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)?
  • 4
    $\begingroup$ Have you looked at the wiki page? en.wikipedia.org/wiki/Quantum_teleportation It has quite a lot of mathematical detail. $\endgroup$ Apr 30, 2013 at 16:03
  • $\begingroup$ This webpage has the best mathematical description of quantum teleportation/entanglement I have ever seen. And it is very accessible. Also, me and a friend just made a youtube video about quantum teleportation, we tried to keep it reasonably fun. Features cats. $\endgroup$
    – Toby
    Jan 27, 2017 at 5:22

3 Answers 3


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.


Teleportation can be explained in terms of locally inaccessible information: information that the Heisenberg picture observables of a system instantiate, but that doesn't change the expectation values of those observables:


Since the information doesn't affect expectation values, it can be sent through decoherent channels. If yoyu have an entagled pair of systems you can access the locally inaccessible information using measurement results from both systems. In teleportation the system to be teleported interacts in the right way with one of an entangled pair of systems. The information required to reconstruct that state is sent in a locally inaccessible form to the other half of the entangled pair where it can be used to reconstruct the state of the system. If you don't track the information carefully it looks like the system was teleported in some mysterious non-local way when in fact the information required to reconstruct the system was just sent in a decoherent channel. The mathematical details are in the paper linked above.


Photon C is in some quantum state near Alice. A and B are entangled photons. Pairs of entangled particles can be described by a wave function that is a superposition of four possible states, each with a prob. of 1/4. Photon B is sent to Bob far away, e.g. via laser bean. Now Alice has A and has C. A wave function describes three particles, C, A and B, again in a superposition of four possible states. Alice makes A and C interact (e.g. by having their beams collide). A-B entanglement is replaced by A-C entanglement; this is the key event! The states of Alice’s two particles, notably including C, have been “projected” or “imposed” onto one of the four states of the entangled Bell or EPR pair. Through entanglement, one of these states emerges in Bob’s particle in the form of one member of a four-part (four state) equation, but Bob does not know which; if he guesses, we'll be right with probability of only 0.25. But Alice can help him out; here is one way (there are others): She does a Bell-state measurement (BSM) by using a CNOT gate (which eliminates two possibilities) and a Hadamard gate (which pins down which state her particle A is in) -- which is the same state Bob's is in. So she tells him by email, phone, laser flash, even smoke signals --some conventional way-- which one state (out of 4) her BSM identified. Her message is in the form of a 2-bit code, 00, 01, 10 or 11. From this code, Bob knows which of four unitary (non-destructive) operations he should do on his particle B. Once he does that, his B will be in the state of the original particle C was in. Teleportation done. It's not cloning because state of C was destroyed by Alice when she allowed interaction. The teleported C-qubit is intact because at no point was a complete quantum measurement done.


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