Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am now studying quantum teleportation. I get what the process is like but I'm wondering why it happens this way.

You've got two entangled particles A and B whose wavefunctions are entangled. You also have a third particle C, which is the one you want to teleport. You get C entangled with either of the two and final result is that the wavefuction of one of the two A and B (depending on which you entangled C with) becomes the same as the C's wavefunction is.

  1. Why does it occur this way (mathematically explained, if possible)? Why does the wavefunction of A or B becomes that of C?
  2. What is the equation explaining this process (addition of wavefunctions, maybe)?
share|improve this question
3  
Have you looked at the wiki page? en.wikipedia.org/wiki/Quantum_teleportation It has quite a lot of mathematical detail. –  joshphysics Apr 30 '13 at 16:03

1 Answer 1

up vote 5 down vote accepted

If I had to summarize quantum teleportation in one equation, I would write $$ |\psi\rangle \otimes |\beta_{00}\rangle = \displaystyle \frac{1}{2} \sum_{z,x \in \{0,1\}} |\beta_{zx}\rangle \otimes X^x Z^z |\psi\rangle $$ You can verify this by explicitly writing out all terms on the right-hand side. Here $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ is an arbitrary qubit state that we want to teleport, $X$ and $Z$ are Pauli matrices and $$ |\beta_{zx}\rangle = \dfrac{|0,x\rangle + (-1)^z |1,\bar{x}\rangle}{\sqrt{2}} $$ are the Bell states for $z, x \in \{0,1\}$ where $\bar{x}$ is the negation of $x$.

Intuitively, this equation says that if Alice has $|\psi\rangle$ and a half of a maximally entangled state $|\beta_{00}\rangle$, then that's the same as her having one of the four Bell states $|\beta_{zx}\rangle$ and Bob having $X^x Z^z |\psi\rangle$. You can think of Bob's state as a "corrupt" version of $|\psi\rangle$. If Alice measures her two qubits in Bell basis, she gets bits $x$ and $z$ which she sends to Bob, who can apply $Z^z X^x$ to recover $|\psi\rangle$. The coefficient 1/2 represents the fact that Alice's bits $x$ and $z$ are uniformly random.

Teleportation becomes less mysterious if you know that one-time pad is it's classical analogue. One-time pad lets you use shared randomness to transmit private random bits over a public channel in the same way as teleportation lets you use shared entanglement to transmit quantum bits over a classical channel.

More precisely, the analogy goes as follows. Alice has a private bit and shares a perfectly correlated pair of random bits with Bob ($00$ and $11$ with probability 1/2). If she XORs her private bit with her half of the shared random bit and sends the result publicly to Bob, he can recover Alice's private bit by XORing the received bit with his half of the shared random bit. Teleportation works very similarly, except we have to send two classical bits, since we are teleporting a pure qubit state which has two degrees of freedom.

You can see my blog post for more details: Teleportation and superdense coding.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.