17 votes

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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12 votes

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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  • 109k
8 votes
Accepted

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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  • 5,194
7 votes

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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  • 143
7 votes
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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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  • 53.9k
5 votes
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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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4 votes
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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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  • 8,602
3 votes

Question About Momentum and Position Operators and the Postulates of Quantum Mechanics

This is a result. No one tried to make momentum eigenstates planewaves. This is a property. The derivation is the following: $$\psi(p)=\langle p| \Psi \rangle = \int\langle p | x \rangle \langle x | ...
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  • 1,160
3 votes

What are the adequate Hilbert spaces for Schrödinger, Schrödinger–Pauli, Dirac equations, and QFT?

One way to proceed is to start with the classical observable and find some suitable Lie algebra containing these as generators. Upon quantization, the observables are promoted to operators, now ...
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3 votes
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Omitting the negative exponential in the plane-wave solution of the Schrodinger equation

I hope by telling you this, things will become clearer. If not, you can always comment. So, to start, first I shall say that both $$\Psi_k(x)=A_ke^{ikx}+B_ke^{-ikx} \ \text{and}\ \Psi_k=A_ke^{ikx}$$ ...
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  • 2,119
3 votes

How to compute the exact universe branches in many worlds interpretation?

This scenario of multiple observers in nested systems is discussed in Everett's thesis starting on page 4. "The question of the consistency of the scheme arises if one contemplates regarding the ...
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3 votes
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How to show that $\psi(x)=-A\exp(-\alpha^2x^2)$ satisfies TISE for $V(x)=\frac 1 2 m\omega_0^2x^2$?

Write the TISE as $$\frac{d^2\psi}{dx^2}=\frac{m^2\omega_0^2x^2-2mE}{\hbar^2}\psi$$ and observe that for large $x$, $2mE$ is negligible compared to $m^2\omega_0^2 x^2$. Therefore, you can write it as $...
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  • 1,074
2 votes

Combine the completeness and linearity of Schrodinger equation's solution, can we say that any wavefunction can solve any Schrodinger equation?

The energy eigenstates are the solutions to the time independent Schrodinger equation, however linear combinations of the energy eigenfunctions are not time independent, so they are not solutions to ...
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2 votes

Do continuous wavefunction form a Hilbert space?

Some say that item 1 is also false. Basically a rigged Hilbert space is more appropriate. See e.g. https://arxiv.org/abs/quant-ph/0502053 or https://doi.org/10.1007/3-540-088431-1.
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  • 1,592
2 votes

What would the wavefunction represent in an observer-less universe?

You probably misunderstood what "observer" means. It has nothing to do with people, consciousness or anything of the sort. In quantum mechanics, an observation, or a measurement, is an ...
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  • 4,323
2 votes

Components of State vector in quantum mechanics

The relationship between wave functions and state vectors is the exact same relationship as between vector components and vectors in Euclidean vector space. When you have an "ordinary" ...
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2 votes

What are the adequate Hilbert spaces for Schrödinger, Schrödinger–Pauli, Dirac equations, and QFT?

Here is another approach. Instead of defining a state space $L^2(\text{something})$ and then defining observables as the self-adjoint operators on that space, you can go the other way around: Choose ...
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  • 275
2 votes
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What is the wave function of a field?

In QFT, fields come into play. How would the wave function of a field be? $\psi(A_\mu)$? Somewhat analogously to the basis of position eigenstates in QM, there is a basis of field eigenstates in QFT. ...
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2 votes

What is the wave function of a field?

I will post an answer that discusses both probabilities in QFT, but also the object in QFT that is analogous to the wave-function in QM. Probabilities : Probabilities and amplitudes are formed with ...
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  • 2,119
2 votes

Question About Momentum and Position Operators and the Postulates of Quantum Mechanics

Instead of explicitly postulating properties of "position space" or "momentum space" or talking about Fourier transforms, it is much more common in axiomatic treatments to - in the ...
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  • 109k
1 vote

If we treat the wave function as a vector, do we treat the wave function in 2d as a tensor?

In 2D, there are two continuous variables ψ(x,y). Do these variables get demoted to indices resulting in a rank 2 tensor? The output of the wavefunction in more than one dimension is still just a ...
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  • 4,531
1 vote

Are there wave functions that are neither symmetric nor antisymmetric?

For simplicity work with two particles. If they were distinguishable, every physical state could be described as $|\Psi \rangle = \sum_{ij}c_{ij} |\psi_i\rangle \otimes |\phi_j\rangle \in H_1\otimes ...
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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, \begin{equation} 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} \...
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1 vote

Does Born's rule guarantee component-wise probabilities?

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,...
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  • 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 ...
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  • 2,119
1 vote
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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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