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The most mathematically general way to write a Fourier expansion is to use complex waves with complex amplitudes. In this case the phase of the waves is represented by the complex phase of the amplitude. You can see this if you write the amplitude in polar form. Here's how it looks for your two-component wave example: $$f(\mathbf x,t) = z_1 e^{\mathbf k_1 ... 0 The article of D J Griffiths assumes that the delta interaction can be approximated by a sequence of even functions and then infers two boundary conditions:$$ \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^-))$$My own ... 4 Let's simplify things, \hbar=m=1 and we put in the \delta^{(n)} as V(x)=V_0\delta^{(n)}/2. Then the problem is$$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$Apart from x=0 the equation gives$$\psi'' = -E \psi $$we want a bound state E<0 which falls off at infinity, so we get a solution \psi_+ = A \exp(-k x) for x>0 and \psi_- = A ... 0 A wavefunction is a function that takes a configuration of all the particles in the universe (if just one particle is in a different place it is a different configuration) and assigns a complex number to that configuration. If you take the square of the length of that complex number you get something people like to call a probability-density. An X-density ... 1 How to get from \left(\frac{-\hbar^2}{2m_e} \Delta_{r_e} + \frac{-\hbar^2}{2M_P} \Delta_{r_p} +V(r) \right)\Psi(\vec r_e,\vec r_p) = E \Psi(\vec r_e,\vec r_p) to: \left(\frac{-\hbar^2}{2(m_e+M_p)} \Delta_{_{R}} + \frac{-\hbar^2}{2\mu} \Delta_{r} +V(r) \right)\Psi(\vec r,\vec R) = E \Psi(\vec r,\vec R) with: \vec R=\frac{m_e \overrightarrow{r_e} + ... 5 Ok, I have a solution for \delta'(x) based on a very crude limit. I'm going to neglect factors of \hbar, m, etc for the sake of eliminating clutter. Let V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}. Then \lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x). We'll solve the Schrodinger equation for finite \epsilon and then take the limit ... 10 What exactly does \Delta_{r_e} mean ? Your wave function isn't a field in space, it is a field on configuration space, i.e. it assigns complex numbers to a configuration. If your electron is at (x_e,y_e,z_e) and your proton is at (x_p,y_p,z_p) then the configuration is the point (x_e,y_e,z_e,x_p,y_p,z_p) in a 6d space. A point in that 6d space ... 1 \Delta_e is called the Laplacian operator and it is defined as$$\Delta_e \equiv \frac{\partial^2}{\partial e_x^2} + \frac{\partial^2}{\partial e_y^2}+ \frac{\partial^2}{\partial e_z^2}$$The Laplacian operator is related to the Del operator \nabla_e$$\nabla_e \equiv \frac{\partial}{\partial e_x}\hat e_x+ \frac{\partial}{\partial e_y}\hat e_y+ ...

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The index of the Laplacian tells you which of the coordinates it acts on, that is, if you write $r = (r_x,r_y,r_z)^T$ and $R = (R_x,R_y,R_z)^T$ as Cartesian coordinates, then \begin{align} \Delta_r & := \frac{\partial^2}{\partial {r_x}^2} + \frac{\partial^2}{\partial {r_y}^2} + \frac{\partial^2}{\partial {r_z}^2} \\ \Delta_R & := ...

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I wanted to bring up the same question as you raised, namely question 2). I agree with the answers to 1) and 3), but am unsatisfied with the proposed answer to 2) of buzhidao and levitopher (the argument given is also used in well-known textbooks such as Shankar, p.176). It is too sloppy in my opinion to say that "at infinity, the function is zero". This is ...

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Probability is the chances of occurrence of an event, say for example the event is to find an electron around the nucleus of an atom. If the atom has an electron around the nucleus then the probability of finding the electron around the nucleus is one(1). but if the electron is to be sort some place away from the vicinity of the nucleus then the probability ...

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As John Rennie has pointed out, the wavefunction $\psi$ itself isn't observable, and any expectation value is calculated squaring the wavefunction $$\langle A \rangle = \int \textrm{d}x \psi^*(x) A \psi(x)$$ Therefore, you are allowed to change the overall phase of the wavefunction without changing the expectation values: $$\psi(x)\to \psi'(x) = e^{i\theta} ... -4 Physical intuition behind negative values for wave function? Take a look at Jeff Lundeen's website. He and a team have done weak measurmeent work where they measure wavefunction directly. See his semi-technical explanation where you can read this: "With weak measurements, it’s possible to learn something about the wavefunction without completely ... 3 The wavefunction contains all the information about the system, but it is not itself observable. Since our intuition tends to be linked to what we have observed in the past, it's not surprising that the wavefunction seems unintuitive. As you say it can be negative, but it can also be a complex number making it even less intuitive (if that's possible!). ... 2 A wavefunction with negative sign works just like any other wave with negative sign. For example, water waves with negative height cancel out with waves of positive height. You can also make a 'negative' wave on a string by pulling the end down and back up, which will cancel with a positive wave. -2 Is the photon's wave function the same as an electromagnetic wave (light)? No, wavefunction is not the same as an electromagnetic wave. However take a look at Jeff Lundeen's material including his semi-technical explanation of weak measurement: "We hope that the scientific community can now improve upon the Copenhagen Interpretation, and redefine the ... 0 Yes, you hit the nail right on the head, with respect to the H-H NBP, as well as Vilenkin's related tunnelling mechanism. But, there is a further problem as to why the H-H proposal simply does not work: Conformal modes lead to the Einstein-Hilbert action not being bounded from below, which in turn implies that the sum over all 4-geometries leads to a sum ... 1 I think it is important to note that the wavefunction of quantum mechanics is not a field like the electric or magnetic field. It assigns complex numbers to configurations. So it is a function where the domain of the function is not space or even spacetime. So there is not a value of the field at a point in spacetime, so there isn't a thing to curve it. ... 1 Unfortunately, that question's answer depends on the theory of quantum gravity that you chose. But roughly here is the situation : If you take a classical spacetime and quantum matter fields (G_{\mu\nu} = \langle T_{\mu\nu} \rangle_\omega), also known as semiclassical gravity, then the stress energy tensor varies with the measurement. The standard ... 0 I think you have a few misconceptions here. You start by talking about the particles in the beam "not interfering with each other" so the "wave function of each particle is lambda/p". There are at least two problems with this statement. I'll take the last part first. It looks like you are confusing "wave function" with "wave length". The wave function ... 1 It’s unrealistic to expect most students to able to derive the energy quantisation of a hydrogen atom on the spot. For hydrogen:$$E_n = \frac{-13.6 }{n^2}\rm eV$$(ignore the minus sign for your problem). For a particle in a 1 D infinite potential well:$$E_n = \frac{n^2h^2}{8mL^2}$$Set n = 1 to obtain the energies of both ground states. From the ... 1 Indeed the cosine function is valid for instance for other boundary conditions$$\psi'(0)=\psi'(L)=0$$Your goal when you look for a set of solutions to the Schroedinger equation is to be able to decompose any general wavefunction as a sum over this set, and to do it consistently your boundary conditions must be the same for all solutions. So it is unlikely ... 1 The separable solutions are exactly the eigenstates of the Hamiltonian which are exactly the ones where the probability density does not change. However you can have a flux or flow of probability even if the probability doesn't change. This is like electromagnetism where you can have current floe through a wire even if the wire has no change of charge ... 0 The particle in a box is subjected to the further restriction that the potential function rapidly tends to infinity at  x=0  and  x=L  This means that the position of the particle is restricted to being within the box. As you said, the general solution to the schrodinger equation for particle in a box is$$ \psi (x)=A\cos kx + B\sin kx $$Because the ... 1 What you've written is only true if \psi(x) is an eigenstate of H. For some general \psi(x) that is not an eigenstate (i.e. H \psi(x) \ne E \psi(x)), then \Psi(x, t) will be more complicated than just the time independent wavefunction multiplied by a phase factor. 0 This would not work for an arbitrary smooth central potential V: \mathbb{R}_+ \to \mathbb{R} with$$V~<~0, \qquad V^{\prime}~>~0,\qquad V(0)~=~-\infty,\qquad V(\infty)~=~0,$$because there might only be a finite number of bound states, and hence no continuous limit of bound states. For instance, one can use WKB methods to argue that if V(r) ... 0 It would be better to specify that V_0 < 0 as otherwise your problem is that of a potential barrier (no bound states) and not that of a finite potential well. The case of a finite potential well is fully developed here (Wikipedia entry), the case of a rectangular potential barrier you can find here (Wiki). 1 The name "1-form" is only used because there are 2-forms and 3-forms and so on. An equivalent but probably better name for 1-forms in this case is that of "dual vector." Any vector space will have associated with it the dual space consisting of all linear functions that map vectors into scalars. This is all a dual vector is: a thing that linearly maps ... 1 Suppose you have a skewed coordinate system, with vectors that have components v^i defined relative to some basis vectors \hat e_i such that \vec v = v^i \hat e_i in the Einstein summation convention (we sum over any index which is repeated once above, once below). The meaning of a skewed coordinate system is that \hat e_i \cdot \hat e_j \ne ... 5 Notation \renewcommand{braket}[2]{\langle #1 | #2 \rangle} First let us define the term "1-form": A 1-form is a linear function which takes a single vector as it's argument. That is it. It is no more confusing or complicated than that single statement. Consider a vector v. Given any other vector w we can form the inner product \braket{v}{w}. ... 0 You are correct that for n = 3 there are 2 non-boundary zero points. Also, the modulus of \psi(x) is highest where V(x) is lowest. Where you are wrong is that \psi(A) is not zero and \psi(B) is not zero either, as you indicate in your schematic. For x < A and x > B, V(x) is not \infty (your well is a finite, not infinite well) and ... 1 The wavefunction \psi(x) satisfies$$ -\frac{\hbar^2}{2m}\psi'' + V_0\left(\frac{x}{L} - 1\right) \psi = E\psi, \quad 0 \leq x \leq L\\ -\frac{\hbar^2}{2m}\psi'' = E\psi, \quad x > L $$Since the bound states have E < 0 let's introduce$$ k = \frac{\sqrt{-2mE}}{\hbar}\\ \varkappa = \frac{\sqrt{2mV_0}}{\hbar} $$Then$$ \psi'' - ...

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Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy ...

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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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