The Question That Leads to Tensor Products
Suppose Alice has one qubit and Bob has another.
Alice’s qubit can be written as
\[|A\rangle=a|0\rangle+b|1\rangle.\]
Bob’s qubit can be written as
\[|B\rangle=c|0\rangle+d|1\rangle.\]
We already know how to describe Alice’s qubit and Bob’s qubit individually. A new question now appears:
How do we mathematically describe Alice and Bob together as one quantum system?
The tensor product is the mathematical operation that allows us to construct this joint description.
Why Two Qubits Need Four Basis States
Alice has two basis possibilities, \(|0\rangle_A\) and \(|1\rangle_A\). Bob independently has two basis possibilities, \(|0\rangle_B\) and \(|1\rangle_B\).
When the systems are considered together, every Alice possibility can occur with every Bob possibility. The joint basis is therefore
\[|00\rangle,\quad |01\rangle,\quad |10\rangle,\quad |11\rangle.\]
For example,
\[|01\rangle=|0\rangle_A\otimes|1\rangle_B.\]
The first position refers to Alice and the second to Bob.
This explains why the dimensions multiply:
\[2\times2=4.\]
More generally, if one system has dimension \(m\) and another has dimension \(n\), their joint tensor-product space has dimension \(mn\).
Calculating a Tensor Product
Suppose two vectors are
\[A=\begin{bmatrix}a\\b\end{bmatrix},\qquad B=\begin{bmatrix}c\\d\end{bmatrix}.\]
Their tensor product is
\[A\otimes B=\begin{bmatrix}ac\\ad\\bc\\bd\end{bmatrix}.\]
The four entries correspond respectively to the four joint basis states
\[|00\rangle,\quad|01\rangle,\quad|10\rangle,\quad|11\rangle.\]
The rule can be understood as taking each entry of the first vector and multiplying it by the entire second vector.
A Simple Tensor Product Example
Take
\[|0\rangle=\begin{bmatrix}1\\0\end{bmatrix},\qquad |1\rangle=\begin{bmatrix}0\\1\end{bmatrix}.\]
Then
\[|0\rangle_A\otimes|1\rangle_B=\begin{bmatrix}1\\0\end{bmatrix}\otimes\begin{bmatrix}0\\1\end{bmatrix}.\]
Calculating the tensor product gives
\[\begin{bmatrix}1(0)\\1(1)\\0(0)\\0(1)\end{bmatrix}=\begin{bmatrix}0\\1\\0\\0\end{bmatrix}.\]
The standard two-qubit basis vectors are
\[|00\rangle=\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\quad |01\rangle=\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\quad |10\rangle=\begin{bmatrix}0\\0\\1\\0\end{bmatrix},\quad |11\rangle=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}.\]
Therefore
\[|0\rangle_A\otimes|1\rangle_B=|01\rangle.\]
Tensor Products of Superpositions
Now suppose
\[|A\rangle=a|0\rangle+b|1\rangle\]
and
\[|B\rangle=c|0\rangle+d|1\rangle.\]
The tensor product expands in the same way as ordinary algebraic brackets:
\[|A\rangle\otimes|B\rangle=ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle.\]
Thus the four amplitudes of the joint state are
\[ac,\quad ad,\quad bc,\quad bd.\]
The tensor product therefore combines the amplitudes of the individual systems into amplitudes for all possible joint outcomes.
The Tensor-Product Space Is Larger Than the Set of Product States
A general pure state of two qubits can be written as
\[|\Psi\rangle=r|00\rangle+s|01\rangle+t|10\rangle+u|11\rangle,\]
where
\[|r|^2+|s|^2+|t|^2+|u|^2=1.\]
An important question now appears:
Can every such four-dimensional state be produced by taking one Alice state tensor one Bob state?
The answer is no.
This is precisely where entanglement enters.
Product States
If the joint state can be written as
\[|\Psi\rangle_{AB}=|A\rangle\otimes|B\rangle,\]
then Alice and Bob each have their own individual pure state.
Such a state is called a product state and is not entangled.
For example,
\[|01\rangle=|0\rangle_A\otimes|1\rangle_B\]
is clearly a product state. Alice has \(|0\rangle\) and Bob has \(|1\rangle\).
Where the Condition ru = st Comes From
Suppose the general joint state
\[|\Psi\rangle=r|00\rangle+s|01\rangle+t|10\rangle+u|11\rangle\]
is actually a product state.
Then there must exist numbers \(a,b,c,d\) such that
\[|\Psi\rangle=(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle).\]
Expanding gives
\[|\Psi\rangle=ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle.\]
Therefore
\[r=ac,\qquad s=ad,\qquad t=bc,\qquad u=bd.\]
Now multiply the outer amplitudes:
\[ru=(ac)(bd)=abcd.\]
Multiply the inner amplitudes:
\[st=(ad)(bc)=abcd.\]
Therefore every pure two-qubit product state necessarily satisfies
\[ru=st.\]
This equality does not cause the qubits to be unentangled. It is simply the algebraic fingerprint produced when four amplitudes can be generated from one Alice vector tensor one Bob vector.
Why ru ≠ st Means Entanglement
If
\[ru\neq st,\]
then the amplitudes cannot possibly have the form
\[ac,\quad ad,\quad bc,\quad bd.\]
Any amplitudes having that product structure would automatically satisfy \(ru=st\).
Therefore, when \(ru\neq st\), there are no individual pure states \(|A\rangle\) and \(|B\rangle\) satisfying
\[|\Psi\rangle_{AB}=|A\rangle\otimes|B\rangle.\]
For a pure two-qubit state, this inability to factor the joint state into individual pure states is exactly what is meant by entanglement.
Example: The Bell State
Consider
\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}.\]
Its amplitudes are
\[r=\frac{1}{\sqrt{2}},\qquad s=0,\qquad t=0,\qquad u=\frac{1}{\sqrt{2}}.\]
Therefore
\[ru=\frac{1}{2}\]
while
\[st=0.\]
Hence
\[ru\neq st.\]
The Bell state therefore cannot be written as one individual pure state for Alice tensor one individual pure state for Bob. It is entangled.
What Entanglement Means Conceptually
The important point is not merely that Alice’s and Bob’s measurements may be correlated.
For a product pure state, the complete joint description can be separated:
\[\text{Alice’s pure state}\otimes\text{Bob’s pure state}.\]
For an entangled pure state, this separation is impossible. The pair has a well-defined joint pure state, but the joint pure state cannot be decomposed into individual pure state vectors for Alice and Bob.
This gives the useful conceptual definition:
For a bipartite pure state, entanglement means that the joint state cannot be factored into a tensor product of pure states of the individual subsystems.
Does ru ≠ st Experimentally Prove Entanglement?
The condition \(ru\neq st\) is a mathematical test, not itself a laboratory measurement.
If the pure state vector is already known, the condition proves mathematically that the state cannot be factored.
In an experiment, however, scientists do not directly look at particles and read the values \(r,s,t,u\). Instead, they prepare many equivalent quantum systems and perform measurements in appropriate bases. The resulting statistics are used to characterize or witness the state and determine whether the observed behavior is compatible with separable states.
Bell-type experiments go further by testing whether observed correlations can be reproduced by local hidden-variable theories satisfying Bell’s assumptions. Violations of Bell inequalities have provided powerful experimental evidence for genuinely quantum correlations.
Correlation Is Not Automatically Entanglement
Two classical systems can also have correlated outcomes. For example, two boxes could be prepared with matching colored cards. Opening one box may allow us to predict the contents of the other without any quantum entanglement.
Therefore simply observing that Alice and Bob often obtain matching results is not enough by itself to establish entanglement.
Entanglement concerns the quantum structure of the joint state and, experimentally, requires measurements capable of distinguishing that structure from appropriate separable or classical alternatives.
The Full Conceptual Chain
The ideas can now be connected in one sequence:
\[\text{one qubit}\rightarrow\text{two qubits}\rightarrow\text{need a joint description}\rightarrow\text{tensor-product space}.\]
For two qubits, the joint basis is
\[|00\rangle,|01\rangle,|10\rangle,|11\rangle.\]
Inside this four-dimensional space are two conceptually different kinds of pure states.
Product states:
\[|\Psi\rangle=|A\rangle\otimes|B\rangle.\]
Entangled states:
\[|\Psi\rangle\neq|A\rangle\otimes|B\rangle\]
for every possible choice of individual pure states \(|A\rangle\) and \(|B\rangle\).
For the general two-qubit pure state
\[r|00\rangle+s|01\rangle+t|10\rangle+u|11\rangle,\]
this distinction can be tested by
\[ru=st\quad\text{(product state)},\]
or
\[ru\neq st\quad\text{(entangled state)}.\]
The central insight is therefore that tensor products do not themselves mean entanglement. Tensor products give us the mathematical space needed to describe multiple quantum systems. Entanglement appears because that joint space contains valid states that cannot be separated into individual pure-state descriptions of the constituent systems.
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- Note type: learning-note
- Subject: learning-notes
- Source: Quantum Computing for Everyone