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I don't get it why among all possibilities we chose the inner product? (1) The position eigenstates are complete: $$1 = \int \mathrm{d}x \, |x\rangle \, \langle x |$$ (2) The state, expressed as a weighted "sum" of position eigenstates where the weights are given by the probability amplitude, i.e., the position wavefunction: $$|\psi(t)\rangle = ... 1 Dan's answer is quite nice but I think the answer to this really depends on the interpretation one chooses to use. The straightforward answer I think is that you probably are confusing collimation with the creation of electrons. In classical mechanics one can independently specify the \vec{x} and \vec{p} very much indepently for the reason that Dan ... 4 The proof is probably not the right word since the expression \Psi(x,t) = \langle{x}|{\Psi(t)}\rangle is actually the definition of position space wave function. We have a complex Hilber space H(\mathbb{C}) and we would like to expand the state vector |\Psi(t)\rangle over a set of eigenvectors of some Hermitian operator which represent some ... 7 There is nothing to prove; this just involves making definitions as follows: Let an element |\psi\rangle of the Hilbert space \mathcal H of a particle moving in three dimensions be given. Let |\mathbf x\rangle denote a simultaneous eigenstate of the position operators X,Y,Z corresponding to eigenvalues x,y,z where \mathbf x = (x,y,z). Then for ... 1 The uncertainty principle's restriction on the minimum spreads in position and momentum is really small. An electron can be confined to a region in both position and momentum space that's extremely small compared to anything human-sized, but still have more than enough spread to obey the uncertainty principle. To give you an idea of how small this is, ... 0 The faster particles move (the more momentum they have), the more they behave like classical "macroscopic" particles. We shoot them in "straight lines" by giving them a lot of energy. I think this can be explained using the path integral formulation of quantum mechanics 1 Notice that$$|\psi|^2 =\psi^*\psi = |A|^2 e^{-x^2/a^2}.$$We are left with a Gaussian integral. 1 The expression |\psi(x)|^2 is the complex modulus squared;$$ |\psi(x)|^2 = \psi(x)^*\psi(x) $$where here the star means complex conjugation. It follows that for any wavefunction of the form$$ \psi(x) = Ae^{ikx} f(x) $$where A and f(x) are real, one has$$ |\psi(x)|^2 = A^2 f(x)^2 $$since |e^{ikx}|^2 = e^{ikx} e^{-ikx} = 1. 1 Well \vec{r} is actually a position vector it points out a particles particular position with reference to some origin, the wave vector \vec{k} actually tells us more as it is related to momentum \vec{p} which gives us a sense of direction of travel of the wave. 1 De Broglie s' matter wave, is, in some sense, the prehistory of the wavefunction, but its original interpretation is false. De Broglies relations makes a link between particles characterics (momentum, energy) and wave characteristics (frequency, wavenumber). De Broglie thought (like Einstein and Schrodinger), that it was a real "matter wave". But, thanks ... 4 All matter is made of waves—at least it can be represented that way, and it behaves that way. Of course, matter also behaves like particles. This is one of the odd-but-true conclusions of quantum mechanics. The de Broglie wavelength gives the wavelength of any "matter wave." These waves aren't waves in the classical sense with amplitude and the like; they ... 3 If you are able to produce multiple copies of the same pure quantum state, then it is possible to reconstruct the wavefunction. In that case, you need a relatively precise experiment, as just measuring the position and building a histogram will only give you the mod-squared of the wavefunction. To get some information of the phase, you might try measuring ... -1 You can measure the amplitude squared of the wave function if you have many copies of the system. You can then make a histogram of the recorded observations of the systems. This histogram will tend to the amplitude squared of the wave function as the number of copies tends to infinity. 1 When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate x must match that at coordinate x+L, so we have:$$\Psi(x)=\Psi(x+L)$$In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be ... 0 Some broadly applicable background might be in order, since I remember this aspect of quantum mechanics not being stressed enough in most courses. [What follows is very good to know, and very broadly applicable, but may be considered overkill for this particular problem. Caveat lector.] What the OP lays out is exactly the motivation for finding how an ... 1 For a free particle, the energy/momentum eigenstates are of the form e^{i k x}. Going over to that basis is essentially doing a Fourier transform. Once you do that, you'll have the wavefunction in the momentum basis. After that, time-evolving that should be simple. Hint: The fourier transform of a Gaussian is another Gaussian, but the width inverts, in ... 3 Within the superposition of the ground and the first excited state, the wavefuncion oscillates between "hump at left" and "hump at right". Maybe you are asked to find the half-period of these oscillations? 2 Yes, I believe you have to think of it as if it were a semiclassical problem; you evaluate with QM the mean square velocity \left< v^2 \right> of the particle, then calculate its square root; this should give you an estimate of the typical velocity of the particle. Once you have it, you divide the length of the well by it and find the time it takes ... 3 You are absolutely right. In fact, the uncertainty relations are a direct consequence of this wave nature and the "conjugated" variables being p \leftrightarrow x and E \leftrightarrow t. I will attempt a simple explanation. Remember that the wavefunction gives you a probability amplitude for finding the particle at some location x at some time t? ... 2 To get it in the momentum representation, one has to do the Fourier transform of this function. This reference can be useful: http://forum.sci.ccny.cuny.edu/Members/lombardi/publications/MOMREP-H-atom.pdf/view At the end, separation of variables after transformation to the momentum space is not trivial, and the mixing of quantum number is presented. 0 1) In general, \psi(\vec{r},t) = {\sf U}(t,0) \psi(\vec{r},0), where {\sf U}(t,0) is the time-evolution operator (a unitary matrix). 2) Given your superposition state at initial time, after time t the wave function would look like$$\psi(r,\theta,\phi,t) = A \left( 2R_{10}Y_{00} e^{-iE_1 t/\hbar} + 4 R_{21}Y_{1,-1} e^{-iE_2 t/\hbar} \right)$$where ... 5 If I had to summarize quantum teleportation in one equation, I would write$$ |\psi\rangle \otimes |\beta_{00}\rangle = \displaystyle \frac{1}{2} \sum_{z,x \in \{0,1\}} |\beta_{zx}\rangle \otimes X^x Z^z |\psi\rangle  You can verify this by explicitly writing out all terms on the right-hand side. Here $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ ...