Recent content by maxverywell

In this Khan Academy video they say that it is ok to break the square root ##\sqrt{a\cdot b}##, with ##a, b \in \mathbb{R}##, into the product of two square roots ##\sqrt{a}\cdot \sqrt{b}##, only when: (1) both are non negative, (2) one of the two is negative and the other is possitive. I...

Do we have to diagonalize the density matrix everytime and then use the formula for the coordinates ##(\sum p_i x_i, \sum p_i y_i,\sum p_i z_i)## ? https://en.wikipedia.org/wiki/Bloch_sphere Not very practical.

What are the coordinates on the 3D Bloch ball of a qubit's mixed state of the form: ##\rho=p_{00}|0\rangle \langle 0|+p_{01}|0\rangle \langle 1|+p_{10}|1\rangle \langle 0|+p_{11}|1\rangle \langle 1|##

No you cannot teleport (FTL) information or energy or matter. In quantum teleportation schemes they "teleport" a quantum state ##\psi##, but a classical channel of communication is always needed in order to complete the teleportation (so the information is not sent FTL, but at the speed of...

Introductory Quantum Optics, C.Gerry and L.Knight, chapter 6.2 (Quantum mechanics of beam splitters), page 141.

You are right for asking this. The problem here is in semantics rather than in physics. What they (physicist in popular science books) mean by "nothing" is absence of matter/fields from space, but not the space and time themselves. Moreover, this absence of matter/fields is observer dependend...

From a textbook on quantum optics: ##|1\rangle_a |1\rangle_b \xrightarrow{BS} \frac{1}{2}\left(\hat{a}^{\dagger}_{c}+i\hat{a}^{\dagger}_{d}\right)\left(i\hat{a}^{\dagger}_{c}+\hat{a}^{\dagger}_{d}\right)|0\rangle_c |0\rangle_d =\frac{i}{\sqrt{2}}\left(|2\rangle_c |0\rangle_d+|0\rangle_c...

What I meant by ##|2\rangle_a |0\rangle_b## is that the state ##|2\rangle## falls onto the beam splitter's port ##a##, while by and ##|0\rangle_b## I represent that at the second port, ##b##, there is nothing.

Can you elaborate on that please? Sending the state ##|2\rangle_a |0\rangle_b## to a 50:50 beam splitter will result on the entangled state ##\propto |2\rangle_c |0\rangle_d-|0\rangle_c |2\rangle_d+2i|1\rangle_c |1\rangle_d##. What will the two detectors at ##c## and ##d## register? On the other...

Thank you all for the clarifications. So it would be more correct to ask the question: what is the difference between the states ##|1\rangle |1\rangle## and ##|2\rangle |0\rangle## (or ##|0\rangle |2\rangle## equivalently)? And now the answer is more obvious. However I have a difficulty in...

What is the difference between two single-photon Fock states ##|1\rangle |1\rangle## and one two-photon Fock state ##|2\rangle## (all in the same mode)? In both cases the mean photon number is 2. How do we distinguish them experimentally?

Not if ##c## is the same for all ##E##, as you have supposed in your proof. In other words, why in your proof ##\psi_0## is unique (same for any ##\psi##)?

I think you presuppose that ##E=m\hbar\omega+c##, where ##m## is integer and ##c<\hbar\omega##. In this case there is indeed a unique ##\psi_0## with energy ##E_0=c##, and any other eigenstate ##\psi## is created by applying ##\hat{a}^n## on ##\psi_0##.

I think that in order to use the formalism of the ladder operators you have to presuppose that the oscillator is quantized.

Well, it follows from the Schrödinger equation for the harmonic oscillator. If you solve (e.g. by power series method) it and apply the boundary conditions and normalization, you find that the energy is quantized, ##E_n=\left(n+\frac{1}{2}\right)\hbar \omega##.