1
$\begingroup$

Suppose I have two bits $x$ and $y$ and a map $M:x,y \to x \veebar y,x \veebar y$ where $x,y$ $\in \{0,1\}$ and $\veebar$ stands for bit xor. If I post-select the case where the output is $x^{'}=0 , y^{'}=1$, no input gives rise to this output. But this is similar to the concept of entanglement. If the $M$ is tensor product of vectors and $x$ and $y$ are qubit states then the an entangled state such as $|\psi\rangle = \frac{1}{2}(|00\rangle+|11\rangle)$ is such one state which cannot be expressed as tensor of two qubits. Does this make the ability to post-select a quantum effect ?

$\endgroup$
1
$\begingroup$

It sounds like your map $M$ sends $|00\rangle$ and $|11\rangle$ to $|00\rangle$ and sends $|10\rangle$ and $|01\rangle$ to $|11\rangle$. As described this can not be done. To implement a dynamic it needs to be reversible. The information must go somewhere.

Usually the information to reverse your computation is sent out of your computer by the exhaust fan and you call it heat and ignore that the details were there to undo your calculation. There is no such thing as a free deletion.

In fact it possible to do arbitrarily complex computation with arbitrarily small amounts of energy, you really just need energy to throw away information (send it elsewhere) not to process it in reversible ways.

So there is no quantum version of your algorithm since it is unphysical until you spell out some other registers to preserve the information.

Entanglement is just a result of linearity. There is no reason to expect or demand factorizability of states. And this isn't related to post selection.

In quantum mechanics there is a concept called post selection. But the point is to start with a subpopulation and after the fact break it into subsubpopulations. It's just grouping things together to compute averages of the groups instead of averages of the whole. Nothing deep.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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