Starting Point: Two Unentangled Qubits
Bernhardt starts with the tensor product
\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}\otimes\begin{bmatrix}1\\0\end{bmatrix}.\]
Because the state is explicitly written as one qubit tensor-product another qubit, the two qubits are initially unentangled.
Why Is the First Qubit a Superposition?
The first qubit is
\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}.\]
Using the standard basis
\[|0\rangle=\begin{bmatrix}1\\0\end{bmatrix},\qquad |1\rangle=\begin{bmatrix}0\\1\end{bmatrix},\]
we can write
\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\end{bmatrix}+\frac{1}{\sqrt{2}}\begin{bmatrix}0\\1\end{bmatrix}.\]
Therefore
\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}=\frac{|0\rangle+|1\rangle}{\sqrt{2}}.\]
The first qubit is therefore in an equal superposition of |0⟩ and |1⟩. If measured in the standard basis, each outcome has probability 1/2.
The factor \(1/\sqrt{2}\) is important. The vector \([1,1]^T\) alone has length \(\sqrt{2}\), whereas a quantum state must be normalized.
The Second Qubit
The second qubit is
\[\begin{bmatrix}1\\0\end{bmatrix}=|0\rangle.\]
So the starting state can be written as
\[\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)\otimes|0\rangle.\]
Calculate the Tensor Product
For two vectors,
\[\begin{bmatrix}a\\b\end{bmatrix}\otimes\begin{bmatrix}c\\d\end{bmatrix}=\begin{bmatrix}ac\\ad\\bc\\bd\end{bmatrix}.\]
Here
\[a=b=\frac{1}{\sqrt{2}},\qquad c=1,\qquad d=0.\]
Therefore
\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}\otimes\begin{bmatrix}1\\0\end{bmatrix}=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\\1\\0\end{bmatrix}.\]
Using the ordered tensor-product basis
\[|00\rangle,\ |01\rangle,\ |10\rangle,\ |11\rangle,\]
this is
\[\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|10\rangle.\]
This makes intuitive sense: the first qubit is in a superposition of 0 and 1, while the second qubit is 0. The possible basis states are therefore |00⟩ and |10⟩.
Send the Pair Through CNOT
Bernhardt uses the CNOT matrix
\[\operatorname{CNOT}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}.\]
Apply it to the four-dimensional input state:
\[\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\\1\\0\end{bmatrix}=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\\0\\1\end{bmatrix}.\]
In ket notation,
\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\\0\\1\end{bmatrix}=\frac{|00\rangle+|11\rangle}{\sqrt{2}}.\]
What Did CNOT Actually Do?
Before CNOT the two nonzero possibilities were
\[|00\rangle\quad\text{and}\quad|10\rangle.\]
CNOT leaves |00⟩ unchanged because the control qubit is 0:
\[|00\rangle\rightarrow|00\rangle.\]
For |10⟩, the control qubit is 1, so CNOT flips the second qubit:
\[|10\rangle\rightarrow|11\rangle.\]
Consequently
\[\frac{|00\rangle+|10\rangle}{\sqrt{2}}\longrightarrow\frac{|00\rangle+|11\rangle}{\sqrt{2}}.\]
Check That the Output Is Entangled
Bernhardt writes a general two-qubit state as
\[r|00\rangle+s|01\rangle+t|10\rangle+u|11\rangle.\]
For a pure two-qubit state in this form, his test is
\[ru=st\quad\Rightarrow\quad\text{unentangled},\]
and
\[ru\neq st\quad\Rightarrow\quad\text{entangled}.\]
For the CNOT output
\[\frac{|00\rangle+|11\rangle}{\sqrt{2}},\]
the 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
\[st=0.\]
Since \(ru\neq st\), the output qubits are entangled.
The Bell State
The resulting state
\[\frac{|00\rangle+|11\rangle}{\sqrt{2}}\]
is a Bell state. If both qubits are measured in the standard basis, the only possible joint outcomes are 00 and 11, each with probability 1/2. The outcomes 01 and 10 have probability zero.
Core Idea
For Bernhardt’s particular input, the process is:
\[\text{superposition on the first qubit}+\text{CNOT}\longrightarrow\text{entangled pair}.\]
More explicitly:
\[\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)\otimes|0\rangle\xrightarrow{\mathrm{CNOT}}\frac{|00\rangle+|11\rangle}{\sqrt{2}}.\]
The important conceptual change is that before CNOT the state can be written as a tensor product of two individual qubit states. After CNOT, the resulting Bell state cannot be separated into one independent state for the first qubit and another independent state for the second.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Quantum Computing for Everyone