle+|–\rangle\right).\]
Similarly,
\[|11\rangle=\frac12\left(|++\rangle-|+-\rangle-|-+\rangle+|–\rangle\right).\]
Adding the two expressions causes the mixed terms to cancel:
\[|+-\rangle-|+-\rangle=0\]
and
\[|-+\rangle-|-+\rangle=0.\]
The surviving terms give
\[|\Phi^+\rangle=\frac{|++\rangle+|–\rangle}{\sqrt2}.\]
This is the same Bell state written in a different basis.
Both Measure in the +/- Basis
The new representation makes the probabilities immediate:
\[P(++)=\frac12,\qquad P(–)=\frac12.\]
Meanwhile,
\[P(+-)=0,\qquad P(-+)=0.\]
So if Alice and Bob both measure in the \(+/-\) basis, their individual results are again random, but their results always agree.
This gives a striking pattern:
Same 0/1 basis → perfect correlation.
Same +/- basis → perfect correlation.
Now Let Alice and Bob Use Different Bases
Suppose Alice measures in the \(0/1\) basis while Bob measures in the \(+/-\) basis.
To analyze this, keep Alice’s basis unchanged and rewrite only Bob’s part.
For the first Bell term,
\[|00\rangle=|0\rangle_A\otimes|0\rangle_B.\]
Using
\[|0\rangle_B=\frac{|+\rangle_B+|-\rangle_B}{\sqrt2},\]
we obtain
\[|00\rangle=\frac{|0+\rangle+|0-\rangle}{\sqrt2}.\]
Similarly,
\[|11\rangle=\frac{|1+\rangle-|1-\rangle}{\sqrt2}.\]
Substituting both into the Bell state gives
\[|\Phi^+\rangle=\frac12\left(|0+\rangle+|0-\rangle+|1+\rangle-|1-\rangle\right).\]
The Four Different-Basis Outcomes
The amplitudes are
\[\frac12,\quad\frac12,\quad\frac12,\quad-\frac12.\]
Probabilities are obtained from squared magnitudes, so
\[\left|\frac12\right|^2=\frac14\]
and
\[\left|-\frac12\right|^2=\frac14.\]
Therefore all four outcomes occur with equal probability:
\[P(0,+)=\frac14\]
\[P(0,-)=\frac14\]
\[P(1,+)=\frac14\]
\[P(1,-)=\frac14.\]
If Alice Gets 0, What Can She Predict About Bob?
Condition on Alice having obtained \(0\).
The compatible possibilities are
\[|0+\rangle\]
and
\[|0-\rangle.\]
They are equally likely.
Therefore
\[P(B=+\mid A=0)=\frac12\]
and
\[P(B=-\mid A=0)=\frac12.\]
The same is true if Alice obtains \(1\):
\[P(B=+\mid A=1)=\frac12,\qquad P(B=-\mid A=1)=\frac12.\]
So Alice’s computational-basis result gives her no predictive advantage for Bob’s \(+/-\)-basis measurement.
The Pattern So Far
We can summarize the behavior of the Bell state as follows:
Alice 0/1, Bob 0/1: perfect correlation.
Alice +/-, Bob +/-: perfect correlation.
Alice 0/1, Bob +/-: all four combinations occur equally, producing 50/50 conditional probabilities.
This shows that measurement statistics depend not only on whether the state is entangled, but also on the measurement bases chosen by Alice and Bob.
Entanglement Test and Measurement Probabilities Are Different Questions
The condition
\[ru\neq st\]
tells us that a pure two-qubit state is entangled.
It does not directly tell us the probabilities of arbitrary measurement outcomes.
Those probabilities depend on how the state is expressed relative to the measurement bases being used.
It is therefore useful to separate two layers:
Layer 1: What kind of state do Alice and Bob share?
For a pure two-qubit state, the factorability condition answers this.
Layer 2: What results will Alice and Bob obtain for particular measurement choices?
This requires expressing the state in the relevant measurement bases and calculating the corresponding amplitudes and probabilities.
Where This Leads Next
So far we have compared measurements in the same basis and in two very different bases.
The next question is more subtle:
What happens when Alice and Bob choose measurement bases separated by intermediate angles?
The correlations then become neither perfectly correlated nor simply 50/50.
Quantum mechanics predicts a specific angle-dependent pattern. Understanding that pattern is the bridge from basic entanglement to Bell’s theorem and Bell inequalities, where the possibility of predetermined local hidden answers can be tested experimentally.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Quantum Computing for Everyone