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Quantum Teleportation

1. Homework Statement
This isn't exactly a problem but rather a problem in understanding the derivation of the phenomenon, or more precisely, one step in the derivation.

In the following we will consider the EPR pair of two spin ##1/2## particles, where the state can be written as
$$ \vert \psi\rangle =\frac{1}{\sqrt{2}}(\vert 0,1\rangle - \vert 1,0\rangle).$$ Now let us assume that Alice and Bob have each one of the two particles of the EPR pair. Alice has another particle with spin ##1/2## in the state ##\vert \phi\rangle##. The state of the whole system, all three particles, is therefore given by
$$\begin{align*}\vert \phi\rangle \otimes \vert \psi\rangle &= \vert \phi\rangle \otimes \frac{1}{\sqrt{2}}(\vert 0,1\rangle - \vert 1,0\rangle)\\
&= \frac{1}{\sqrt{2}} (\vert\phi,0\rangle \otimes \vert 1\rangle - \vert\phi,1\rangle \otimes \vert 0\rangle). \end{align*}$$ Now Alice can measure her two particles, for example using ##P_i= \vert \chi_i\rangle\langle \chi_i\vert, i\in \{1,2,3,4\}## and
$$\begin{align*}
\vert\chi_1\rangle &= \frac{1}{\sqrt{2}}(\vert 0,1\rangle - \vert 1,0\rangle)\\
\vert\chi_2\rangle &= \frac{1}{\sqrt{2}}(\vert 0,1\rangle + \vert 1,0\rangle)\\
\vert\chi_1\rangle &= \frac{1}{\sqrt{2}}(\vert 0,0\rangle - \vert 1,1\rangle)\\
\vert\chi_1\rangle &= \frac{1}{\sqrt{2}}(\vert 0,0\rangle + \vert 1,1\rangle).
\end{align*}$$
Up until this point I understand the definitions and the idea. The problem arises when I try to calculate for example
$$P_1 \vert \phi\rangle\otimes\vert\psi\rangle = \frac{1}{2} \vert \chi_1\rangle \otimes (-\vert 1\rangle\langle 1\vert\phi\rangle - \vert 0\rangle\langle 0\vert\phi\rangle )$$

2. Homework Equations
All given above.

3. The Attempt at a Solution
We first need to figure out how ##P_i## acts on the tensor product of the states. Expanding the state gives
$$ P_1 \vert \phi\rangle\otimes\vert\psi\rangle = \frac{1}{\sqrt{2}} P_1(\vert\phi\rangle \otimes\vert 0\rangle \otimes \vert 1\rangle - \vert\phi\rangle \otimes\vert 1\rangle \otimes \vert 0\rangle). $$ Form this we can conclude that ##P_i## is of the form ##P_i = A\otimes B \otimes C##, where ##A,B,C## can be any operator. I tried to compute now the follwoing:
$$\begin{align*}\vert \chi_1\rangle\langle \chi_1\vert
&= \frac{1}{2}(\vert 0,1\rangle - \vert 1,0\rangle)(\langle 0,1 \vert -\langle 1,0\vert)\\
&= \frac{1}{2} (\vert 0\rangle \otimes\vert1\rangle - \vert 1\rangle \otimes\vert0\rangle)(\langle 0\vert\otimes\langle1 \vert -\langle 1\vert\otimes\langle0\vert),
\end{align*}$$
but can't really proceed from here since I don't really know how to calculate this... I suspect that after finishing this calculation I could define ##A\otimes B := \vert \chi_1\rangle\langle \chi_1\vert ##. Then I would only need to find ##C##, but I'm not really sure how to do that...

Am I doing something completely wrong here, or is this the right approach?
 

DrClaude

Mentor
6,969
3,140
Using that Alice has particles 1 and 2 and Bob particle 3, and particles 2 and 3 are the EPR pair, then the ##P_i## are projection operators for particles 1 and 2 only, with an identity operation for particle 3,
$$
(P_i)_{12} \otimes I_3
$$
For example, if the three particles are in the state
$$
| \Psi \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle_1 |1 \rangle_2 |0 \rangle_3 + |1 \rangle_1 |0 \rangle_2 |1 \rangle_3 \right)
$$
then
$$
P_1 | \Psi \rangle = \frac{1}{\sqrt{2}} | \chi_1 \rangle_{12} \left( |0 \rangle_3 - |1 \rangle_3 \right)
$$
 
Thank you very much for the suggestion, but I'm still not sure if I'm doing the math right... Could you maybe just glance over this and tell me if the individual steps work?
$$\begin{align*}P_1\vert\phi\rangle\otimes\vert \psi\rangle &= \left(\vert \chi_1\rangle\langle\chi_1\vert\otimes I\right) \left( \vert\phi\rangle\otimes\vert\psi\rangle\right) \\ &= \left(\vert \chi_1\rangle\langle\chi_1\vert\otimes I\right) \frac{1}{\sqrt{2}} (\vert\phi,0\rangle \otimes \vert 1\rangle - \vert\phi,1\rangle \otimes \vert 0\rangle) \\
&= \frac{1}{\sqrt{2}} \left(\vert \chi_1\rangle\langle\chi_1\vert \phi,0\rangle \otimes \vert 1\rangle- \vert \chi_1\rangle\langle\chi_1\vert\phi,1\rangle \otimes \vert 0\rangle\right)\\
&= \frac{1}{2} \left([ \underbrace{\langle 0,1\vert \phi,0\rangle}_{=0} - \langle 1,0\vert \phi,0\rangle] \vert \chi_1\rangle \otimes \vert 1\rangle- [ \langle 0,1\vert \phi,1\rangle - \underbrace{\langle 1,0\vert \phi,1\rangle}_{=0}]\vert \chi_1\rangle \otimes \vert 0\rangle\right)\\
&= \frac{1}{2}\left(-\langle 1\vert\phi\rangle \vert\chi_1\rangle\otimes \vert 1\rangle -\langle 0\vert\phi\rangle \vert\chi_1\rangle\otimes \vert 0\rangle\right)\\
&= \frac{1}{2}\vert\chi_1\rangle \otimes \left( -\langle 1\vert\phi\rangle \vert 1\rangle -\langle 0\vert\phi\rangle \vert 0\rangle \right) \\
&= \frac{1}{2}\vert\chi_1\rangle \otimes \left( -\vert 1\rangle\langle 1\vert\phi\rangle -\vert 0\rangle\langle 0\vert\phi\rangle \right)
\end{align*}$$
 

DrClaude

Mentor
6,969
3,140
$$\begin{align*}P_1\vert\phi\rangle\otimes\vert \psi\rangle &= \left(\vert \chi_1\rangle\langle\chi_1\vert\otimes I\right) \left( \vert\phi\rangle\otimes\vert\psi\rangle\right) \\ &= \left(\vert \chi_1\rangle\langle\chi_1\vert\otimes I\right) \frac{1}{\sqrt{2}} (\vert\phi,0\rangle \otimes \vert 1\rangle - \vert\phi,1\rangle \otimes \vert 0\rangle)
\end{align*}$$
This is very hard to follow. It is not clear which are single-particle kets and which are two-particle kets. Otherwise, it looks correct.
 

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