New answers tagged operators
1
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Understanding the parabolic state of a quantum particle in the infinite square well
As explained in previous answers, your wave function belongs to the full Hilbert space $L^2$, and the domain of $H$, but not to the domain of $H^2$. In general, unbounded operator (like $H$ or $H^2$) ...
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10
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Understanding the parabolic state of a quantum particle in the infinite square well
Before we can answer the question, we need a little bit more mathematical clarity here. However, we won't be fully rigorous and leave out details which are not of interest for the question.$^\ddagger$
...
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-1
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Dirac-delta-functions as eigenbasis of the position operator - pure nonsense? Or can more be said?
I begin with a definition of the $f$ representation and $q$ representation of a wave function [1]. Next, in Proposition 1, I use the definition to show that the use of Dirac-delta distributions as ...
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5
votes
Understanding the parabolic state of a quantum particle in the infinite square well
Barring the (now corrected) mistake of the omission of part of eigenvectors,
the point is quite trivial: here one is referring to two different notions of energy observable but using the same name!
...
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4
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Understanding the parabolic state of a quantum particle in the infinite square well
Let's be clearer with the essence of your problem, now that the other answerer brought to your attention that you were missing the cosines.
Your state is $\left<x|\psi\right>=\sqrt{\frac{15}{16\...
- 9,907
7
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Understanding the parabolic state of a quantum particle in the infinite square well
There's a simple error in your $\psi_n$ which is screwing you up. In general, the form of the eigenstates $\psi_n(x;x_o,L)$ for a well of length $L$ starting at $x_o$ is
$$
\psi_n(x;x_o,L) \propto \...
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4
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Accepted
When a function of the position operator is self-adjoint?
Necessary and sufficient conditions on $f: \mathbb{R} \to \mathbb{C}$ to make $f(A)$ selfadjoint if $A$ is selfadjoint (generally unbounded and defined in a dense domain) are that both
(a) $f$ is real ...
- 75.6k
3
votes
Accepted
Existence of Primary operators in CFT
The correct statement is that the scaling dimension of all local operators should be bounded below, and the issue is that $K_\mu {\cal O}(0)$ is not a local operator. You are correct that $K_\mu$ can ...
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2
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Why does the Schriefer-Wolff transformation works for phonons?
Schrieffer-Wolff is an approximate way of diagonalizing electron-photon Hamiltonian. On the one hand it is inspired by by an exact polaron transformation (see, e.g., this thread), while on the other ...
- 61.8k
0
votes
Decomposing operator exponentials
Equation $$e^Ae^B=e^{A+B+\frac{1}{2}[A,B]}$$ is valid for the case when $[A,B]$ is a number, i.e., commutes with both $A$ and $B$. So reversing the equation is straightforward:
$$e^{A+B}=e^Ae^Be^{-\...
- 61.8k
0
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How can you prove that the squares of the expected values of the three components of spin sum to 1?
The chapter initially states:
There is an important theorem that you can try to prove.
So what follows next is generally a consequence of that (unproven) theorem.
When the author states:
On the ...
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4
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Explicit Expression of $S_2(\zeta)$ on a general Fock State?
The operators $\hat a\hat b\sim K_-$ and $\hat a^\dagger \hat b^\dagger\sim K_+$ are ladder operators of the $su(1,1)$ algebra so one possibility is to use the $SU(1,1)$ disentanglement formula
$$
D(\...
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3
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Explicit Expression of $S_2(\zeta)$ on a general Fock State?
I think that in case of general $p$, $q$ it may be impossible to obtain simple expressions and relations for $T(m,n,p,q)$. The operator $S_2(\zeta)$ is the exponential of the quadratic boson operator, ...
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Why we have $|P\rangle =\int dp \space |p\rangle p \langle p| =-i\hbar \int dx |x\rangle \frac{d}{dx} \langle x|$?
The other answers and comments have made it clear that you can define momentum operator to satisfy $$\tag1\left<x|\hat p|\psi\right>=-i\hslash\frac{\partial\ }{\partial x}\left<x|\psi\right&...
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3
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Accepted
Why we have $|P\rangle =\int dp \space |p\rangle p \langle p| =-i\hbar \int dx |x\rangle \frac{d}{dx} \langle x|$?
The momentum operator is usually defined by its matrix elemenets, $\left\langle x\right|P\left|\psi\right\rangle :=-i\hbar\frac{d\psi\left(x\right)}{dx}$ where $\left|x\right\rangle$ is a position ...
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Proving quantum operator relationship used in derivation of Radial Schrodinger Equation
I have managed to find a solution but I would appreciate some verification if the logic of this solution is correct ...
So I can make it work if I should interpret any vector operator "square&...
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Proving quantum operator relationship used in derivation of Radial Schrodinger Equation
Your mistake happens in the last line. When you use the commutator. You get $r^2p^2=rppr+i\hbar rp$.
But you still have it on the right side of your equation. Push it to the left so you get the minus ...
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1
vote
Accepted
Occupation number doubt in QFT
Every position in a wave function/state vector corresponds to a mode - that is to a specific value of index $k$ in $a_k, a_k^\dagger$. In this sense, writing $|n_1,n_2,...\rangle$ is confusing, since ...
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2
votes
The negative value of the hopping integral in the tight-binding model Hamiltonian
Hopping integral is essentially a non-diagonal matrix element in a Hamiltonian. The principal constraint is that Hamiltonian should be Hermitian, but nothing prevents the non-diagonal elements from ...
- 61.8k
3
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What' the intuition behind Shankar's postulate II?
I don't know if it's the intuition, but here's one way in which one could arrive at the postulate.
When you are first introduced to quantum mechanics, you usually don't work in an abstract Hilbert ...
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4
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Accepted
What' the intuition behind Shankar's postulate II?
In classical physics the evolution of the position and momentum of a particle are described by functions $x(t),p(t)$ whose values are the value you would get if you measured the position and momentum ...
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1
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What' the intuition behind Shankar's postulate II?
You don't rationalize postulates. Postulates are the "starting point", and the rest of the theory builds from them. Then we test the validity of the set of postulates and the theory by ...
- 58.1k
0
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Ambiguity in path integral approach to 2D Liouville gravity, versus operator formalism
Lets review some basic material on Vertex opertors in the path integral setting.
For a Euclidean Bosonic scalar field in the whole of ${\mathbb R}^2$
we have the Gaussian functional integral ...
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0
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Ambiguity in path integral approach to 2D Liouville gravity, versus operator formalism
It seems I have found an answer to my question in the articles https://arxiv.org/abs/1712.00829 , and https://arxiv.org/abs/2404.02001 . The answer is that the construction I provided is not rigorous ...
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Diagonalize the Swap operator $\mathbb{S}_\text{AB} = \sum_{i,j=1}^{d} |i_\text{A}j_\text{B}\rangle \langle j_\text{A}i_\text{B}| $
For two qubits, the swap opreator acts on basis vectors as:
\begin{align}
\mathbb{S} &|0 0 \rangle = |0 0 \rangle, \\
\mathbb{S} &|0 1 \rangle = |1 0 \rangle, \\
\mathbb{S} &|1 0 \rangle = ...
0
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Ambiguity in path integral approach to 2D Liouville gravity, versus operator formalism
Are you aware that the fields $\varphi(z_i)$ are inserted at different points $z_i\neq z_j$ on the sphere. So you cannot simply sum the exponents together. The primary fields of quantum Liouville ...
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-1
votes
What is the mathematically precise definition of raising and lowering operators?
There is a simple way in which I understand this concept of raising and lowering operators. I'll try to explain it in two stages.
First, given the assumed relation
$$ [\hat{a},\hat{a}^{\dagger}] = \...
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1
vote
What is the mathematically precise definition of raising and lowering operators?
There are two main different ways in which one can generalize the notion of ladder operators, but neither is particularly useful in practice:
Naive ladder operators. The nice thing about linear ...
- 127k
1
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What is the mathematically precise definition of raising and lowering operators?
In general, a ladder operator $O_{\pm}$ is an operator defined for some states $|n\rangle$ as
$$O_{\pm}|n\rangle = C_{\pm}(n)|n\pm1\rangle \ \ \mathrm{and} \ \ O_+^\dagger
=O_- .$$
The term $C_\pm$ is ...
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1
vote
Accepted
A question about operators in QM and notation
The notation here is a bit loose. We have that
$$
\hat{a}|n\rangle = \sqrt{n}|n-1\rangle
$$
and hence
$$
(\hat{a}|n\rangle)^{\dagger} = \langle n| \hat{a}^{\dagger} = \sqrt{n}\langle n-1|.
$$
However, ...
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2
votes
Accepted
Expectation value of angular momentum operator
Total angular momentum is a vector,
$$
\begin{align}
\mathbf J &= \begin{bmatrix}
J_x \\
J_y \\
J_z
\end{bmatrix}
\end{align}$$,
so an expectation ...
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0
votes
Expectation value of angular momentum operator
It most likely refers to the magnitude of the total angular momentum, $J \equiv \sqrt{<\vec{J}^2>} = \sqrt{<J_x^2 + J_y^2 + J_z^2>}$.
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How are rotations invariant by time-reversal?
The sign of the magnitude is not preserved in reversed-time; only the magnitude itself is unchanged in 4D during acceleration.
Flipping the sign of the magnitude is what makes it reverse time.
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1
vote
Multiplication of operators defined by commutation relations
Algebraic structures can be defined somewhat abstractly. The Lie bracket is not necessarily represented as $[A,B] = AB-BA$, though this is the standard choice for operators as it satisfies the various ...
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6
votes
Multiplication of operators defined by commutation relations
Multiplication of generators is a feature of the representation, not the Lie algebra.
Indeed, you are asking about the unital associative universal enveloping algebra of generators which defines the ...
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6
votes
What is the commutator of the lowering operator $J_-$ and the exponential of $J_z$, arranged so that the lowering operator is always to the right?
The other answer is perfect, but a less direct simpler method is application of Hadamard's lemma on the adjoint action, a trick you'll may well be using once a month,
$$
e^{iaJ_z}J_- e^{-iaJ_z}= J_-+...
- 65.1k
5
votes
Accepted
What is the commutator of the lowering operator $J_-$ and the exponential of $J_z$, arranged so that the lowering operator is always to the right?
It is convenient to consider this problem in the basis of eigenvectors of the operator $J_z$. For any vector $|m\rangle$ such that
$$
J_z|m\rangle = m|m\rangle,
$$
the following equalities are true
$$
...
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1
vote
How are Schwinger functions defined as moments if they are actually operators?
The correspondence between the operator/Hilbert space langauge and the measure theoretic language is the essence of the Feynman-Kac isomorphism between Euclidean quantum field theory (the analytic ...
- 55.2k
3
votes
Accepted
How are Schwinger functions defined as moments if they are actually operators?
Perhaps this is not what you were asking, but just in case: The measure-theoretic formalism that you are describing is that of Euclidean field theory. It is not quantum mechanics, and the field is not ...
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0
votes
Derivation of momentum operator from kinetic energy operator
Your question really is: how can I define the square root of an operator? The answer is diagonalise, replace the eigenvalues by their roots, transform back.
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1
vote
Tensor Operators and Spherical Harmonics
So the last question:
"then how is it true that in quantum mechanics, they can be both an eigen basis for wave functions and also an operator?"
so look at ordinary cartesian coordinates $(x, ...
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1
vote
Tensor Operators and Spherical Harmonics
The spherical harmonics can be considered as tensor operators acting by multiplication on other spherical harmonics because
$$
Y_{\ell_2}^{m_2}(\theta,\phi)Y_{\ell_1}^{m_1}(\theta,\phi)=
\sum_{\ell} C^...
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