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## Hot answers tagged wavefunction

12

The definition is $$\psi(x)=\langle x| \psi\rangle, ~~~\leadsto \\ |\psi\rangle= \int dx ~~\psi(x) | x\rangle , ~~\leadsto \\ |\Psi\rangle= \int dx ~~ A e^{-ikx}| x\rangle .$$ Wavefunctions are coefficients of coordinate kets. NB You may also then check $$\langle p|\Psi\rangle= \int dx ~~A e^{-ikx} \langle p|x\rangle ={A\over \sqrt{2\pi \hbar}} \int dx ~e^... 9 If you want to use the theory of probability, a necessary condition for a wavefunction to be physically meanigful is$$\psi \in L^2(\mathbb{R}^3,d^3x)\:.$$That is because, as a basic postulate of QM, we have that: \qquad\qquad\qquad\qquad |\psi(x)|^2 is the probability density to find the particle at x, and the total probability must be 1:$$\int_{\...

7

A ket is an abstract vector which describes the state of your system. A wavefunction is a representation of that vector in a particular basis, usually the position basis (although you will sometimes also hear about, e.x. "momentum-space wavefunctions"). The way the position-space wavefunction $\psi$ is defined is by letting $\psi_\alpha(x)$ denote ...

7

Your problem -- as far as I can see -- is a simple calculation error: while you are correct that $\sin(k_0 x) = \mathcal{I}(e^{ik_0 x})$, it does not follow that $$\mathcal{I}(e^{i(k_0-k) x}) = \sin(k_0 x)e^{ikx} \quad \text{(Wrong!)},$$ as you should quickly be able to see since the left-hand side should be real but the right hand side has an imaginary part....

7

The first of your two suggestions doesn't make sense, "the probability amplitude of finding the particle at position $x$ in the state $|\psi\rangle$". The particle is either in the state $|\psi\rangle$ or at position $x$ (in which case it would be in the state $|x\rangle$), assuming $|\psi\rangle\neq|x\rangle$. The quantity $\langle x|\psi\rangle$ ...

6

This is where the textbooks, in a way, lie to you. The operator $\hat{x}$ (and its counterpart, $\hat{p}$) is not a "good" quantum operator for a number of reasons, including the fact that these operators do not have normalizable eigenvectors, as you have seen. In particular, $$|x\rangle$$ is not a sensible eigenvector as it is not normalizable. ...

5

The problem with my question was I assumed that this below postulate of quantum mechanics (given in R. Shankar chapter 4 as 3rd postulate) to be literally true If the particle is in a state $|\psi\rangle$, measurement of the variable (corresponding to) $\hat{\Omega}$ will yield one of the eigenvalues $\omega$ with probability $P(\omega)\propto |\langle \... 5 I think what you just discovered is that every (separable*) infinite-dimensional Hilbert space$\mathcal H$is isometrically isomorphic to$\ell^2$. The isomorphism is given by mapping a vector$|\psi\rangle \in \mathcal H$to the coefficient list$( \langle e_n | \psi \rangle )_n$. In particular,$L^2(\mathbb R^d)$is isomorphic to$\ell^2$. For example, ... 4 A single electron does not experience Coulomb repulsion with itself. The potential and wavefunctions you describe involve only one-electron, so the state energy cannot be associated (even as an approximation) with the Coulomb repulsion between two electrons. On the other hand, the double delta potential can be seen as a very crude approximation of the ... 4 (1) Nothing wrong there. Plane waves are states of infinitely precise momentum and cannot be properly normalized in position space due to having infinite spread from Heisenberg uncertainty. In practice they still help e.g. in the scattering matrix formalism to get an amplitude for reflection and an amplitude for transmission, and to settle e.g. the basic ... 4 The relationship of the values that wave functions take to "kets" is the same as that of vector components to vectors - which makes sense since "kets" are a type of vector, just one in a space that is more general and quite easily (very!) infinite-dimensional. In Euclidean space, you have a vector of the form $$a_1 \mathbf{e}_1 + a_2 \... 4 Scattering experiments are exactly the sort of experiments you describe. Consider the experiments done at the Large Electron–Positron Collider where electrons were scattered off positrons. In the lab frame it is true that both the electrons and positrons are moving, but there are no absolute velocities so it is just as valid to analyse the results in a frame ... 4 What you've uncovered is motivation for the statement, "relativistic quantum mechanics is intrinsically a many-body theory". Standard model operators like creation and annihilation do precisely what you describe - have a particular particle's wavefunction cease to exist after a certain time. Spacetime normalisation is the only Lorentz-invariant way ... 4 It's tempting to try to interpret the mathematics of QM or QFT too literally, however this can be a dangerous game to play. Quite often you will cause more problems for yourself than you will solve by asking what exactly particles are (are they points, are they waves, are they fields, are they strings etc.), not least because different theories will give you ... 3 All experiments are consistent with the point particle nature of the electron. You are mixing up particle shape, potential function and wave function. 3 For the first part, you would use Bloch's Theorem. For the second part, why don't you plug it into the Schrodinger Equation and see what happens? 3 The \delta potentials may be the main reason why this is a problem. In reality, the nuclei have 1/r potentials. The potential for a single electron is lower between the nuclei than on either side. When the electron wave function has a node between the nuclei, the negative charge density is low. The positive nuclei also feel each other's 1/r potential ... 3 There is no a-priori sense in having an infinitely large potential, you can't calculate with infinite numbers (except for some very special mathematical considerations). So, the physical meaning that is given to such a potential must be implicitely contained in the accompanying text that explains the infinite potential well. An this meaning is commonly: &... 3 In one improper basis as |\mathbf{r}\rangle we have the resolution of the identity in the following form$$\int d^3\mathbf{r} |\mathbf{r}\rangle\langle \mathbf{r}|=\mathbf{1}\tag{1}.$$Now since this is just the identity operator you can insert it wherever you want, that being one very common manipulation in QM. In particular you may insert it in between \... 3 Emission and absorption of a photon aren't instantaneous processes as long as you don't perturb the system by measuring its state. If you do, the system collapses in a non-reversible way. Let me explain how it evolves in the case you don't measure: Spontaneous emission An atom in the excited state | a \rangle emits light in the same spatial pattern as a ... 3 Nothing went wrong, the difference between both results is just a global constant phase, which has no meaning. If \psi_n(x) is an eigenfunction of the Hamiltonian, so is e^{i \alpha}\psi_n(x). In your case, e^{i\alpha}=(-1)^m. 2 The position eigenstates are not actually part of any Hilbert space at all. A position eigenstate wave function with eigenvalue x_{0} is \psi(x)=\delta(x-x_{0}). If we try to normalize this, we need to calculate$$\int_{-\infty}^{+\infty}dx\,|\psi(x)|^{2}=\int_{-\infty}^{+\infty}dx\,[\delta(x-x_{0})]^{2}=\delta(0)=\infty.$$So this basis state is ... 2 Yes, the product of two antisymmetric wavefunctions is symmetric. However, such a system is not a fundamental particle, it is a composite system (of an even number of fermions). Such systems do exist: they are called composite bosons (in order to distinguish them from elementary bosons like the photon or the Higgs boson). Examples include Cooper pairs in ... 2 This answer is meant as a supplement to Valter Moretti's one. Quantum systems are defined by their Hamiltonian, however, this is normally not their single observable. We have the basic observables stemming from the space-time context (coordinate, momentum, spin), plus the Hamiltonian, plus other observables stemming from other symmetries (orbital / total ... 2 It is a mathematical phenomenon, it happens due to time-dependent Schroedinger equation. It is similar to diffusion equation, localized peaks in tend to dissipate according to both equations. So any localized wavepacket spreads unless that spreading is stopped by a potential rise. This behaviour can be most easily shown and quantified for free particle. If ... 2 Wavefunctions in quantum mechanics are elements of the Hilbert space L^2(\mathbb{R}). This is usually described as the vector space of square-integrable functions with the inner product:$$\langle \psi,\phi \rangle = \int_\mathbb{R} \psi^*(x)\phi(x) dx$$However one should note that one of the axioms for an inner product is that the inner product of any ... 2 Periodic boundary conditions state$$\psi(x)=\psi(x+L)\tag{1'}.$$Goodstein says "[...] it (the particle) has equal probability of being anywhere on the line", which would not be true if we take$\psi(0)=\psi(L)=0$. You can see that with the condition ($1'$),$k$takes the values that he claims. Edit: In this particular case, I don't know what are ... 2 Elliptic regularity implies that$\psi$admits weak derivatives of every order which are$L^2$locally. Now, Sobolev's lemma implies that these derivatives must be standard derivatives. In summary, every weak solution is just the standard solution$\psi_m$. 1 For a state of definite energy, the Schrodinger equation says that the time dependence is supposed to be$e^{-i(E/\hbar)t}$, where$E$is an eigenvalue of the Hamiltonian. For most Hamiltonians, it is not going to be true that$-E$is also an eigenvalue, so the complex conjugate$e^{i(E/\hbar)t}$will not be a solution of the Schrodinger equation. Therefore ... 1 Are there experiments that show that the electron wave function can be point-like? The wave function is not a measurable quantity, so there are no experiments to measure it. What is measurable is the probability distribution for a given experimental situations,$Ψ^*Ψ\$, which can be compared with a large number of measurements with the same boundary ...

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