# Tag Info

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It won't affect the measurement, due to the fact, that identity operator commutes with everything. In Schroedinger picture: $$\langle \psi(t) | A | \psi(t) \rangle = \langle \psi | e^{-i (H - \lambda I) t} A e^{i (H - \lambda I) t} | \psi \rangle =$$ $$= e^{-i \lambda t }\langle \psi | e^{-i H t} A e^{i H t} | \psi \rangle e^{i \lambda t} = \langle \... 1 Let |\Psi \rangle be some state in which the system is. Then the energy expectation value is:$$\langle \Psi|(H-\lambda I)|\Psi \rangle = \langle \Psi|H|\Psi \rangle - \lambda.$$Note that |\Psi \rangle is arbitrary so we see that this just corresponds to an overall shift of the energy. For comparing energy levels this does not matter but for energy ... 0 MRI is a good example. The human body is largely composed of water (H2O), and the hydrogen nuclei - protons - are particles with spin 1/2. These spin 1/2 protons are then subjected to a large static external magnetic field which sets up a "preferred" orientation, which you can call the "z-axis." Furthermore, this also establishes ... 0 Determinants and permanents are not the only way. There is a general result for the permutation group that, if you want to construct a fully symmetric wavefunction by combining the spatial and spin parts, then both parts must have the same permutation symmetry (i.e. must transform by identical irreps). This applies also for multidimensional representation ... 0 The fact is that all states of an N-particle system can be written in this way. This is because the Hilbert space of such a system is a subspace of the n-fold tensor product of the one-particle Hilbert space. However, there are certainly other methods of describing states in this Hilbert space. A particularly useful one is in terms of occupation number. ... 0 Here’s an alternate derivation that uses the expansion of unity:$$ \mathbb{1}=\sum_c \vert \psi_c\rangle\langle \psi_c\vert rather than explicit hermiticity. From \begin{align} \langle\psi_b\vert H_0 r\vert\psi_a\rangle= \langle\psi_b\vert H_0\mathbb{1} r\vert\psi_a\rangle =\sum_c\langle\psi_b\vert H_0\vert\psi_c\rangle\langle \psi_c\vert r\vert\psi_a\... 0 Start with conservation of 4-momentum. Before scattering, the 4-momentum of the photon was p_B^\mu = (h\nu_0/c, h\nu_0/c, 0, 0) and of the electron P_B^\mu = (mc, 0, 0, 0). After the scattering, we have the photon's 4-momentum p_A^\mu = (h\nu/c, h\nu\cos\theta/c, h\nu\sin\theta/c, 0) and electron's 4-momentum P_A^\mu = (\varepsilon/c, P\cos\phi, P\sin\... -1 Your post already mentions the Ehrenfest theorem. Just have a look at it and think about which operator's expectation value gives the energy. Plug it in and you're finished. The only thing you need to know for that is that the Hamilton-operator is time independent. 1 The reason is that H_0 as an observable is hermitian H_0 = H_0^\dagger. That means you can use\langle\Psi_b|H_0 r|\Psi_a\rangle = (\langle\Psi_a|r^\dagger H_0^\dagger|\Psi_b\rangle)^\ast = (\langle\Psi_a|r^\dagger H_0|\Psi_b\rangle)^\ast$$You might also just say that it is part of the definition of the bra-ket notation, that operators A inside act ... 0 It is a difference between open and closed systems. A quantum system S undergoing measurement is an open system. The full system SAE (System+Apparatus+Environment) is a closed system that undergoes unitary evolution. But the SAE Hamiltonian couples the subsystems, so S alone evolves non-unitarily. The measurement process will never be described by a ... 1 The discussion seems to be finished long ago, but I'll put in my 5 cents anyway. This is a very good question that I have been trying to answer in order to understand Matsubara's Green function formalism. Surprisingly, this question is not addressed in the textbooks that I am aware of. The condition [\hat{\mathcal{O}}, \hat H] = 0 is not sufficient to be ... 0 From the point of view of the Hilbert space H=L^2(\mathbb{R}^3) itself the wave function is allowed to blow up not just in the origin r=0, but also in other points, as long as it is square integrable. Now if the wavefunction should satisfy the TISE there are further restrictions. If the TISE is spherically symmetric it is often useful to use spherical ... 1 Yes. Provided that the resulting wavefunction is normalizable. The point is that that, in polar coordinates the radial part of the Laplacian is a singular point of the equation at r=0. Depending on the form of the potential, such singular points can be in Weyl's limit point or limit circle case. In the latter case there is a one-parameter family of ... 4 I was wondering if quantum particles do actually exists in two different states simultaneously and if it has been proven they do indeed exists in a superposition of states This is a common misconception. "Superposition of states" does not mean "existing in multiple states". Quantum systems are only in one state at a time, and this state ... 2 We need to make a careful distinction here between what is happening mathematically and what we are justified in interpreting physically. It is true that in quantum mechanics the state of a system at any given moment is represented mathematically by a vector^1, and that vector may be written as a linear combination of basis vectors, each of which may ... 1 It is sometimes said that the delta distributions \delta(x-x_0) are the eigenstates of the position operator \hat x. However, as its name may imposes, those are not classical functions and precisely not elements of the Hilbert space x is defined on. The same is true for the momentum operator and the functions e^{i\langle x,k\rangle}. They are not ... 2 Collapse of the wave-function, as required by the Copenhagen interpretation indeed requires a non-unitary evolution of the system. This, however, is also widely viewed as problem. It is one of the elements that makes the measurement problem into a problem. The modern take on this problem involves quantum decoherence. In a nut-shell, quantum decoherence is a ... 1 Eigenvalues are the values that are measured in the experiment, i.e. eigenvalues of \hat{x} are the values of the position obtained when measuring it. Every measurement will produce a different result, i.e. a different eigenvalue, unless the system was prepared in an eigenstate of the measured quantity - this is the statistical nature of quantum mechanics. ... 0 The reason for mass difference is because a proton contains 2 down quarks and 1 up quark and a neutron contains 2 up quarks and 1 down quarks(you were right about that).And since an charge of up quark is$$+\frac{2}{3}$$and the charge of a down quark is$$-\frac{1}{3}$$they are totally different particles so they'll have different masses (even though they ... 1 Since you're asking about intuition my answer will be in that direction. You shouldn't mentally associate rotations with vectors on \mathbb{R}^3. Instead, you should associate them with the group SO(3). Moreover, you should think of SO(3) as an abstract group, not as a set of 3\times 3 matrices. The set of 3x3 matrices is rather merely a ... 0 There's certainly a lot that QED can bring to quantum chemistry. Coulomb's law is an approximation to a much more complicated law described using QED that does not follow an instantaneous, spin-independent interaction. For instance, there are other terms like the Breit- and Gaunt-coupling terms that are certainly relevant when doing calculations on heavy ... 3 The Pauli and Dirac matrices are basis vectors of Clifford algebras of 3d Euclidean space and 3+1d Minkowski space respectively. If you want to understand spinors, you'll probably need to understand Clifford algebras. In Clifford algebras, reflections through the origin are represented by unit vectors (think of them as the surface normals of mirrors). The ... 4 A rotation of a spinor \psi (looks like a complex 2-vector) by an angle \phi around the unit axis \hat n is but$$ \psi \mapsto e^{i {\phi\over 2} \left(\hat{n} \cdot \vec{\sigma}\right)} \psi= \left (I\cos {\phi\over 2} + i (\hat{n} \cdot \vec{\sigma}) \sin {\phi\over 2}\right ) \psi , $$where \vec \sigma are the three Pauli matrices, twice the ... 1 The spin at position n does get counted twice but it changes the value of S_i^z S_{i+1}^z from +\frac{1}{4} to -\frac{1}{4}, that's a change of -\frac{1}{2}. Multiply by 2 because two elements of the sum get changed and you get your answer. 1 The replacement channel is a trivial example. For input state \sigma, the channel gives you output \Lambda(\sigma) = \text{Tr}(\sigma)\rho. 0 I’ll give an EE’s perspective from the RC properties from my experiences since 1975. Zeners always have an exponentially continuous reducing resistance above knee voltage threshold that flattens to a fixed bulk electrode size dependent resistance as in all common diodes . ON semi quotes threshold resistance as Zzt and “on” knee resistance as Zzk which is ... -1 Calculate <\psi|X|\psi> = \int \psi^*(X; c_0, c_1) X \psi(X; c_0, c_1) dX with \psi^* as the Hermitian conjugate. Keep the c_0 and c_1 as parameters. Then find the minimum of the output function. You'll want to use the c_0^2 + c_1^2 = 1 constraint to get it down to a single free parameter. For the integrals, you'll need to ... 0 By solving the TISE and wondering why its solutions depend on the energy, you are begging the question. The TISE reads$$\hat H\lvert\psi\rangle = E\lvert\psi\rangle$$where E is the energy eigenvalue, i.e. the energy of the system when it is in the state \lvert\psi\rangle. In other words, the \psi that you have obtained above came from a process of ... 2 Sometimes the expansions are not obvious. For example The harmonic oscillator time-dependent Schr"odinger equation$$ i\partial_t \psi = -\frac 12 \partial^2_x \psi +\frac 12 \omega^2 x^2 \psi $$has a breathing'' solution$$ \psi(x,t)= \left(\frac{\omega}{\pi}\right)^{1/4}\frac 1{\sqrt{e^{i \omega t} +R e^{-i\omega t}}}\exp\left\{ - \frac \...

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Go to a more complicated system, the orbitals of the hydrogen atom which give the solution for the possible locations in (x,y,z) of an electron trapped around a proton. 1,2,3 is the $n$ , the quantum number connected with the energy level, and there is a difference in space for the probability of the electron. Why? That's what the solution gives.

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Any observable quantity is a real number, which means that it corresponds to a real eigenvalue. This is why they are described by Hermitian operators. Real number here means that it is not an imaginary number, not that it is "real" in some other philosophical or mathematical sense.

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In your example, if you had considered the non-Hermitian operator $i\hat a$, you would have obtained imaginary expectation value: $$\left<\psi\right|i\hat a\left|\psi\right> = i\sum_{j=2}^k\sqrt{j+1}$$

0

$\newcommand{\ket}{|#1\rangle}$ The time-independent Schrödinger Equation written in Dirac notation is often expressed as $$H\ket{\psi_n} = E_n\ket{\psi_n},$$ which is know in Mathematics as an eigenvalue equation. It is "more proper" to see the dependencies of the energy eigen-value as having dependencies on $n$, instead of the $n$ in the $n$-...

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In order to observe a specific path one needs to design an appropriate observable. Clear discussions on this subject are usually appear in the context of the two-slit experiment, although it is not the first thing that comes to mind when one is dwelling on all the mathematical difficulties of the path integrals. Some specific realizations of the two-slit ...

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Don't do it. It doesn't make any sense. Study warp drives instead - which are gravity drives. For instance, the recent improvements on equations to the Alcubierre general relativistic warp drive equations, as written about here: https://phys.org/news/2017-01-alcubierre-warp.html and perhaps here: https://bigthink.com/technology-innovation/alcubierre-drive?...

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From Wikipedia: For example, in quantum mechanics, the one-dimensional time-independent Schrödinger equation is an S-L problem. The time-independent Schrödinger equation: The Sturm-Liouville equation (see here): If you put $p(x)=1$ and $y=y(x)$, this reduces to: $$\frac{d^2y}{dx^2} +q(x)y(x)=-\lambda\omega (x)y(x).$$ So if we put $y=\psi_n(x)$, $q(x)=V(x)... 2 Essential knowledge about the Schrödinger-equation SE (I limit my post to the time-independent equation) is that it is an eigen-value equation. I assume that you know the basics of linear algebra so if you take a, for instance a 2x2 matrix$A$, an eigen-value problem can be set up: Let's assume $$A= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\... 5 There are two equations commonly called the Schrödinger equation. Most physicists I know would prefer to call*:$$i\hbar \frac{d}{dt}|\psi\rangle = \hat{H}|\psi\rangle$$'the' Schrödinger equation. This is also called the 'time dependent' Schrödinger equation. It is the quantum mechanical replacement for Newton's second law:$$ m \frac{d^2}{dt^2}\vec{r} = F(\... 8 The Schrödinger equation is just another way of writing the conservation of energy, right? Wrong. The original motivation for Erwin Schrödinger can be read in his 1926 paper An Undulatory Theory of the Mechanics of Atoms and Molecules (behind paywall, but freely readable copies can easily be found on the web). Namely, the ansatz is the Hamilton-Jacobi ... 3 The eigenstates$\Psi_n$of the Harmonic Oscillator are the Hermite polynomials (with a Gaussian weight) which form a complete set, and so you can write any initial state $$\Psi(x,0) = \sum_{n=0}^\infty c_n \Psi_n(x),$$ and solve for the$c_n$s. Once you've done this, you just need to tack on the time dependence for each eigenstate (of the form$e^{-iE_n/\...

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Note that from $a^\dagger |n \rangle = \sqrt{n+1} |n+1 \rangle$ it follows $\langle n | a = \sqrt{n+1} \langle n+1 |$ so $a$ acting on the left is essentially a creation operator.

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Let me rephrase it in one sentence. What current knowledge of science concludes about light and electron(example of particle) Let us be clear. Nature means observation , measurements and experiments. This double slit experiment with light, where the intensity of the laser is brought to single photon at a time shows the dual nature for photons/light. Light ...

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The energy barrier for the inversion is $24.2$ kJ/mol or about $0.25$ eV. At room temperature the thermal energy $kT$ is about $0.02$ eV or about a factor of $13$ times less than the energy required to get through the barrier. So the proportion of molecules with enough energy to go over the barrier and invert would be: $$P = e^{-13} \approx 2.28 \times 10^{-... 1 Resonators change time scales and coupling strengths by confining the electromagnetic field. Say you have your circuit or waveguide and you want to couple to an artificial atom. If you just let them do their thing, your coupling strength will be weak (an excitation is more likely to pass by than to excite the artificial atom). If you employ a resonator, you ... 1 The smooth BE distribution does not apply to the lowest state. That has to be taken out and counted separately, as the assumption that the energy spectrum of the states can be treated as a continuum breaks down. At sensible temperatures this state is unoccupied so it doesn't matter. At low temperatures the occupancy of the lowest state becomes significant ... 0 Imagine a series of quantum states of a single particle: In the first state the particle has a certain position but an uncertain momentum In the last state the particle has an uncertain position but a certain momentum Also imagine all other possible (obeying the Uncertainty principle) states in between What is the mathematical space that describes all those ... 9 I'll start with a generic perspective, and then I'll apply it to the question about Hilbert space. Here's the generic perspective: Sometimes we use one mathematical thing to represent another mathematical thing. Here, a representation is a mapping from A to B that preserves the essential structure of A, where A and B are both mathematical things. ... 1 As was mentioned in the other answer, a Hilbert space is not necessarily a space of continuous functions. This is just one example of a Hilbert space. The term Hilbert space applies to any set of mathematical objects that satisfy the axioms of a vector space over either \mathbb{R} or \mathbb{C}, that have an inner product defined on the space, and that ... 1 The single-particle Hilbert space is the space of functions \psi: \, \mathbb{R} \to \mathbb{C} [with value \psi(\vec{r})] with finite L^2 norm (normalizable functions). Furthermore boundary conditions may limit this set. In any case it is indeed not the space of all continuous functions. Fock space is the many-particle generalization, the space of ... 1 So a complex number a+bi acts upon the 2D space (x,y)\leftrightarrow x+yi by a matrix we could write as$$\begin{bmatrix}ax-by\\bx+ay\end{bmatrix}=\begin{bmatrix}a&-b\\b&a\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}, and in fact all of the strange operations of complex numbers are actually very straightforward operations of matrix addition and ...

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