74
votes
Accepted
Normalizable wavefunction that does not vanish at infinity
Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.
- 2,740
29
votes
Normalizable wavefunction that does not vanish at infinity
Let
$$ \psi(x)
= \begin{cases} 1 & \exists\, n \in \mathbb N: x \in [n, n+\frac 1 {n^2}]\\ 0 & \text{otherwise.} \end{cases}
= \sum_{n \in \mathbb N} \mathbf 1_{[n,n+\frac 1 {n^2}]}(x) , $$
...
- 7,356
24
votes
Accepted
Getting particles from fields: normalization issue or localization issue?
Might as well collect my comments, most deleted, in this memo answer.
Essentially, QFT does not want you to go near position eigenstates of the style of QM. The eigenstate of the momentum operator, $|...
- 65k
19
votes
Accepted
How should Dirac notation be understood?
Physicists usually generously relax the condition that the norm should be finite and they sometimes say that $|\vec r\rangle,|\vec p\rangle$ belong to the "Hilbert space". It's exactly the same "...
- 180k
11
votes
A common standard model Lagrangian mistake?
It is a mistake. It's true, as FrodCube said, that you can work around it by redefining the field, but no one actually defines the field that way.
The origin of it is a CERN photo session of John ...
- 28.1k
10
votes
Accepted
Partition function in classical thermodynamics
The factor of $V$ comes from integration over $\mathrm d\vec x$ (as we assume that the Hamiltonian is position independent). On the other hand the factors of $2\pi$ and $\hbar$ are essentially ...
10
votes
Normalize wave function with respect to time instead of space
For a quantum system with one degree of freedom on the closed interval $I$, the Hilbert space is $L^2(I)$. In this case the $I$ is the range for the spatial coordinate $x$, so that normalisation ...
- 9,599
10
votes
Accepted
The use of $a^\dagger(\mathbf{k}) = -i \int d^3x e^{ikx}\stackrel{\leftrightarrow}{\partial}_0 \phi(x)$ in the derivation of the LSZ-formula
OP asks good questions. Let us try to sketch the logic of the LSZ reduction formula.
In the Heisenberg picture, a free real field $\hat{\varphi}(x)$ has a Fourier expansion
$$ \hat{\varphi}(x)~=~\...
- 208k
10
votes
Normalization of vacuum state in field theory
The vacuum state is denoted $|0\rangle$ in the analogy with the $0$-quanta ground state $|0\rangle$ of a quantum SHO, and in both cases $\langle 0|0\rangle=1$. Unsurprisingly, this implies $3$-...
- 25k
9
votes
Accepted
Weinberg QFT 1 Normalization one 1 particle states p. 66
I had basically the same questions as you! In my case at least, it stemmed from not properly understanding the previous arguments, so I'll try to explain everything in an organized way. It took many ...
- 1,593
9
votes
Why do different quantum field theory books have different conventions regarding the normalization?
Some books wish to have a simple-looking commutator for the annihilationa and creation operators:
$$
[\hat a_p,\hat a^\dagger_q]= (2\pi)^3 \delta^3(p-q)
$$
while other prefer the normalization
$$
[\...
- 55.2k
9
votes
A common standard model Lagrangian mistake?
It's not really a mistake, just a non-canonical normalization for $\psi$. The physics with and without the $h.c.$ is literally the same.
The canonical way of normalizing $\psi$ is such that
$$\mathcal ...
- 2,174
8
votes
Accepted
Can momentum never be zero in quantum mechanics?
You are not wrong, but it is worth noting that the same thing is true of any momentum eigenstate (or closely related unbound eigenstate of a Hamiltonian with a potential well in it). Explicitly
$$
-i\...
- 9,261
7
votes
Accepted
Why is $\langle x| x' \rangle=\delta(x-x')$?
Isn't it just from the sifting property?
$$f(x) = \int\mathrm{d}x'\;f(x')\,\delta(x - x')$$
That is, if you accept the above and if you accept that
$$|\psi\rangle = \int \mathrm{d}x'\,\psi(x')\,|x'\...
- 59.9k
6
votes
Problem with physical application of Dirac Delta
The ambiguity is resolved once you think about the dimension of the $\delta$-function. It's actually your method of non-dimensionalization that led you astray here. Given $\delta(f(x))$, the $\delta$-...
- 127k
6
votes
Accepted
Meaning of localization of wave function?
"Wave packets are formed by localisation" is a weird statement. Let me try to make sense of it.
If you have a system (such as a particle in non-relativistic quantum mechanics), quantum mechanics ...
- 15.7k
6
votes
Accepted
Normalization of eigenfunction to Dirac-delta function
For a function $f(x)$ the Fourier transform is defined as:
\begin{equation}
\overset{\boldsymbol{\sim}}{f}\left(k\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\...
- 16.2k
6
votes
Vector space in quantum mechanics
The set ${\cal L}^2(\mathbb{R}^3)$ of square integrable functions $\psi:\mathbb{R}^3\to \mathbb{C}$ is a $\mathbb{C}$-vector space, and hence includes $0$.
The set $\{\psi\in{\cal L}^2(\mathbb{R}^3)\...
- 208k
6
votes
Accepted
Dirac Delta Function and Position
This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis.
For an operator with discrete eigenvalues $n$ with eigenvectors $|n\rangle$,
...
- 58.1k
6
votes
Why do we refer the cross section ratios to muons?
Muons are easy to measure. They're extremely long-lived relative to most other particles, and they're pretty massive compared to electrons, which means they have substantial penetrating power that is ...
- 35.7k
6
votes
Accepted
Coefficients of the wave function - a free particle in a box
The normalisation constant $A$ could depend on $n$ (and indeed in most problems it does, for example, the Harmonic Oscillator and the Hydrogen atom, to name two famous ones). It just turns out that it ...
- 11.4k
6
votes
Normalization of momentum eigenstates in QFT
If you're prepared to keep track of extra factors in the LSZ formula and such, it seems to me that you can change the normalization however you want. But the reason why the zeroth component appears in ...
- 8,371
5
votes
Accepted
Linearly dependent quantum state vectors
In Quantum Mechanics, a physical state is always normalized by definition. So, in fact, you have to normalize your linearly dependent vectors, obtaining the same normalized vector, and, in fact, both ...
- 1,198
5
votes
On Griffith Quantum example 2.1: normalization of wave function in time.
In the cross terms that are time-dependent, the integral is $0$ due to the orthogonality of the wave functions (when integrated over $x$).
- 46.8k
5
votes
Accepted
Clashing definition of rays in Weinberg and Sakurai and Born interpretation without normalizability
The space of states of a quantum system is the space of rays of some Hilbert space $\mathcal{H}$. That is, given the equivalence relation $\psi\sim z\psi$ for any $z\in\mathbb{C}-\{0\}$ and $\psi\in\...
- 4,703
5
votes
Accepted
Normalize wave function with respect to time instead of space
Time isn't a quantum-mechanical observable; it's a label. To understand the difference we must consider classical mechanics, in which canonical coordinates are functions of a time label. In particular,...
- 25k
5
votes
Accepted
Expectation value of position of eigenstate of position
The formula for the expectation value $\langle A\rangle=\langle\psi|\hat{A}|\psi\rangle$ is given for the normalized states $\langle\psi|\psi\rangle=1$. You can generalize it as
\begin{equation}
\...
- 8,629
5
votes
Accepted
Normalization of states and bracket notation
Try commuting $a_p$ and $a_q^\dagger$ and keep in mind that $a_n|0\rangle = 0$.
- 641
5
votes
Transition from an initial/final position state to the ground state in the path integral
Let us first of all mention that Ref.1 is considering QM as opposed to QFT, i.e. no particle creation & annihilation are allowed. Eq. (6.19) can alternatively be written as
$$\begin{align} \langle ...
- 208k
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