Quantum Information HT24, Composite systems and entanglement
Flashcards
What is a product state in a Hilbert space $\mathcal H _ {AB}$?
A tensor product of states from $\mathcal H _ A$ and $\mathcal H _ B$.
Suppose
\[\vert \Phi^+\rangle = \frac{ \vert 0\rangle \otimes \vert 0\rangle + \vert 1\rangle \otimes \vert 1\rangle}{\sqrt{2}
}\]
Why are states like these called an “entangled” state?
Because they are not product states, i.e. that cannot be written as the tensor product of two unit vectors.
Suppose we have orthonormal bases
\[\\{ \vert \alpha _ m\rangle \mid m = 0,1,\ldots,d _ A - 1\\}\]
and
\[\\{ \vert \beta _ n\rangle \mid n = 0,1,\ldots,d _ B - 1\\}\]
What is the product basis?
Suppose we have orthonormal bases
\[\\{ \vert \alpha _ m\rangle \mid m = 0,1,\ldots,d _ A - 1\\}\]
and
\[\\{ \vert \beta _ n\rangle \mid n = 0,1,\ldots,d _ B - 1\\}\]
Then the product basis is an orthonormal basis
\[\\{ \vert \alpha _ m\rangle \otimes \vert \beta _ n\rangle \mid m = 0,\ldots,d _ A - 1, n = 0,\ldots,d _ B - 1\\}\]
Is every orthonormal basis in the composite system a product basis?
No.
Suppose $U _ A$ and $U _ B$ are unitary matrices in the Hilbert spaces $\mathcal H _ A$ and $\mathcal H _ B$. How can we construct a unitary matrix in the Hilbert space $\mathcal H _ {AB}$?
If states in $\mathcal H _ {AB}$ that cannot be written as the product of two states in $\mathcal H _ A$ and $\mathcal H _ B$ are called entangling, what are the two types of gates $U$ (or equivalently, unitary matrices) that cannot be written as the tensor product of two unitary $U _ A$ and $U _ B$ matrices?
- Interaction gates in general
- Entangled gates if they transform product states into entangled states
Describe the action of the CNOT gate on the computational basis $ \vert a\rangle \otimes \vert b\rangle$.
Given the following equations
\[\begin{align*}
\text{CNOT} \vert +\rangle \otimes \vert 0\rangle &= \frac{ \vert 0\rangle \otimes \vert 0\rangle + \vert 1\rangle \otimes \vert 1\rangle}{\sqrt{2}\\,} = \vert \Phi^+\rangle \\\\
\text{CNOT} \vert -\rangle \otimes \vert 0\rangle &= \frac{ \vert 0\rangle \otimes \vert 0\rangle - \vert 1\rangle \otimes \vert 1\rangle}{\sqrt{2}\\,} = \vert \Phi^-\rangle \\\\
\text{CNOT} \vert +\rangle \otimes \vert 1\rangle &= \frac{ \vert 0\rangle \otimes \vert 1\rangle + \vert 1\rangle \otimes \vert 0\rangle}{\sqrt{2}\\,} = \vert \Psi^+\rangle \\\\
\text{CNOT} \vert -\rangle \otimes \vert 1\rangle &= \frac{ \vert 0\rangle \otimes \vert 1\rangle - \vert 1\rangle \otimes \vert 0\rangle}{\sqrt{2}\\,} = \vert \Psi^-\rangle
\end{align*}\]
where $\Phi^+$, $\Phi^-$, $\Psi^+$ and $\Psi^-$ represent the Bell basis, how can you measure an arbitrary state $ \vert \Psi\rangle$ on the Bell basis using only a product basis? Illustrate the case of determining the probability of $ \vert \Psi\rangle$ taking outcome $\Phi^+$
- First apply a CNOT gate to $ \vert \Psi\rangle$
- Measure on the product basis $\{ \vert +\rangle\otimes \vert 0\rangle, \vert +\rangle\otimes \vert 1\rangle, \vert -\rangle\otimes \vert 0\rangle, \vert -\rangle\otimes \vert 1\rangle\}$
In this particular example,
\[\begin{aligned} \vert \langle+ \vert \otimes\langle0 \vert \text{CNOT} \vert \Psi\rangle \vert ^2 &= \vert \langle\Phi^+ \vert \Psi\rangle \vert ^2 \end{aligned}\]Suppose $\mathcal H _ A$ and $\mathcal H _ B$ are the Hilbert spaces of two quantum systems, and $A$ undergoes the reversible process $U _ A$ and $B$ undergoes the reversible process $U _ B$. What happens to the system $AB$?
It undeergoes the reversible process $U _ A \otimes U _ B$.
Assuming $\alpha \ne 0$, how could you show that the state
\[\frac{1}{\sqrt{1 + \vert \alpha \vert ^2}
} \Big( \vert 0\rangle _ A \otimes (\beta \vert 0\rangle _ B + \alpha \vert 1\rangle _ B) + \alpha \vert 1\rangle _ A \otimes \vert 0\rangle _ B\Big)\]
is entangled?
Consider measuring in the computational basis. Then $p(11) = 0$, but $p(10) = \alpha$. Hence the outcomes are correlated, and it can’t be a product state.
Suppose
\[\vert \psi\rangle := a \vert 00\rangle + b \vert 01\rangle + c \vert 10\rangle + d \vert 11\rangle\]
What’s a quick trick for working out if $ \vert \psi\rangle$ is entangled?
Form the coefficient matrix
\[\begin{pmatrix} a & b \\\\ c & d \end{pmatrix}\]Then if the determinant is zero, the state is a product state.
What does it mean for two vectors $ \vert \alpha\rangle, \vert \beta\rangle$ to be proportional to one another, i.e.
\[\vert \alpha\rangle \propto \vert \beta\rangle\]
?
There exists $\lambda$, $\mu \in \mathbb C \setminus \{0\}$ such that
\[\lambda \vert \alpha\rangle = \mu \vert \beta\rangle\]Suppose $ \vert \psi\rangle _ {AB}$ is a state in a composite quantum system $AB$. Can you give a helpful necessary and sufficient condition for it to be a product state?
$ \vert \psi\rangle _ {AB}$ is a product state iff for all vectors $ \vert \alpha\rangle _ A, \vert \alpha’\rangle _ A \in A$
\[({} _ A \langle \alpha \vert \otimes I _ B) \vert \psi\rangle _ {AB} \propto ({} _ A\langle \alpha' \vert \otimes I _ B) \vert \psi\rangle _ {AB}\]Quickly prove that $ \vert \psi\rangle _ {AB}$ is a product state iff for all vectors $ \vert \alpha\rangle _ A, \vert \alpha’\rangle _ A \in A$
\[({} _
A \langle \alpha \vert \otimes I _
B) \vert \psi\rangle _
{AB} \propto ({} _
A\langle \alpha' \vert \otimes I _
B) \vert \psi\rangle _
{AB}\]
Product state implies proportional: Suppose $ \vert \psi\rangle _ {AB}$ is a product state, so $\exists \vert \phi\rangle _ A, \vert \omega\rangle _ B$ such that
\[\vert \psi\rangle _ {AB} = \vert \phi\rangle _ A \otimes \vert \omega\rangle _ B\]Then $\forall \vert \alpha\rangle _ A, \vert \alpha’\rangle _ A$, we have
\[({} _ A \langle \alpha \vert \otimes I _ B) \vert \psi\rangle _ {AB} = {} _ A\langle \alpha \vert \phi\rangle _ A \vert \omega\rangle _ B\]and
\[({} _ A \langle \alpha' \vert \otimes I _ B) \vert \psi\rangle _ {AB} = {} _ A\langle \alpha' \vert \phi\rangle _ A \vert \omega\rangle _ B\]so they are both propotional.
Proportional implies product state: Write $ \vert \psi\rangle _ {AB}$ in its general form as
\[\vert \psi\rangle = \sum _ {m, n} c _ {mn} \vert m\rangle \otimes \vert n\rangle\]Then the condition that for all vectors $ \vert \alpha\rangle _ A, \vert \alpha’\rangle _ A \in A$
\[({} _ A \langle \alpha \vert \otimes I _ B) \vert \psi\rangle _ {AB} \propto ({} _ A\langle \alpha' \vert \otimes I _ B) \vert \psi\rangle _ {AB}\]means we have that $\forall m, m’$,
\[({} _ A \langle m \vert \otimes I _ B) \vert \psi\rangle _ {AB} \propto ({} _ A\langle m' \vert \otimes I _ B) \vert \psi\rangle _ {AB}\]so
\[\sum _ n c _ {mn} \vert n\rangle \propto \sum _ n c _ {m'n} \vert n\rangle\]And hence $\exists \vert \beta\rangle$ such that $\forall m$, $\exists \lambda _ m$ so that
\[\sum _ n c _ {mn} \vert n\rangle = \lambda _ m \vert \beta\rangle\]Then
\[\begin{aligned} \vert \psi\rangle &= \sum _ m \vert m\rangle \otimes \sum _ n c _ {mn} \vert n\rangle \\\\ &= \sum _ m \vert m\rangle \otimes \lambda _ m \vert \beta\rangle \\\\ &= \left(\sum _ m \lambda _ m \vert m\rangle\right) \otimes \vert \beta\rangle \end{aligned}\]So it is a product state.
Consider the following situation. Alice and Bob share two qubits in the Bell state
\[\vert \Phi^+\rangle = \frac{1}{\sqrt 2} ( \vert 0\rangle \otimes \vert 0\rangle + \vert 1\rangle \otimes \vert 1\rangle)\]
But instead want to share two qubits in the state
\[\vert \psi\rangle := \alpha \vert 0\rangle\otimes \vert 0\rangle + \beta \vert 1\rangle \otimes \vert 1\rangle\]
Suppose that Alice is only allowed to send one classical bit to Bob (no quantum channels), and both Alice and Bob are allowed to measure systems they control, prepare additional qubits, or apply quantum gates.
Describe a protocol that can be used to do this.
Alice prepares a new qubit in the state $\alpha \vert 0\rangle + \beta \vert 1\rangle$. Then the system is in the state
\[\frac{1}{\sqrt 2} (\alpha \vert 0\rangle \otimes \vert 0\rangle \otimes \vert 0\rangle + \alpha \vert 0\rangle \otimes \vert 1\rangle \otimes \vert 1\rangle + \beta \vert 1\rangle \otimes \vert 0\rangle \otimes \vert 0\rangle + \beta \vert 1\rangle \otimes \vert 1\rangle \otimes \vert 1\rangle)\]Then she applies a CNOT gate with control on the second qubit. This gives:
\[\frac{1}{\sqrt 2} (\alpha \vert 0\rangle \otimes \vert 0\rangle \otimes \vert 0\rangle + \alpha \vert 1\rangle \otimes \vert 1\rangle \otimes \vert 1\rangle + \beta \vert 1\rangle \otimes \vert 0\rangle \otimes \vert 0\rangle + \beta \vert 0\rangle \otimes \vert 1\rangle \otimes \vert 1\rangle)\]Then she measures in the computational basis. If she measures $ \vert 0\rangle$, then system $AB$ is in state
\[\alpha \vert 0\rangle \otimes \vert 0\rangle + \beta \vert 1\rangle \otimes \vert 1\rangle\](Hooray!). If she measures $ \vert 1\rangle$, then the system is in state
\[\alpha \vert 1\rangle \otimes \vert 1\rangle + \beta \vert 0\rangle \otimes \vert 0\rangle\](Boo.)
If she measures $ \vert 0\rangle$, she sends “$0$” to Bob. If she measures $ \vert 1\rangle$, she sends “$1$” to Bob.
If Bob receives $0$, he does nothing. If Bob receives $1$, both him and Alice can apply the $X$ gate to their systems to obtain
\[\alpha \vert 0\rangle \otimes \vert 0\rangle + \beta \vert 1\rangle \otimes \vert 1\rangle\]as required.