8.12 Quantum Teleportation, Superdense Coding and No-Cloning Theorem

Quantum Teleportation

If Alice performs a measurement in the n direction, it will dictate Bob's post-measurement state to align with the opposite \(n\) direction.

Quantum teleportation enables Alice to transmit an unknown quantum state \(|\psi\rangle\) to Bob.

Initial state: \(|\psi^{A'}\rangle|\Psi_-^{AB}\rangle\), \(|\psi^{A'}\rangle = a|0\rangle+b|1\rangle\) Alice and Bob don't know a and b

Then \(|\psi^{A'}\rangle|\Psi_{-}^{AB}\rangle = (a|0\rangle^{A'}+b|1\rangle^{A'})(\frac{1}{\sqrt{2}} (|01\rangle^{AB}-|10\rangle^{AB}))\) \(= \frac{1}{\sqrt{2}}(a|001\rangle - a|010\rangle + b|101\rangle - b|110\rangle)\)
We use Bell Basis of \(\mathbb{C}^2 \otimes \mathbb{C}^2\): \(|\Psi_\pm^{A'A}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle) \quad |\Phi_\pm^{A'A}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)\)
Then \(|00\rangle^{A'A} = \frac{1}{\sqrt{2}}(|\Phi_+^{A'A}\rangle + |\Phi_-^{A'A}\rangle)\) and \(|11\rangle^{A'A} = \frac{1}{\sqrt{2}}(|\Phi_+^{A'A}\rangle - |\Phi_-^{A'A}\rangle)\)
\(|01\rangle^{A'A} = \frac{1}{\sqrt{2}}(|\Psi_+^{A'A}\rangle + |\Psi_-^{A'A}\rangle)\) and \(|10\rangle^{A'A}= \frac{1}{\sqrt{2}}(|\Psi_{+}^{A'A}\rangle - |\Psi_{-}^{A'A}\rangle )\)

Then \(|\psi^{A'}\rangle|\Psi_-^{AB}\rangle = \frac{1}{2}(a|\Phi_+^{A'A}\rangle|1\rangle^B + a|\Phi_-^{A'A}\rangle|1\rangle^B - a|\Psi_+^{A'A}\rangle|0\rangle^B - a|\Psi_-^{A'A}\rangle|0\rangle^B + b|\Psi_+^{A'A}\rangle|1\rangle^B - b|\Psi_-^{A'A}\rangle|1\rangle^B - b|\Phi_+^{A'A}\rangle|0\rangle^B + b|\Phi_-^{A'A}\rangle|0\rangle^B)\)
\(= \frac{1}{2}( |\Phi_+^{A'A}\rangle(a|1\rangle^B-b|0\rangle^B) + |\Phi_-^{A'A}\rangle(a|1\rangle^B+b|0\rangle^B) + |\Psi_+^{A'A}\rangle(-a|0\rangle^B+b|1\rangle^B) + |\Psi_-^{A'A}\rangle(-a|0\rangle^B-b|1\rangle^B) )\)
Here we change the state in Alice to Bob

Then we can do a measurement: \(\Pi_0^{A'A} = |\Psi_-\rangle\langle\Psi_-|\), \(\Pi_1^{A'A} = |\Phi_-\rangle\langle\Phi_-|\)
\(\Pi_2^{A'A} = |\Phi_+\rangle\langle\Phi_+|\), \(\Pi_3^{A'A} = |\Psi_+\rangle\langle\Psi_+|\)
Then clearly: \(\sum_{i=0}^{3} \Pi_{i}^{A'A}= I^{A'A}\), \(\Pi_x\Pi_y = \delta_{xy}\Pi_x, \forall x,y \in \{0,1,2,3\}\)
Thus \(\{\Pi_x\}_{x\in\{0,1,2,3\}}\) is a projective von-Neumann Measurement.

For \(|\psi^{A'}\rangle|\Psi_-^{AB}\rangle = |\varphi^{A'AB}\rangle\), we do measurement.

\(x=0\), \(P_{0} = \Pr(x=0) = \langle\varphi|[\Pi_{0}^{A'A}\otimes I^{B}]|\varphi\rangle^{A'AB}\)
Post-Measurement state: \(|\varphi_{0}^{A^{\prime}AB}\rangle:=\frac{1}{\sqrt{P_{0}}}\Pi_{0}^{A^{\prime}A}\otimes I^{B}|\varphi\rangle^{A^{\prime}AB}\)
\(x=1\), ......
\(x=2\), ......
\(x=3\), ......

Generally, \(P_{x}=\Pr(x)=\langle\varphi|\Pi_{x}^{A^{\prime}A}\otimes I^{B}|\varphi \rangle^{A^{\prime}AB}\)
Post-Measurement state: \(|\varphi_{x}^{A^{\prime}AB}\rangle:=\frac{1}{\sqrt{P_{x}}}\Pi_{x}^{A^{\prime}A}\otimes I^{B}|\varphi\rangle^{A^{\prime}AB}\)

Example:
\((\Pi_{0}^{A'A}\otimes I^{B})|\varphi^{A'AB}\rangle = -\frac{1}{2}|\Psi_{-}^{A'A} \rangle(a|0\rangle+b|1\rangle)\)
\(\langle\varphi^{A'AB}|(\Pi_0^{A'A} \otimes I^B)|\varphi^{A'AB}\rangle = \frac{1}{4} = P_0\)
\(|\varphi_{0}^{A^{\prime}AB}\rangle=\frac{1}{\sqrt{1/4}}(\Pi_{0}^{A^{\prime}A}\otimes I^{B})|\varphi^{A^{\prime}AB}\rangle=-|\Psi_{-}^{A^{\prime}A}\rangle\underbrace{(a|0\rangle+b|1\rangle)} _{|\psi^{B}\rangle}\)

Check: \(P_0=P_1=P_2=P_3=1/4\) \(\psi^{B}:= |\psi^{B}\rangle\lang\psi^{B}|\)

Outcome	Simplification	Post-Measurement State
\(x=0\)	\(\Psi_{-}^{A'A}\otimes \psi_{0}^{B}\)	\(\\|\psi_{0}^{B}\rangle=\sigma_{0}\\|\psi^{B}\rangle=\\|\psi^B\rang\)
\(x=1\)	\(\Phi_-^{A'A} \otimes \psi_1^B\)	\(\\|\psi_1^B\rangle := a\\|1\rangle+b\\|0\rangle = \sigma_1\\|\psi\rangle\)
\(x=2\)	\(\Phi_+^{A'A} \otimes \psi_2^B\)	\(\\|\psi_{2}^{B}\rangle := a\\|1\rangle-b\\|0\rangle = -i\sigma_{2}\\|\psi\rangle\)
\(x=3\)	\(\Psi_+^{A'A} \otimes \psi_3^B\)	\(\\|\psi_3^B\rangle = a\\|0\rangle-b\\|1\rangle = \sigma_3\\|\psi\rangle\)

where we denote by \(\{\sigma_x\}_{x=0,1,2,3}\) the identity matrix \(\sigma_0 = I_2\), and the three Pauli matrices \(\sigma_1, \sigma_2\), and \(\sigma_3\).
Hence, up to a global phase, Bob's state after outcome \(x\) occurred is \(\sigma_x|\psi\rangle\).
After Alice sends (via a classical communication channel) the measurement state \(\sigma_{x}(\sigma_{x}|\psi\rangle)=\sigma_{x}^{2}|\psi\rangle=|\psi\rangle\)

Therefore, by using shared entanglement, and after transmitting two classical bits (cbits), Alice was able to transfer her unknown qubit state \(|\psi\rangle\) to Bob's side.

Super dense coding

\(|\Phi_{+}^{AB}\rang = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\)

After encoding, Alice send this state to Bob, then Bob has two states

\(|\psi_0^{AB}\rangle = \sigma_0 \otimes I |\Phi_+^{AB}\rangle = |\Phi_+^{AB}\rangle\)
\(|\psi_1^{AB}\rangle = \sigma_1 \otimes I |\Phi_+^{AB}\rangle\) \(= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes I (\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle))\) \(= \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle)\) \(= |\Psi_{+}^{AB}\rangle\)
\(|\psi_2^{AB}\rangle = (\sigma_2 \otimes I) |\Phi_+^{AB}\rangle\) \(= -i|\Psi_-^{AB}\rangle\)
\(|\psi_3^{AB}\rangle = (\sigma_3 \otimes I) |\Phi_+^{AB}\rangle = |\Phi_-^{AB}\rangle\)

Bob perform the measurement
\(\{|\underbrace{\psi_{x'}\rangle\langle\psi_{x'}}_{\pi_{x'}}|\}_{x\in\{0,1,2,3\}}\) where \(\Pi_x \Pi_{x'} = \delta_{xx'} \Pi_x\) and \(\sum =I\), then it's projective von-Neumann Measurement
\(\Pr(x') := \langle\psi_2^{AB}|\Pi_{x'}^{AB}|\psi_2^{AB}\rangle = \delta_{x'2}\), thus 100% we get result \(x'=2\)

The same for other \(x'\)
We perform one measurement, many possible outcomes e.g. energy

Generally

use one qubit to transmit two cbits (d=2)

No cloning theorem

\(A \rightarrow A \otimes A\)

Theorem

Quantum Clonning is not possible.

If \(\langle\psi_1|\psi_2\rangle = \frac{1}{2}\), then \(\langle\psi_1|\psi_2\rangle = \langle\psi_1|\psi_2\rangle^2\), then \(\frac{1}{2}= (\frac{1}{2})^{2}\) is impossible

Only orthogonal state can be cloned

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