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2

The important thing is the relative sign between the potential and the Laplacian. Otherwise there are two square roots of $-1$ namely $\pm i.$ You could write $j=-i$ and then you have two perfectly good square roots of $-1$ you can use $i$ or you can use $j$. For square roots of positive numbers you get things like $\pm\sqrt 2$ and there is a way to pick ...


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In response to your edit: The real part of $(A^*(S)A(T))$ is equal to the real part of $(A(S)A^*(T))$, since they are just complex conjugates of each other. So he could have written either one. Concrete example: let $(A^*(S)A(T))=a+ib$ for some $a,b$. Then $(A(S)A^*(T))=a-ib$, since it's just the complex conjugate. Then $(A^*(S)A(T))+(A(S)A^*(T))=2a$. We ...


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$\mathfrak{R}e$: real part. $A^*$: complex conjugate of probability amplitude $A$.


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The two amplitudes need to match because the wave function has to be continuous. If it were not continuous then you would have discontinuities in the probability function $\lvert \psi (x) \lvert ^2$ which leads to an absurd situation in which a point in space can accumulate or vanish particles. I think that Feynman used the term of subsidiary waves just to ...


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Your integration is wrong. The probability density function measures the probability of finding the particle between $x$ and $x+dx$. If you integrate over $0 \leq x \leq L$ you don't get a function of $x$ but a number instead (one for this interval). If you state that $P_n$ is the integrated density, then it should depend on the interval in which you did the ...


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The discreteness of the allowed orbitals is a consequence of quantum mechanics, which was conceived precisely to explain this observation, among other things. However, the orbitals are not orbits - there is no "motion" in the classical sense going on, and an electron in an orbital does not have a fixed distance to the nucleus (it may even have non-zero ...


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It must not be greater than 1. To find the probability function you must integrate the probability density, $\psi^* \psi$, over the region in which you want to calculate the probability: $$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$


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Strangely enough, when you do a quantum mechanics experiment, you get a result that says something about what you already know. If you place a detector at one (or both) of the slits, you watch individual particles go through the slits, and you get a particle result from your detector (e.g., no interference pattern). If you don't know which slit the ...


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No, if you observe which slit they traveled through then there is NOT an interference pattern. The act of observing, or more accurately, the need for the location of the electron to be resolved causes it to take on a definite position and then continue on from that position as a particle. If it is not observed or interacted with in some way that would make ...


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Electron as a standing wave Yes, the electron is a standing wave. See atomic orbitals on Wikipedia: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". I couldn't understand how come Bohr who interpreted electron as a particle, formulated an equation for electron's angular ...


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I now come to my point, why one restricts a particle's motion to some discrete set of distances? Is it to provide a theory on the particle's stability? An attempt at an answer to your first question anyway. The electrons surrounding an atom need to obey energy level (and other) rules. As you mention distance, if you imagine that the further away the ...


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A wavefunction of a quantum system is the system's state written in a particular form - no more nor any less than that. Here the word state has an exactly analogous meaning to the state of a classical system insofar that the state at any time uniquely defines the state at any other time and contrariwise. The state's evolution with time in either a quantum or ...


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The "wave function of an orbital" is just referring to "the wave function for electrons the corresponding state". When speaking of electron levels in an atom, it is commonly accepted the Copenhagen interpretation of Quantum Mechanics, where a wave function of an isolated system, contains all the information relevant for its full description. Rigorously you ...


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In quantum mechanics we use a wave function to describe the quantum state that an electron is in. It basically tells you where you are likely to find a particle. Notice how it is typically introduced as a function of position. That is because by taking the value at a certain location and squaring it we get the probability of finding it there.


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Just extending Anna's answer. In order to observe the interference pattern which your probing electron(s) make, its associated de Broglie wavelength has to be several tens of nanometers at least. The wavelength of the electrons consisting the wall is approximately the distance between the atoms of the wall which is at least a few orders of magnitude lower ...


3

To get a wave function one has to solve the quantum mechanical equation for the boundary conditions of the experiment:"electron impinging on two slits". In the usual description one is using approximations : the incoming electron is a plane wave, the effective potential of the electrons in the matter of the slits is high. Then the distance between slits and ...


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$$\int_{-\frac{d}{2}}^{\frac{d}{2}}\psi^*(x)\psi(x)dx$$ is the probability of finding particle between $-\frac{d}{2}$ and $\frac{d}{2}$. Expectation value is : $$\langle x\rangle=\langle \psi|\hat{x}|\psi\rangle$$ $|\psi\rangle$ is the summation of probability amplitude times given basis kets $$|\psi\rangle = \sum_i c_i |{e_i}\rangle$$ ...


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The definition of the expectation value of an operator A is \begin{equation} \langle A\rangle=\int{\psi^* (x) A(x) \psi (x) dx} \end{equation} (because it represents "the value of the variable" $A(x)$ times "the probability of being in that configuration" $P(x)=\psi^* (x) \psi (x)$) and for the particular case of the expectation value of the position ...


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Yes, it is important to take in attention the surface electrons from edges. Electrons, shoot far enough to the edge, as well as photons, interact with their electric field with the electric field of the surface electrons. The difference is, tha the first intensity fringe from electrons makes the real shadow wider. In case of photons the first fringe makes ...


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Yes, to the notion that it is really interactions between charged particles with quantum natures. The "walls" of the slit are really the outer electrons of the surface atoms of the substance that are either covalently or metallically bonded to each other. Those bound electrons form a potential surface that is not exactly square, but when viewed from the ...


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The first 2-D image you posted is a typical simplification for teaching purposes. In it, they use the height of the sine wave to represent magnitude, and the directions of the sine waves to show how the fields point relative to each other. The light itself however is not itself at all cone-like. You have to imagine this sine wave existing at multiple points ...


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If your measurement would give you an exact upper and/or lower bound, but no more information (i.e. a probability distribution), the state would collapse into the projection onto the subspace of possible values, so it would still be a superposition. More generally and realistically, we would measure a value, and assign decreasing (classical) probabilities ...


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Did you cover the uncertainty principle? In quantum mechanics there is and uncertainty between energy and time: $$ \Delta E \Delta t > \frac{h}{4\pi}$$ this means that if you try to measure Energy with perfect accuracy you will have a great uncertainly in time (actually an infinity uncertainty). I guess this is what the professor was referring to, and ...


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I) Let us here phrase the problem in the context of some position operator $\hat{q}$ of QM for simplicity. The generalization to QFT can formally be achieved by replacing the position operator $\hat{q}$ with a quantum field $\hat{\psi}({\bf x})$. We know that the overlap with Minkowski (M) signature is given as a path integral ...


1

I was also puzzled with the way Zettili defined Δx and Δk to derive uncertainty relations. His definition is not the same as the well known FWHM=2.3548σ. First we need to realize that |ψ(x)|^2 and |ϕ(k)|^2 have the standard Gaussian form (1/σ√2π) Exp[−(1/2) (x/σ)^2] with a^2=2σ^2 --not |ψ(x)| and |ϕ(k)|. He defines the half maxima (0.6 to be precise) of ...


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You're not missing anything. You are right, $k=\omega/c$. The argument $\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in the Bessel function is the projection of the wavevector onto the radial direction. The use of Bessel functions beclouds what's going on a bit. Recall that a plane wave with wavevector $\vec{k}$ has the functional variation $\psi(\vec{r}) = ...


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In general you need both kinds of functions $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) + i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ and $H_n^{(2)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) - i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ in any superposition, where $(r\,\phi)$ are the polar co-ordinates and $k$ the wavenumber. If the homogeneous region in question contains the point $r=0$ ...


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But how can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? I interpret that your question basically asks how do we ...


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So I see your whole question as this: How can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? This is a really basic ...


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The form of the solution shown by Griffiths is not unique. That means that there exist cases where a basis $\{\psi_n(x)\}$ will reproduce $\Psi$ as $$ \Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}, $$ but there exists a second, different basis $\{\varphi_n(x)\}$ which (with different coefficients) also reconstructs $\Psi$: $$ ...


0

You are touching upon the measurement problem in quantum mechanics. The wave-function changes in two distinct manners: Unitary evolution by the Hamiltonian in the Schrodinger equation. Non-unitary measurement by Born's rule, i.e. what is often called "collapse of the wave-function." To many people, the latter remains puzzling, though this is a ...



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