New answers tagged operators
2
votes
Correlation functions zero for repeated creation or annihilation operators
Okey, the zero has to come from computing the trace
$$
{\rm Tr}(R_0 b_j^\dagger b_k^\dagger)=0,
$$
but I don't see easily why this is the case.
Here $R_0$ is the density matrix of the bath composed ...
- 20k
1
vote
Translation operators and positive-semidefinite condition
Your working looks correct, but there is a much simpler approach. It is easy to see that your operator is unitary. This implies that it has a complete set of eigenstates. Furthermore all its ...
- 9,121
1
vote
Coordinate-free calculations in bra-ket notation
I'm interested if bra-ket notation can be used without having to introduce a basis or coordinates in calculations and proofs.
If you are asking if it can be used without a basis, yes, it can.
There ...
- 20k
4
votes
Question about the identity operator and the bosonic ladder operators
The answer is no. Below I give a proof for the $n=1$ case that the only operators obeying these constraints are of the form
$$B=b P_1+ b^{-1} P_2 \quad , \tag 1$$
where $P_1=\sum\limits_{n=0}^\infty |...
- 7,931
3
votes
Question about the identity operator and the bosonic ladder operators
I am a bit confused by your condition that $B$ should be self-adjoint, which would mean $B^{\dagger}=B$. I suspect you meant that $B$ was to be unitary, $B^{\dagger}=B^{-1}$. However, it seems to ...
Buzz♦
- 16.1k
1
vote
Where does the complex conjugate term generally come from in a Hamiltonian?
This type of Hamiltonian is a second quantisation one. The most important part is defining your operators $a$ and $a^\dagger$. For example, if you have two interacting dipoles the interaction energy ...
4
votes
Accepted
Where does the complex conjugate term generally come from in a Hamiltonian?
The simplest justification is probably that the Hamiltonian has to be Hermitian. After all, its eigenvalues are interpreted as the possible energies of the system, and hence they need to be real. This ...
- 17.7k
9
votes
Accepted
Determining Bound States from Møller Operator
If $H=H_0+V$, where $V$ is a so-called Kato potential, then the point spectrum of $H$ corresponds to the bound states, while the scattering states correspond to the continuous spectrum (Ruelle's ...
- 7,931
1
vote
Why does the Pauli objection not disqualify the existence of the position operator?
According to the Pauli objection (see for example here or the answer to this question) there can be no time operator $\hat{T}$ canonically conjugate to the Hamiltonian $\hat{H}$ of a physical system ...
- 20k
2
votes
Accepted
Equivalent definitions of Wick ordering
For what it's worth, given a family $(\hat{A}_i)_{i\in I}$ of operators $\hat{A}_i\in{\cal A}$ in a (super) operator algebra ${\cal A}$ there seems to be an implicit/tacit assumption that the $n$th ...
- 202k
3
votes
What are single-, double- and multi-trace operators in AdS/CFT?
Briefly speaking, on the CFT boundary side of the holographic principle/AdS-CFT correspondence, there is a (super) YM theory with an $SU(N)$ color group. A boundary observable ${\cal O}(x)$ [build ...
- 202k
2
votes
Pauli matrix exponentials
am I correct that it follows that $e^{-i\frac{\theta}{2}\hat{X}}|+\rangle = e^{-i\frac{\theta}{2}}|+\rangle$ and $e^{-i\frac{\theta}{2}\hat{X}}|-\rangle = e^{i\frac{\theta}{2}}|-\rangle$ since $\hat{X}...
- 20k
0
votes
Choice of spacetime foliation while quantising a conformal field theory
A Hilbert space is assigned to a foliation of the manifold. More precisely, we assign a Hilbert space to every slice in the foliation. The Hilbert spaces on two different slices in the same foliation ...
- 26k
2
votes
Exercise on self-adjointness of Hamiltonian
Thanks to @ZeroTheHero who brought me onto the right track, I was able to find the solution to the problem myself.
The Hamiltonian with respect to the basis $\{ | \psi \rangle, |\phi \rangle, |\Gamma\...
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2
votes
How do we know at the operator-level that the tadpole $\langle\Omega|\phi(x)|\Omega\rangle=0$ vanishes in scalar $\phi^4$ theory?
I think you are basically there:
Let $\hat{U}$ be the unitary operation you defined, $\hat{U} \hat{\phi}(x) \hat{U}^\dagger \equiv -\hat{\phi}(x)$. This is a definition. We then assume (or check) that ...
- 1,568
-1
votes
Extracting the dimension of an operator from algebra
Can we extract the dimension of $P$
and $K$
from the commutation relations?
The answer is no as, dimensionally, $P$ and $K$ cancel out of those equations.
- 1,965
2
votes
Extracting the dimension of an operator from algebra
Yes, the dila(ta)tion operator $D/i=x^{\mu} \partial_{\mu}$ is a Euler vector field that literally counts the length dimension for operators made from only $x^{\nu}$ and $\partial_{\lambda}$.
- 202k
0
votes
Generator of two-qubit quantum gate
I'm not sure I fully understood your question, but the thing you exponentiate to get a controlled $Y$ gate,
$$ \mathsf{CY} = \frac{1}{2} \left( \mathbb{1} \otimes \mathbb{1} + Z \otimes \mathbb{1} + \...
- 874
0
votes
Dictionary between interpretations of field operators
Perhaps there is a simple confusion. As you may notice yourself your points 1. and 2. are not compatible with each other. What you can say is that the quantization of a real scalar field is given by ...
- 1,965
2
votes
Accepted
Dictionary between interpretations of field operators
Assuming that you are interested in the physical interpretation of the hermitean field operator of a free relativistic scalar theory in the Heisenberg picture, its Fourier decomposition (omitting the ...
- 6,073
0
votes
Total momentum operator of the Klein-Gordon field (before limit to the continuum)
When you related the quadrature operator $q$ to the ladder operators $a$'s in the "solution," there is an implied commutation relation for the ladder operators that you can derive from the ...
- 14.6k
3
votes
Accepted
Ladder operators and creation & annihilation operators - different between $a$, $b$ and $c$
When we talk about the quantum harmonic oscillator, $a$ and $a^{\dagger}$ are usually used.($a$ may stands for "annihilation".) Note that we have not discuss second quantization here, so $a$ ...
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2
votes
Ladder operators and creation & annihilation operators - different between $a$, $b$ and $c$
Someone can tell me the different between the notation
$a$ or $b$ or $c$?
The name of these operators doesn't matter.
The important thing are the commutation relations
between these two operators (...
- 39.5k
3
votes
Where does the "arbitrary constant" in the $L_{0}$ Virasoro operator come from?
Starting from the classical Poisson algebra of $L_m$ quantities (which form a Witt algebra), we would like to quantize it, i.e. define operators $\hat{L}_m$. Due to the operator ordering ambiguity, ...
- 202k
1
vote
How to find the expectation value of momentum operator?
I am going to rewrite your formulas because the placement of some of your quantities is personally confusing.
$$\left \langle \hat{p} \right \rangle = \int_{-\infty}^{\infty}dx \ \psi^{*}(x)\left(-i\...
- 2,114
1
vote
Why do QM books point out that $S^2$ commutes with $S_x$, $S_y$, and $S_z$?
$\newcommand\ket[1]{|{#1}\rangle}\newcommand\up{\uparrow}\newcommand\down{\downarrow}$To expound on one of the comments: $S^2$ is not always so simple, as a matrix. For a system with a fixed spin, yes,...
- 4,113
4
votes
What can we say about the eigendecomposition of quantum channels?
As you observe correctly, $\mathbb N$ is a linear map. Thus, the same holds as for any eigendecomposition of linear maps.
In particular, there need not be a complete basis of eigenvectors (there can ...
- 20.4k
2
votes
Why do QM books point out that $S^2$ commutes with $S_x$, $S_y$, and $S_z$?
The 2 x 2 spin matrices are for the (fundamental) spinor representation: two component spinors.
All two component spinors have a spin quantum number $s=\frac 1 2$, and a fixed angular momentum:
$$ S = ...
- 33.7k
2
votes
Why do QM books point out that $S^2$ commutes with $S_x$, $S_y$, and $S_z$?
Your QM book stresses these facts because it means that this operator is something called the "Casimir invariant", which must be proportional to the identity operator and commute with all ...
- 874
4
votes
Accepted
What can we say about the eigendecomposition of quantum channels?
Generally speaking, no. Here is a quantum channel which provides a counter example. Let $|i\rangle$, $i = 1, \dots, n$ be an orthonormal basis on an $n$ dimensional Hilbert space, and define
$$A_i = |...
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