Suppose there are two quits called q1 and q2. They are in two different superposition states. If we represent their states on the Bloch sphere surface, q1 is located at point_1 and q2 is placed at point_2. Please note that point_1 is different from point_2 and we know where point1 and point2 are. Now, my question consists of two parts:   How to make q1 and q2 entangled? After making them entangled, how to check if they are entangled indeed?
81184
The Bell state $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ is an entangled state. But why is that the case? How do I mathematically prove that?
106947
Can someone show to me the steps to derive the joint state at the bottom of this image, please?   I tried to follow his explanation but I didn't get the same results… This is taken from the lecture notes of Ronald de Wolf in case it may help
109211
Is there a machine or instrument which will notify us instantly when the wave-function has collapsed if we have access to only one of the entangled particles? Note that the wave function collapse will be triggered by observing the particle which we do not have. Or in other words is there a way to identify (instantaneously) if the entangled pair of a particle has been observed and the wave function has collapsed.
92568
I'm curious about QM and spent the last 3 days thinking about Delayed Choice Quantum Eraser (DCQE) experiment, but I couldn't solve this issue:   Assume we do the DCQE experiment so that for the whole experiment (let's say, 1000 particles) finishes before the first particle reaches the detectors 1-4 (using the notation from ). Assume also we're using a switch to control if the idler particle goes to which-way-detector or to the eraser.  shows what we should see in D0, even though we don't know yet R01-04 Now, knowing D0, use the switch to send the 100 particles that reached the most-left side of D0 to the eraser detectors (D1 and D2). It seems much more likely that they reach D2 than D1, because it's on peak on D2 (and on rest on D1) even though it's going through a half-silvered mirror. This looks like a contradiction to me.   So, am I missing something or QM is just that weird?
69284
When quantum entanglement is explained in "layman's terms", it seems (to me) that the first premise, that we have to accept on faith, is that a particle doesn't have a certain property (the particle is not IN one state or another) until that property is measured.    For example, I have read that a particle may have "spin up" or "spin down" but -- here's where I get lost -- it's not that we don't KNOW which spin the particle has until we measure it, but actually the particle has NEITHER spin until we measure it.  Now, if we accept that premise, then we can see how spooky it is, when two particles are created with correlated spins -- if one has spin up, the other has spin down, but neither particle "has" either spin until one of them is measured.  At that instant, the other particle "assumes" the other spin.   At least that's how I interpret things I have read.  Certainly, if we accept that a particle doesn't actually HAVE the property until it is measured, but rather the particle exists in a superposition of states until the measurement is made, then we "regular citizens" can understand what is meant by "spooky action at a distance".  (As soon as one particle is measured, the other particle assumes the "other" spin, no matter how far away it is.  How did the second particle know the first particle was measured?  That, I think, is the spooky action at a distance (and the information transfer can be faster than light).)  However, from a layman's perspective, I want to cry out "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured!  Duh!"  Now, I can't believe that all of the physics community hasn't thought of that objection -- but -- here's my point -- although the spooky action at a distance can be explained to us, once we accept the premise of superposition -- Why isn't the fact of superposition explainable also, in terms mortals can understand?  I have looked, but I haven't found a layman's explanation of why we should accept the fact that a particle doesn't have a particular spin, or other property, until it's measured.  It sure seems that the particles "have" the properties, even though we haven't measured them yet.    I know that much of quantum mechanics is not "intuitive".  If anyone can explain why particles don't have a definite property, even before we measure them, I would be grateful.
80498
Consider altitude $AH$ of $\Delta ABC$. $B_1$ and $B_2$ are points on side $AB$ such that $HB_1 \perp AB$ and $HB_2 \parallel AC$. $C_1$ and $C_2$ are points on side $AC$ such that $HC_1 \perp AC$ and $HC_2 \parallel AB$. Prove that $BC$, $B_1C_1$,  $B_2C_2$ are concurrent.     I tried letting $B_1C_1 \cap B_2C_2 = \{A'\}$ and tried to prove that $\widehat{AA'B} = \widehat{AA'C}$ but I don't see the light at the end of the tunnel.
59509
recently i was watching a  on quantum computing where the narrators describes that quantum entanglement information travels faster than light!  Is it really possible for anything to move faster than light? Or are the narrators just wrong?  Regards,
58352
To take an example,           Let's think of an electron which ban be in either one of two positions [...] There are two possible states of definite energy for the electron.  Each state can be described by the amplitude for the electron to be in each of the two basic positions.  In either of the definite-energy states, the magnitudes of these two amplitudes are constant in time, and the phases vary in time with the same frequency.  On the other hand, if we start the electron in one position, it will later have moved to the other, and still later will swing back to the first position.   (Feynman derives this behaviour from first principles in chapter 8-6 by solving equations for the state of an ammonia molecule, but that's a nice summary).  This sounds like the uncertainty principle - that you can know energy or position but not both.  Is it possible to derive the uncertainty principle from such an analysis?  Or is the uncertainty principle somehow axiomatic in the way this behaviour is derived?
121205
According to various sources on the web, it seems like the general concensus is that there isn't actually any wave-particle duality with quantum particles. For example,  implies that diffraction patterns in  were interpreted as wave interference due to apparatus limitation at the time they were first performed.  Does this mean that all those sources and animations showing two waves interfering are simply incorrect, classical conclusions which don't have anything to do with (quantum) reality?  What's actually the most confusing is that most sites which state that it is now possible to pass individual photons through these slits, also claim that these individual photons somehow interfere with themselves resulting in the observed patterns. That seems like a rather thin explanation, doesn't it?  So, is there actually any need to use wave interference to explain the phenomena, or can we simply state that the pattern is probabilistic in a certain way, without involving the "spooky" explanations?
2369
Let's consider an infinite square well. In the first exited state there is a node at the middle of the well (i.e. wave function and thus probability of finding the particle is zero there). If I measure the position right now, I may find the particle on the left side of the node, and if I measure the particle after some time, I may find the particle on the right side of the node. But must  a particle on the left side to go to right side not pass through the node point? Thus the particle will spend some time at that point, so there should be some probability of finding the particle there, but the wave function is zero there, so what's wrong with the arguement? Does the particle move here to there, without passing through the space in between? But that will just violate continuity equation?
11971
Suppose I have a system say SHO in a superposition of energy eigenstates $|{n_1}\rangle$ and $|{n_2}\rangle $ given by $|{\psi}\rangle = \frac{1}{\sqrt{2}}|{n_1}\rangle + \frac{1}{\sqrt{2}}|{n_2}\rangle $.  If I measure the energy of system there is equal probability of getting $(n_1 + \frac{1}{2})\hbar\omega$ or $(n_2 + \frac{1}{2})\hbar\omega$.  The question then arises where does rest of the energy go or come from? On one good day I measure system to be in state $|{n_1}\rangle$ while on other I measure it to be $|{n_2}\rangle$. It seems to me that it violates the energy conservation. Does this have to do with the measuring device? If yes then why do we never talk about it in formulating states?
43609
Say we have some state $$|\psi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)$$ it is in a quantum superposition of $|0\rangle$ and $|1\rangle$. Its density matrix is $$\rho=\begin{pmatrix}\frac 1 2 &amp; \frac i 2\\ -\frac i 2 &amp; \frac 1 2\end{pmatrix}$$ If we measure $\rho$ we get some classical density matrix of the form $$\rho=\begin{pmatrix}\frac 1 2 &amp; 0\\ 0 &amp; \frac 1 2\end{pmatrix}$$  Is there some nice intepretation of what these cross terms mean, other than just arrising from the phase of the state?
56669
So I've been reading a bit and watching some videos about the double slit experiment, and therefore the wave particle duality; I've also read this "implies" that a particle can be in two places at the same time. Can you clarify this, and how could that be possible?
47725
Let's say I have eigenstates $|x\rangle$ associated with measurement of position. I know that the eigenstates corresponding to their respective eigenvalues form a basis, let's call it $A$. Now let's say I want to expand a generic state vector $|ϕ\rangle$ in that basis formed by eigenstates. Now, what does it mean when physicists say that:     "$ϕ(x)$ is the coordinate of the state $|ϕ\rangle$ in the basis $A$?"    Any explanation or reference to similar kind of questions would be helpful.
84124
Suppose there're two systems $A$ and $B$, if we're in a pure state $|\psi\rangle\in\mathbb{H}_a\otimes\mathbb{H}_b$. Let $\hat A$ be an operator acting on $\mathbb{H}_a$ and $|\psi\rangle=\sum_{i,j}\psi_{i,j}|a_i\rangle|b_j\rangle$. The expectation value of A with respect to the state $|\psi\rangle$ can be calculated as $$ \langle A\rangle_{\psi}= Tr_{\mathbb{H}_a\otimes\mathbb{H}_b}(A|\psi\rangle\langle\psi|)=\sum_{i,j}\langle a_i|\langle b_j|A\otimes\mathbb{1}_B|\psi\rangle\langle\psi|a_i\rangle|b_j\rangle=... $$ I'm very confused about where these two equalities coming from? In particular, where does the term $|\psi\rangle\langle\psi|$ in the sum come from? Can I understand its function as generating a $\psi$ matrix with different constants $\psi_{i,j}$?
40798
A few years ago, I did research in quantum mechanics, specifically dealing with generalized displacement operators. In such musings, BCH lights (or gets in, depending on your viewpoint) the way. A question that struck me today was: does $\exp(A+B) = \exp(A)\exp(B)$ hold if and only if $A$ and $B$ commute? Clearly, if they do commute this is true but I have not seen anything detailing the opposite direction. Given the complexity of BCH, I would be inclined to think that it's not simple to prove if it is true. I've thought about it on my own but haven't been able to come to any sort of conclusion one way or the other.
43958
I have two question concerning information content of qubit.  Question 1: How many classical bits are needed to represent a qubit:  A qubit can be represented by a vector $q = \begin{pmatrix}\alpha \\\beta \end{pmatrix}, ~~ \alpha, \beta \in \mathbb{C}$. So, we need four real numbers to represent it. But when facts that (i) $|\alpha|^2+|\beta|^2 = 1$ and (ii) two qubits which differ in global phase only are indistinguishable, are taken into account, only two real numbers are necessary (angles on Bloch sphere). Moreover, we can choose a precission of the qubit representation (i.e. number of decimal places in $\alpha$ and $\beta$ or Bloch sphere angles) which influence number of classical bits needed to described the qubit.   So, it seems to me that the qubit representation cannot be used for measuring qubit information content but only memory consumption in simulation. Am I right?  Question 2: What is an (effective) information content of qubit:  A superdense coding enable us to send two classical bits via one qubit. But on the other hand you need two entangled qubits prepared in advance.  Given these facts, what is an information content of qubit? One or two classical bits? Or do I need to use another angle of view given the fact that qubits are "continuous" (i.e. described by complex numbers $\alpha$ and $\beta$)?