# Tag Info

### Do continuous wavefunction form a Hilbert space?

Item 2 is simply false if wavefunction is understood as a generic state vector of the system. This is compatible with the answer NO to your question. With the above interpretation wavefunctions are ...
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### What are the adequate Hilbert spaces for Schrödinger, Schrödinger–Pauli, Dirac equations, and QFT?

Ordinary QM has essentially a fixed Hilbert space of $L^2(\mathbb{R}^n)\otimes S$, where $n$ is the number of spatial dimensions and $S$ some representation of the rotation group. This is due to the ...
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### Components of State vector in quantum mechanics

A state vector is just a wavefunction where the domain doesn't need to be $\mathbb{R}^d$. Similarly, you can call wavefunctions vectors because the functions used in quantum mechanics (square ...
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### Do continuous wavefunction form a Hilbert space?

Item 2 is false. The Hilbert space $\mathcal H=L^2(\mathbb R)$ consists of all square integrable functions, including discontinuous ones. Even though the Hamilton operator $H=-\Delta+V$ is a ...
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### Is a normalized wave function in the position basis automatically normalized in the momentum basis?

For a particle on a line (with Hilbert space $L^2(\mathbb R)$), both position and momentum eigenstates are non-normalizable and therefore unphysical. When we talk of a "normalized" momentum ...
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### Question About Momentum and Position Operators and the Postulates of Quantum Mechanics

Out of the 4 statements you've given, (3) and (4) are the most "fundamental"... they are equivalent to just the commutation relation $[x,p]$ which is the most common to take as a postulate ...
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### Why is the eigenwave function not periodic in the lattice?

At the same time, in the lattice, for any function $f(\mathbf{r})$, we have: $$f(\mathbf{r}+\mathbf{R}_n) = f(\mathbf{r})$$ No. This is not true. Why would this be true for "any function" ...
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1 vote

### Omitting the negative exponential in the plane-wave solution of the Schrodinger equation

As Schris38 pointed out $\Psi_k$ is not the most general solution either way. But even so, A e^{ikx}+B e^{-ikx}=(A+iB) \cos(kx)+(A-iB)\sin(kx)\equiv C_1\cos(kx)+C_2\sin(kx) \tag{1} \...
1 vote

Firstly, $c^* c$ will always be positive and real, because for any complex number $c=a+bi$ which is equal to $a^2+b^2\ge 0$. Secondly, your notation looks a little confused! A vector $\vec\psi=(\psi_a,... • 432 1 vote ### If arbitrary wavefunctions can be expanded as energy eigenfunctions of a schodinger equation, is it mean that it can solve schodinger equation anyway? This means that any function that respects the boundary conditions of the solutions of the Schroedinger equation indeed solves the Schroedinger equation, yes. But beware, as this restriction is ... • 2,119 1 vote Accepted ### Why Airy disk (Fraunhofer diffraction) is first order of the Bessel function of the first kind? It's not the Bessel function itself, instead it's proportional to a ratio of the Bessel function and its argument. This turns the first-order zero of the$J_1$at$x=0$into a constant as$x\to0,\$ i.e....
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