# Tag Info

## New answers tagged wavefunction

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In the first case I get points where there is an oscillation in time, while in the second one, as stated, the points "do not oscillate": where there is a minimum, that minimum stays there in time and the same happens for maxima. This is not true. There is identical time dependence, of frequency $\omega$, in both situations; as I said in my answer to your ...

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The Schrodinger equation provides the definite energy space an electron can occupy inside a shell in the atom. The wave function, one of its variables, actually gives the probable location of an electron in space.

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Let us label the state spaces clearly as $\mathcal{H}_1$ and $\mathcal{H}_2$ for the first and second particle respectively and denote the canonical isomorphism sending a state in $\mathcal{H}_1$ of the first particle to the exact same state of the second particle by $\phi : \mathcal{H}_1\to\mathcal{H}_2$. Let us further denote the canonical "flip ...

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To normalize this function you have $$1~=~\int_0^{2\pi}\Phi^*\Phi d\phi ~=~|C|^2\int_0^{2\pi} e^{-im\phi}e^{im\phi}d\phi~=~|C|^2\int_0^{2\pi}d\phi~=~2\pi|C|^2.$$ The result is then obvious.

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This experiment has been done, first by Birgit Dopfer in 1998, then later by Dr. John Cramer of the University Of WA. In Dopfer's experiment, there was a "coincidence detector" which is basically an AND gate to filter out only the entangled pairs. By moving the detector in the beam of photons not going to the double-slit, the information about the photon's ...

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In the last integral $x^2 -y^2$ will integrate to zero. You can show that by splitting the integral into two pieces and changing variables in one of the integrals as $x\leftright y$

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From the list of Spherical Harmonics here, I can rewrite your wavefunctions as : $$\psi_1(\vec r) \propto Y_1^{-1}(\theta,\phi) r g(r) \\ \psi_2(\vec r) \propto Y_1^{0}(\theta,\phi) r g(r) \\ \psi_3(\vec r) \propto Y_1^{1}(\theta,\phi) r g(r) \\$$ Use the fact that $Y_{lm}$'s are a orthogonal set of functions. $$\int Y_l^m \ Y_{l'}^{m'} \text d \Omega = \... 0 I can show this backwards and then explain the motivation. The uniqueness of the solution follows from the constraint on the transformation to be unphysical. Say, we take your equation and transform \psi:$$ i \partial_t \psi = \left( \frac{\Pi^2}{2m} + e \phi \right)\psi $$becomes$$ i \partial_t \left(e^{i\frac{e}{c} \Lambda}\psi\right) = \left( \...

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This answer is motivated by the Aharonov-Bohm effect and proves what the OP asks for, but in the special case $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A} =\boldsymbol{0}=\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}' \quad \text{that is} \quad \mathbf{B} =\boldsymbol{0} \tag{01}$$ To simplify the expressions we : set ...

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Short (but cryptic) answer: complex numbers arise in quantum mechanics because we would like find solutions to the differential equation $$\frac{\partial}{\partial x}f(x) = cf(x)$$ which don't blow up as $x\to \pm\infty$. Long answer: Fundamentally, the shift from classical mechanics to quantum mechanics is replacing functions (observables) and numbers (...

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It depends on which wavefunctions are meeting. There are actually several different scenarios in your question which will change how they are treated: 1) wavepackets, or portions of a wavefunction that describe the same single particle system add linearly, as you have guessed. This is because (most) field equations in free space are linear field operators:...

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One example for a quantum mechanical system is "one free electron in 3-dimensional space". The state "the electron is at the point $\vec x_1$" is described by one wave function $\psi_1(\vec x)$, the state "the electron is at the point $\vec x_2$" is described by another wave function $\psi_2(\vec x)$, and a third wave function $\psi_3(\vec x) \sim \psi_1(\... 3 We reject non-normalizable bound states because they are unphysical. A bound state is (by one definition) simply an eigenstate belonging to the discrete spectrum of the Hamiltonian. The discrete part of the spectrum always has associated eigenstates within the Hilbert space (that is, within the space of normalizable wave functions). The non-normalizable "... 1 From yuggib's answer: "…we consider physically meaningful only wavefunctions such that there exist a continuous and differentiable representative in its equivalence class. However, also this requirement is not physical." Not quite. A set of countable pointwise discontinuities may be tolerable, at least at first sight, but there is actually a very good ... 3 If a particle is a wavefunction describing a probability amplitude distributed through space, what happens when two wavefunctions meet? The particle exists, the wavefunction is a solution of the quantum mechanical equation specifically defined for the boundary values of the observation. Its complex conjugate square gives the probability density of ... -1 I think when two particles (with associated wavefunctions) with wavefunctions$\psi_1\left(x\right)$and$\psi_2\left(x\right)$overlap (i.e. they interact in vacuum) they will interfere and the resultant wavefunction will be$\psi\left(x\right)=\psi_1\left(x\right)+\psi_2\left(x\right)$with some normallization constant. If there is constant phase ... 0 Imaginary numbers are very useful for calculations, but as their name suggests - they are imaginary, not real. So if you measure something in the real world, you can only expect to get real numbers. This is true for both electrodynamics and QM, expectation values of quantum mechanics observables (i.e. measurements) will always turn out real. Imaginary ... 0 It probably depends what interpretation of quantum physics you subscribe to. That sounds approximately right for the Copenhagen interpretation, in which you aren't allowed to analyze where the wave function comes from. For those who appreciate more what de Broglie, Einstein, Bell and others have put into quantum physics, there's always the interpretation ... 6 There is no physical requirement for a wave function to be smooth, or even continuous. At least if we accept the common interpretation that a wavefunction is nothing else than a "probability amplitude". I.e. it represents, when multiplied with its complex conjugate, a probability density. Now a probability density can be discontinuous or even undefined on ... 0 Discontinuity in first derivative of wave function implies that the wave function experiences a sudden force that changes its momentum instantly. Thus physically speaking this is not possible as there are no dirac delta potentials. There are potentials very close to dirac delta and thus in the dirac delta approximation the wavefunction will have a first ... 4 The equation will become $$- \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \bigg) \psi(x,y,z) = (E-V_0) \psi(x,y,z)$$ And the solutions are the same: $$\psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}).$$ And ... 2$\hat U$is an operator, and an operator is very different from a scalar. Just think about this: every operator can be expressed as a matrix in some basis and every state as a vector. So the difference between $$\exp \left(-\frac{i \hat H t}{\hbar} \right) \mid \psi \rangle$$ and $$\exp(-i \phi) \mid \psi \rangle$$ where$\phi$is a real number, is the ... 2 Purely formal considerations suggest that the Lie bracket of quantum observables should be proportional to the Poisson bracket of their classical counterparts. (In particular, the classical Poisson bracket satisfies$ \lbrace u,wx\rbrace v +u \lbrace v,wx\rbrace =\lbrace uv,w\rbrace x + w \lbrace uv,x\rbrace$, and the proportionality follows from this and a ... 1 The position operator also appears in the Hamiltonian, as soon as you have a potential. However, when using the position representation (the Schroedinger's "wave equation") it then appears as a multiplication of the wave-function by a real number. Don't put too much in the wave part of "wave equation". This is just one of the multiple representation of the ... -1 Use the momentum operator with$\hat p = -i\hbar\partial/\partial x$on the wave$\psi(x,t) = Ae^{-c|x - y|^2}e^{ikx}$and compute the velocity as $$v\psi = -\frac{i\hbar}{m}\frac{\partial}{\partial x}Ae^{-c|x - y|^2}e^{ikx} = -\frac{\hbar}{m}(k - 2ic|x - x_0|)\psi$$ The$x_0$is the start condition and so I would have the initial velocity$v = \hbar k/m$.... 0 Your question is the most basic concept of quantum mechanics. Classically what you are thinking is right but when you have a quantum system it can occupy as many quantum states as possible. There is a certain probability that the system will be in certain state. The quantum state is destroyed if you make a measurement and you will then get a definitive end ... 1 My answer will be very non-technical, but hopefully will convey some basic ideas about what the quantum state (or wavefunction) is about. One intuitive way to picture the nature of the quantum state of a system is to see it as the interference (hence the "wave" idea) of every different changes it could possibly undergo while it is not being messed with. ... 3 There are 37.2 trillion cells in a typical human body, (probably a good few more in mine ;), then in each cell there are 20 trillion atoms, then you have to obtain the wave function for each of the electrons....... Actually, it may well be that you cannot describe a wavefunction for a macroscopic object, like a human body. In the study of quantum ... 1 Like gonenc pointed out your assumption that normalizing your wave function does not imply continuity. And yes you'll probably won't need the normalization factor in your further calculations. The reason for you doing this could be consistency with the Interpretation of the wave function squared as a probability amplitude: $$P = \int|\psi(r,t)|^2 dr$$ 2 Diffractive optics in the Fresnel (paraxial) approximation is exactly the same as the quantum mechanics of a single particle when thickness along the optical axis is replaced by time, refractive index is replaced by mass and the inverse angular frequency of the monochromatic light is replaced by Planck's constant. Here is a brief sketch. Classical ... 5 Even mathematically, Schrödinger's equation cannot be derived from the principle of least action because it only depends on the first derivatives of time,$\psi' = \partial \psi / \partial t$. This proves that$\psi' $would have to appear in the action, but then$\psi''$would unavoidably appear in the Euler-Lagrange equations, too, unless the action had ... 2 The Huygen's principle is that a wave amplitude$A(t_0)$, usually a plane wave, is modified into a spherical wave with an amplitude $$A(t,r) = \frac{A(t_0)e^{ikr + i\delta}e^{-i\omega t}}{r}.$$ The radial distance at$t_0$is$r \simeq \lambda$and the phase$\delta$is set so that$k\lambda = \delta$, which is set to$2\pi\$. Now consider a Taylor ...

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Essentially, separation of variables in the time-independent Schroedinger equation amounts to diagonalizing the Hamiltonian. One can see this easily by considering the case where the Hilbert space is finite-dimensional, and the Hamiltonian is a Hermitian matrix. In case of a partially continuous spectrum one gets the same, except that the sum must be ...

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I cannot be sure, but I suspect that you can get analytical solutions of the Pauli equation by taking a non-relativistic limit of analytical solutions of the Dirac equation. The latter can be found in many books, say Bagrov, Vladislav G. / Gitman, Dmitry, The Dirac Equation and its Solutions (http://www.degruyter.com/view/product/177851) (you can find a ...

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