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This paradox is an artifact of describing the mirror classically. Quantum mechanics is itself free from paradoxes, and it is easy to see that this paradox vanishes the moment you describe the mirror as a quantum mechanical object. A simple way to see what goes wrong is to apply the uncertainty relation. Suppose that we have a freely floating mirror and we ...


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An entangled state is a joint state for both the particles. For instance if you had electrons then when you write $\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle$ for instance the minus sign in the middle there is a definite phase between the two parts. And the joint state itself has an overall phase. So for instance ...


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The suggestion of Kevin Zhou, on other hand, has the advantage that we could try to control all the poles of the original well and the halved one. The polology of the square well was studied in Nussenzveig "The poles of the S-matrix of a rectangular potential well or barrier" Nuclear Physics, 11:409–521, 1959. and revisited in his book "Causality and ...


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I'd say that the non adiabatic problem is ill-defined. Lets see. All the initial solutions $\Psi_n(x) = A_n \sin(n \pi x / 2a)$ have a probability current $\Psi'^* \Psi - \Psi^* \Psi'$ equal to zero everywhere, so in principle you can cut them at any point without having a leak of probability. The problem is that the boundary conditions for the domain of ...


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The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic ...


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Entanglement is a quantum mechanical phenomenon. It is a shorthand to saying " aspects of the wavefunction for these particles are completely known", i.e. the particles are entangled by the wavefunction describing their probabilistic behavior. A laser beam emerges from zillions of coherent photons, i.e. their phases with respect to each other and the beam ...


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Entanglement experiment performed in one frame of reference guarantees that the two measured results are synchronized in this frame of reference. If we try to perform the same entangled experiment while the two ends of the same length fiber optic lines are attached to two frames of references moving with a constant relative velocity, the two measured results ...


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The mass is given by the Casimir operator $\hat{P^2}$, the square of the momentum, generated by translation invariance. This will give you the actual mass of the particle, as opposed to the bare mass that you will find in the theory, as the actual mass can be affected by the particle's interactions.


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To add to Ernie's Answer: Actually, tides on Earth are tantamount to an observation of a Moon-like object. Bernard Schutz in his book "A First Course in General Relativity" imagines an instance of your scenario whereby Earthlings lived under skies permanently shrouded completely and impenetrably by clouds: "The true measure of gravity on the Earth are ...


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What's an observation? I think your question delves into the nature of "consciousness", a term which has never to my knowledge been satisfactorily defined. The seas observe the moon, and therefore there are tides. Whether or not a "conscious" being observes the seas is of no consequence in physics.


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The electron crawls along very slowly. And long before it gets to the diodes other electrons flow through the diodes which then both emit light. Why? Because the electrons in front of it feels the field from the electron before the electron gets to it and they move forwards and when they move the electrons in front of them move, etc. So some other electrons ...


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I'm not sure your question is as well posed as you think it is. In order to be 100% sure to have an entangled spin state, one would have to measure it, but can entangled states be eigenvectors of Hermitian operators (= results of measurements) other than the trivial one? If you know something is in one of various orthogonal states then in principle the ...


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Indeed using a rigged Hilbert Space is almost certainly what you should use and there are tutorials on it on arxiv (I'll add them when/if/I can find them again). However it isn't necessarily going to help you understand what an average physicist is doing on the one hand, and on the other hand there were already many things you were doing that not rigorous ...


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No. The particles have a joint spin state, $\left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow\right\rangle.$ When you send the left particle through a Stern-Gerlach device then the device changes from $\left|0\right\rangle$ to $\left|1\right\rangle$ when the left particle is detected as down. So $\left(\left|\uparrow \downarrow ...


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Imagine you have a pair of coins. Whenever you throw them, each of them is fully random, but their outcomes are opposite. Now imagine you throw the two coins. You look at the left coin. When it is head, you discard both coins and start again, when it is tail, you keep it. Since you have never looked at the right coin, it should still be completely random. ...


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Your claim that $p(n)$ is the $\lvert c_n \rvert^2$ from the decomposition $\lvert\psi\rangle = \sum_n c_n \lvert \psi_n \rangle$ is incorrect. In statistical quantum mechanics, we must differentiate between a pure state of the system $\lvert \psi \rangle$ and a mixed state, which is given by a collection of states $\lvert \psi_n\rangle$ and the probability ...


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Let's say $\Phi$ is a delta function, $\Phi(k)=\delta(k-k_0)$. Presumably, you want this to be an eigenstate of the momentum operator with momentum $\hbar k_0$. With the convention you've chosen, we can convert this to a real-space wavefunction (I'm ignoring normalization for convenience): $$ \Psi(r)= \int dk \delta(k-k_0)e^{ikr}=e^{ik_0 r} $$ We can then ...


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You claim that there seem to be quantum jumps and ask why this does not refute non-collapse theories. To answer the question of whether there is evidence that rules out no-collapse theories, you first have to work out what would happen is no collapse took place. So suppose that you have an atom $a$ in an excited state $|e\rangle_a$ that has some half life ...


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The no collapse people would say that if you write down the continuous evolution equations for the actual experimental setup in question that it predicts a perfectly continuous evolution of the wave vector into two parts that don't overlap. For instance a Stern-Gerlach device takes an incoming beam with a spin state and that beam branches into two. So the ...


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As all the answers are oriented in the theoretical side let me remind everybody of the Balmer series, an experimental observation fitted with a mathematical series, which was first modeled with the Bohr model of the hydrogen atom, and then was the cornerstone of building quantum mechanics, as it came out from the Schrodinger equation. So in this sense a ...


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A simple answer is to look in the Heisenberg picture with the Heisenberg equation of motion. In this picture, operators evolve instead of states as in the Schrodinger picture. Here the evolution of the operator is chosen so that its expectation value has the same time evolution as in the Schrodinger picture. To do so its time evolution is governed by: $ ...


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Generically, given a representation $V$ of a group, the tensor representation $V\otimes V$ will decompose into the symmetric and antisymmetric parts $$ V\otimes V = \Lambda^2 V \oplus S^2 V$$ and in the case of the rotation group (or its universal cover), the symmetric 2-tensors have a certain invariant under rotations - their trace! So when $j_1 = j_2$, the ...


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First let us cut away some of the meat and the bones of your question. Let us forget about thermodynamics for a minute and classical physics; however we will need GRAVITY (not Newton’s version). Instead of calling it Gravity let us call it General Relativity or (GR). We can still use the term Gravity, but when discussing Quantum Mechanics it’s better to ...


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I suggest the American Journal of Physics as a good reference for undergraduates. The readership is Physics teachers so usually covers the topics that are of interest to students and the parts of a research subject that are of most interest to students. Plus it tries to make thing accessible to people that aren't in the specific subfield (in case they are ...


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The article of D J Griffiths assumes that the delta interaction can be approximated by a sequence of even functions and then infers two boundary conditions: $$ \Psi'(0^+)-\Psi'(0^-)= (-1)^n {m c \over \hbar^2} (\Psi'^{(n)}(0^+)+\Psi'^{(n)}(0^-))$$ $$ \Psi(0^+)-\Psi(0^-)= (-1)^{n-1} {m c \over \hbar^2} n (\Psi'^{(n-1)}(0^+)+\Psi'^{(n-1)}(0^-))$$ My own ...


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See some basic QM textbook as mention above, @Todd R, Briefly, $S$ which in $3$ dimensional case can be represented in $x, y,z$ basis as \begin{align} \vec{S} = S_x \hat{x} + S_y \hat{y} + S_z \hat{z} \end{align} If we choose $z$ direction as (consider $S_z$ and state as a eigenvalue problems : Usually many textbook choose $z$, actually choice of ...


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The answer is no, I am afraid. As you may well know, the self-adjoint Laplace operator $-\Delta$ on $L^2(\mathbb{R})$ has purely absolutely continuous spectrum $\mathbb{R}^+$. Now let $V\in L^{\infty}(\mathbb{R},\mathbb{R}^+)$ be an arbitrary bounded positive function. Then $-\Delta_x +V(x)$, where $V$ acts as a multiplicative operator is self-adjoint and ...


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1) Your perturbation operator does not conserve the particle number of the phonons, so only even powers of it will contribute to equilibrium expectation values. Since you are interested only in the ground state, which doesn't have any phonons excited, this means that you have to create a phonon first. After that either another phonon can be created or the ...


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The rotation group, its representations, and their carrier spaces are fundamental parts of quantum mechanics. Every object in the universe is either a spin=0, 1/2, 1, 3/2, 2,… object. For the integer spin objects, the rotation group is O(3), and the rotation matrices contain only real numbers. However, there are half integer spin particles in the world, ...


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Just to close shop here. $H$ and $\sigma_i p_i$ do not commute, because $H$ does not necessarily include $\sigma_i p_i$, because $\alpha$ is free. Under the constraint that $\alpha$ is purely imaginary, the term does turn into $\sigma_i p_i$ and the commutation is valid. So, both of you are right. $H$ and $\sigma \cdot p$ should commute, which they only do ...


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Let's simplify things, $\hbar=m=1$ and we put in the $\delta^{(n)}$ as $V(x)=V_0\delta^{(n)}/2$. Then the problem is $$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$ Apart from $x=0$ the equation gives $$\psi'' = -E \psi $$ we want a bound state $E<0$ which falls off at infinity, so we get a solution $\psi_+ = A \exp(-k x)$ for $x>0$ and $\psi_- = A ...


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It's important to remember that the quantum state is physical, and not merely a description of our knowledge or ignorance of a system. At every point during Alice's experiment, the pair has a single quantum state (insofar as the measurement apparatus can be treated as classical), but Alice, Bob and Charlie have different degrees of knowledge about what that ...


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As the OP asks to try to avoid Schrödinger equation, I think that it could be worthwhile to exhibit some variation of my answer from the similar post http://physics.stackexchange.com/a/202298/1335: The great thing of the exponential measure in Feynman path integral is that when we evaluate the probability it transforms to a sort of derivative of the action, ...


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Maybe you should take into account that $(\sigma_i p_i)^2=p_i p_i$.


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If a particle either "is" or "isn't", then its count is either 1 or 0. Even in quantum mechanics it's not possible that half a particle exists. It is possible to detect it with 50% probability, but if you set about counting all the particles one at a time, you necessarily end up with an integer answer.


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Remembering the double slit (where the observation of the electron in one slit destroys the superposition |left slit>+ |right slit>) I am almost convinced that the interaction between the spin superposition |up>+|down> and the magnetic field if not observed would not lead to destruction in it. Like in double slit where the superposition is not destroyed ...


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No. The other photon might even be forbidden to produce a pair over by itself all by itself since there might be no nucleus over by it. The other photon doesn't have to copy what the first one does. But many things could happen to the entanglement. And that is partly because there are many ways the photons could have been entangled. For instance, you could ...


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When I was asking this question, I didn't understand the relation between the commutativity of two operators and their eigenspaces: If an operator $A$ commutates with another operator $B$, then $A$ leaves the eigenspaces of $B$ invariant: $$ B\psi = \epsilon\psi \implies BA\psi = AB\psi = \epsilon A\psi $$ But this does not imply that $\psi$ is an ...


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No. Being entangled does not mean they mirror everything the other does. It only means certain properties are in an inseparable state. Destroying one of the pair would end this state


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At a time when my hammer was the Dirac's delta distribution, I conjectured that the answer was Feynman Integral is a generalization of a Dirac's delta, the use of this delta being to find the extreme of the action. Given a function $f(x)$, find a Dirac measure $\delta_f$ concentrated in the critical points of $f$. The answer is obviously $ \delta(f'(x))$, ...


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You seem to have confused the CNOT gate with the controlled phase gate. There's no CNOT gate involved in the implementation of a C-phase-gate (let's denote it CPG for short), as all we do with it, is multiplying by a phase factor. For example on two qubits it is defined as: $$ U_{CPG}(\phi) |xy\rangle = \exp(i(x\land y)\phi)|xy\rangle $$ Where $\land$ is ...


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Any set of commuting stabilizers can be mapped to any other set of commuting stabilizers by applying a unitary from the Clifford group, see http://arxiv.org/abs/quant-ph/9807006. The Clifford group is generated by CNOT, Hadamard, and the phase gate $\left(\begin{smallmatrix}1\\&i\end{smallmatrix}\right)$, so your $U$'s indeed have a special form. ...


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A wavefunction is a function that takes a configuration of all the particles in the universe (if just one particle is in a different place it is a different configuration) and assigns a complex number to that configuration. If you take the square of the length of that complex number you get something people like to call a probability-density. An X-density ...


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Firstly, you cannot simply 'neglect' $a_0$ (as the *.pdf actually shows). In some test books the following substitution is used: $\rho=\frac{r}{a_0}$, so the $a_0$ doesn't have to be 'carried around'. As your expression for $P(\rho)$ is made of three factors, each factor can be evaluated for extrema individually. $\rho=0$ is an obvious minimum from the ...


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The instructor hasn't thrown out $r^4$. If you do the calculations properly you will get the desired result. What the instructor is telling is that while simplifying and taking commons out for $\frac{\text{dP(r)}}{\text{dr}}=0$ you get,$$(\text r^3)(6-\frac{\text r}{\text a_0})(f(\text r))=0$$which gives, $\text r=0$ and $6\text a_0$ from the first two ...


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A quantum system is described by a suitable $*$-algebra of observables. The quantum states are functionals of the observables, that when applied on observables yield the average value of it in the state. So, given an observable $A$, and a state $\omega$, $$\omega(A)$$ is the evaluation of $A$, i.e. its average value on the state. Now the state (or ...


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Having worked with elementary particles all my working life I can assure you that particles have a trajectory. Here is proof Another proof is the existence of accelerators which create the beams that we can scatter against targets, as in picture, or against each other and study the results statistically. That is how the standard model of particle ...


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Most people would say that $|\Psi(a)|^2$ is the probability density that the particle would be found in some small neighborhood of $a$. Most people would not say it is the probability that it is in that location. This is because because that same kind of thinking (that it has a probability of having a property and that a measurement merely reveals the value ...


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The three generators of right-handed spinor rotations are given by $\left\{- i\sigma_x,-i\sigma_y,-i\sigma_z\right\}$, see for instance Peskin & Schroeder page 44, and the rotation matrix for a spinor rotation over an angle $\phi$ around a unit vector $\hat{s}$ is given by: $R~=~ \exp\left(-i\frac{\phi}{2}~\hat{s}\cdot\vec{\sigma}\right) ~=~ ...


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How to get from $\left(\frac{-\hbar^2}{2m_e} \Delta_{r_e} + \frac{-\hbar^2}{2M_P} \Delta_{r_p} +V(r) \right)\Psi(\vec r_e,\vec r_p) = E \Psi(\vec r_e,\vec r_p)$ to: $\left(\frac{-\hbar^2}{2(m_e+M_p)} \Delta_{_{R}} + \frac{-\hbar^2}{2\mu} \Delta_{r} +V(r) \right)\Psi(\vec r,\vec R) = E \Psi(\vec r,\vec R)$ with: $\vec R=\frac{m_e \overrightarrow{r_e} + ...



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