When solving one dimensional problems in quantum mechanics it is often assumed that the first derivative of the wave function is continuous. What justifies this assumption?
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Is there anything in the physics that enforces the wave function to be $C^2$? Are weak solutions to the Schroedinger equation physical? I am reading the beginning chapters of Griffiths and he doesn't mention anything.
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How does a wavefunction collapse into one state? More specifically, what conditions cause a wavefunction for a quantum particle to collapse?    Does this have to do with density matrices?    Please try to give a mathematical explanation that is simple and concise...
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The general formulation of quantum mechanics is done by describing quantum mechanical states by vectors $|\psi_t(x)\rangle$ in some Hilbert space $\mathcal{H}$ and describes their time evolution by the Schrödinger equation $$i\hbar\frac{\partial}{\partial t}|\psi_t\rangle = H|\psi_t\rangle$$ where $H$ is the Hamilton operator (for the free particle we have $H=-\frac{\hbar^2}{2m}\Delta$).  Now I have often seen used spaces like $\mathcal{H}=L^2(\mathbb{R}^3)$ (in the case of a single particle), but I was wondering whether this is correct or not.  In fact shouldn't we require to be able to derivate $\left|\psi_t\right&gt;$ twice in $x$ and thus choose something like $\mathcal{H} = H^2(\mathbb{R}^3)$?   If we treat directly $\psi(t,x) := \psi_t(x)$, shouldn't we require them to be in something like $H^1(\mathbb{R};H^2(\mathbb{R}^3))$? i.e., functions in $H^1(\mathbb{R})$ with values in $H^2(\mathbb{R}^3)$, e.g. the function $t\mapsto\psi_t$.
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If a particle is totally localized at $x=0$, its wave function $\Psi(x,t)$ should be a Dirac delta function $\delta(x)$. Accordingly, its Fourier transform $\Phi(p,t)$ would be a constant for all $p$, thus the particle's momentum is totally uncertain. I guess this is what the Uncertainty Principle told us.  But in another hand, since the particle is totally localized at $x=0$, it is not moving and therefore static. Its velocity should be 0, so is its momentum, which contradicts the conclusion above.  So, what's wrong with my thinking?  Is it that there shouldn't be a TOTALLY STATIC particle, or it's in a non-normalizable state, so the uncertainty principle is not applicable?
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Let $u$ solve the initial-value problem or the wave equation in one dimension: $$\begin{cases}u_{tt}-u_{xx}=0 &amp; \text{in } \mathbb{R} \times (0,\infty) \\ u = g, u_t = h &amp; \text{on } \mathbb{R} \times \{t=0\}. \end{cases}$$ Suppose $g,h$ have compact support. The kinetic energy is $k(t) := \frac 12 \int_{-\infty}^\infty u_t^2(x,t) \, dx$ and the potential energy is $p(t) := \frac 12 \int_{-\infty}^\infty u_x^2(x,t) \, dx$. Prove (a) $k(t)+p(t)$ is constant in $t$, (b) $k(t)=p(t)$ for all large enough times $t$.  This is Chapter 2, Exercise 24 of PDE Evans, 2nd edition. I am only doing part (a) right now; my work is shown below: Define $$e(t):=k(t)+p(t)=\int_{-\infty}^\infty u_t^2+u_x^2 \, dx.$$ Then $$e'(t)=\frac 12 \int_{-\infty}^\infty 2u_tu_{tt}+2u_{x}u_{xt} \, dx= \int_{-\infty}^\infty u_tu_{tt}+u_{x}u_{xt} \, dx.$$ Now, I want to get $e'(t)=0$ so that $e(t)$ is constant. How can I go about doing this? I do know that I can use $u_{tt}-u_{xx}=0$, if $t &gt; 0$. With $t&gt;0$ into mind, do I have to consider the $t=0$ case separately? Or can I treat both cases together as $t \ge 0$, since $g$ and $h$ have compact support?
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Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?
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I stumbled on this question while answering the following question:  Given a general wavefunction $\psi(x)$ and an operator $\hat Q$, with eigenvalue spectrum $\{q_r\}$ and corresponding eigenfunctions $\{u_r\left(x\right)\}$, show that the expectation value of $\hat Q$ is given by $$\langle\hat Q\rangle = \int_{-\infty}^\infty{\psi^*\hat Q \psi \,dx}$$  I started of with the ususal definition of the expected value and inserted the identity operator in position representation: $$\langle\hat Q\rangle = \langle \psi|\hat Q|\psi\rangle = \langle \psi|\left(\int dx|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\right)|\psi\rangle$$  Taking $\langle\psi|$ and $|\psi\rangle$ into the integral gives: $$\left(\int dx\langle \psi|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\psi\rangle\right)$$  Pulling the 2nd integral into the first:$$\int dx\int  {dx}'\langle \psi|x\rangle\langle x'|\psi\rangle\langle x| \hat Q|x'\rangle$$  This is the point where I'm stuck. If $\langle x|\hat Q|x'\rangle = \hat Q \delta(x-x')$, then the last expression would evaluate to the answer. However I  don't know why this should be true and even if it is true I wouldn't know how to prove this equality.
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I've been studying the postulates of QM and seeing how to derive important ideas from them. One thing that I haven't been able to derive from them, however, is the identity of the momentum operator.  For simplicity, I'm only thinking about no relativistic effects, no spin, no time-dependent potentials, and one spatial dimension. Also I'm assuming the position operator is simply multiplication by $x$, as in, I'm in position space. So the Hamiltonian operator is  $ H = -\frac{\hbar^2}{2m}\nabla^2+V$.  I know that the momentum operator is $p = -i\hbar \frac{\partial}{\partial x}$.  But how do I get there from the postulates? I know that it makes sense, as it results in the Ehrenfest Theorem, the De Broglie wavelength hypothesis, the Heisenberg Uncertainty Principle (for $x$ and $p$), the momentum operator being the generator of the translation operator, and possibly many other desirable theorems, and correlations with classical momentum.  But none of these are postulates (at least, not in the various formalisms I encountered), so you can't derive $p = -i\hbar \frac{\partial}{\partial x}$ from them. Rather, they are consequences of it. You need to know the operator beforehand to see that they are correct. Yes, this is just semantics, but that is the core issue for me:  Regardless of how much sense it makes, is the identity $p = -i\hbar \frac{\partial}{\partial x}$ (under the assumptions I made) a Postulate, meaning that you can't derive it from other postulates, or can it in fact be obtained from them? And in the latter case, could you show me how?  Note: I know that there are many different and equivalent sets of postulates for QM. But in none that I saw did they name it as a postulate nor properly derived it.
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The $\delta(x)$ Dirac delta is not the only "point-supported" potential that we can integrate; in principle all their derivatives $\delta', \delta'', ...$ exist also, do they?  If yes, can we look for bound states in any of these $\delta'^{(n)}(x)$ potentials? Are there explicit formulae for them (and for the scattering states)?  To be more precise, I am asking for explicit solutions of the 1D Schroedinger equation with point potential,  $$- {\hbar^2 \over 2m} \Psi_n''(x) + a \  \delta'^{(n)}(x) \Psi(x) \ = E_n \Psi_n(x) $$  I should add that I have read at least of three set of boundary conditions that are said to be particular solutions:   $\Psi'(0^+)-\Psi'(0^-)= A \Psi(0)$ with $\Psi(0)$ continuous, is the zero-th derivative case, the "delta potential". $\Psi(0^+)-\Psi(0^-)= B \Psi'(0)$ with $\Psi'(0)$ continuous, was called "the delta prime potential" by Holden. $\lambda \Psi'(0^+)=\Psi'(0^-)$ and $\Psi(0^+)=\lambda\Psi(0^-)$ simultaneusly, was called "the delta prime potential" by Kurasov   The zero-th derivative case, $V(x)=a \delta(x)$ is a typical textbook example, pretty nice because it only has a bound state, for negative $a$, and it acts as a kind of barrier for positive $a$. So it is interesting to ask for other values of $n$ and of course for the general case and if it offers more bound states or other properties. Is it even possible to consider $n$ beyond the first derivative?  Related questions  (If you recall a related question, feel free to suggest it in the comments, or even to edit directly if you have privileges for it)  For the delta prime, including velocity-dependent potentials, the question has been asked in   In the halfline $r&gt;0$, the delta is called "Fermi Pseudopotential". As of today I can not see questions about it, but  seems to be the same potential.  A general way of taking with boundaring conditions is via the theory of self-adjoint extensions of hermitian operators. This case is not very different of the "particle in 1D box", question  A general take was the question  A related but very exotic question is  because obviosly the point-supported interacctions have a peculiar scaling  Comments  Of course upgrading distributions to look as operators in $L^2$ is delicate, and it goes worse for derivatives of distributions when you consider its evaluation $&lt;\phi | \rho(x) \psi&gt;$. Consider the case $\rho(x) = \delta'(x) = \delta(x) {d\over dx}$. Should the derivative apply to $\psi$ only, or to the product $\phi^*\psi$?
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Let $f$ be a continuous function on the interval $[a,b]$ such as for all x in $(a,b)$ : The right hand derivative equals $0$. Is $f$ constant?   The obvious reflex one would have is to try to find a counter-example, I tried, many times and I always failed. I assumed it to be true and tried to prove it, I tried manipulating sequences, it never worked. I tried calculating the left hand derivative of a point by using a sequence, it seemed to work at first, but it didn't as well. All in all, I think $f$ may not be necessarily constant.
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The collapse of the wave function by measurements is one of the most mysterious properties of quantum mechanics.    At what scale does the wave function collapse? What are the conditions for a collapse?  Can it be made precise?  Does the collapse happen instantaneously, or is it spread out in time?  What is the precise equation for the collapse?  Why is the collapse time-asymmetric, and can this explain the arrow of time?
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To take an example,           Let's think of an electron which ban be in either one of two positions [...] There are two possible states of definite energy for the electron.  Each state can be described by the amplitude for the electron to be in each of the two basic positions.  In either of the definite-energy states, the magnitudes of these two amplitudes are constant in time, and the phases vary in time with the same frequency.  On the other hand, if we start the electron in one position, it will later have moved to the other, and still later will swing back to the first position.   (Feynman derives this behaviour from first principles in chapter 8-6 by solving equations for the state of an ammonia molecule, but that's a nice summary).  This sounds like the uncertainty principle - that you can know energy or position but not both.  Is it possible to derive the uncertainty principle from such an analysis?  Or is the uncertainty principle somehow axiomatic in the way this behaviour is derived?
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What is the underlying reason that the same pairs of conjugate variables (e.g. energy &amp; time, momentum &amp; position) are related in  (e.g. time symmetry implies energy conservation) and likewise in QM (e.g. $\Delta E \Delta t \ge \hbar$)?
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Wave equations take the form:  $$\frac{ \partial^2 f}  {\partial t^2} = c^2 \nabla ^2f$$  But the Schroedinger equation takes the form:  $$i \hbar \frac{ \partial f}  {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$  The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation?  And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?
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When an observer causes the wave function of a particle to collapse, how can we know that the wave function was not collapsed already before the measurement?  Suppose we measure the z-component of the spin of an electron. After the measurement, it is entirely aligned along the measured direction, e.g. the +z-direction. Before the measurement we need to assume that a probability distribution proportional to $|\Psi|^2$ of the two allowed directions is present.  If we repeat the measurement with many identically prepared electrons, we should see such a distribution finally. For example, we could measure 40% spin-down and 60% spin-up.  However, it seems we could also assume that all of these particles have a defined spin-direction before we measure them.  What is an intuitive (being aware that quantum phenomena as such are rarely intuitive) explanation for why we cannot simply assume that the spin was already aligned completely in that measured direction?    With regards to the suggestion that this two-year old question is a duplicate , I would like to point out that my question isn't limited to entanglement, but asks about a very fundamental principle in quantum mechanics, and as such is not a duplicate.
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The Schrödinger equation in non-relativistic quantum mechanics yields the time-evolution of the so-called wavefunction corresponding to the system concerned under the action of the associated Hamiltonian. And this wavefunction is, in general, complex, and its modulus squared yields the probabilities observed experimentally. Though, perhaps, this question has been asked many times, I am wondering if there is a direct physical interpretation - something that physically corresponds to - the wavefunction. Or is it just an intermediate calculational tool to arrive at the appropriate predictions for experimental outcomes, and nothing more? Of course, things like superposition and interference effects follow from the complex nature of the probability amplitude. So there must be something physical about it. What is it? Or are we not supposed to ask that question? Is it because the probability amplitude is complex that we have difficulty in relating it to something physical? Can we do quantum mechanics without complex numbers?
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I want to ask if it is reasonable that I use the Dirac-Delta function as an initial state ($\Psi (x,0) $) for the free particle wavefunction and interpret it such that I say that the particle is exactly at x=0 during time t=0? If I use this initial state, can I also use it to predict how the wavefunction should evolve in time? That is, if $\Psi (x,0) = \delta(x) $, then, $$ \phi(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \delta(x)e^{-ikx} dx=\frac{1}{\sqrt{2 \pi}} $$ then, $$ \Psi(x,t)= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \phi(k)e^{i(kx-\omega t)} dk $$ or, $$ \Psi(x,t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty}e^{i(kx-\omega t)} dk. $$  Is the above wavefunction a valid expression to see how a particle that is initially localized in the origin evolves in time?