5 votes
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Outer product as an operator in an infinite dimensional Hilbert space

For any $\psi\in H$, you can define an operator $P_\psi$ by $P_\psi \phi:=\langle \psi, \phi\rangle_H \psi$ for all $\phi\in H$. It is easily verified that $P_\psi$ is a linear bounded operator, which ...
Tobias Fünke's user avatar
5 votes
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Hermiticity of a projection operator

Indeed, you missed that $\sigma_{00}^0=I$. \begin{equation} \hat{U}=I+\sigma_{00}(e^{i\omega}-1), \hspace{6 mm} \hat{U}^{\dagger}=I+\sigma_{00}(e^{-i\omega}-1), \end{equation} and it is now easy to ...
Ruben Campos Delgado's user avatar
4 votes

Deduction of Kinetic energy operator in quantum mechanics

Once a differential operator represents the momentum $p$, $p^2 f$ should be intended as the operator $p$ acting on $pf$. Therefore, the second expression is the correct one. Edit after some exchange ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
4 votes

What does the state $a_k a_l^\dagger|0\rangle$ represent?

As $a_k a_\ell^\dagger |\text{vac}\rangle= [a_k, a_\ell^\dagger] |\text{vac}\rangle=\delta_{k\ell} |\text{vac}\rangle$, the result is $0$ for $k\ne \ell$ and $|\text{vac}\rangle$ for $k=\ell$.
Hyperon's user avatar
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4 votes
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Are Hermitian operators Hermitian in any basis?

A linear operator on a Hilbert space $A:\mathcal H\to\mathcal H$ is Hermitian if for all $v,w\in\mathcal H$, $$\langle Av, w\rangle = \langle v,Aw\rangle$$ This definition makes no reference to any ...
Er Jio's user avatar
  • 1,191
4 votes

Difference between real operators and Hermitian operators in quantum mechanics

In order to avoid unnecessary mathematical complications, let us consider a finite-dimensional complex Hilbert space $\mathcal H$, i.e. a complex vector space with dimension $n\lt \infty$ and a (...
Hyperon's user avatar
  • 4,751
4 votes
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Energy levels of a translating quantum harmonic oscillator

OP seemingly wants to understand whether or not there is a contribution to the energy from the overall translation of the system with velocity $v$. Of course there should be, and we should expect ...
hft's user avatar
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3 votes

Energy levels of a translating quantum harmonic oscillator

$\hat H_𝑣=\exp(−𝑖𝑣𝑡\hat p/ℏ)~\hat H\exp(𝑖𝑣𝑡\hat p/ℏ)$, so the two hamiltonians are canonically equivalent and have the same spectrum. You may also see this from $[\hat x -vt,\hat p]=i\hbar$, so ...
Cosmas Zachos's user avatar
3 votes

Are Hermitian operators Hermitian in any basis?

An operator is not Hermitian "in a basis". An operator $A$ on an inner product space is defined to be Hermitian iff it equals its own adjoint, $A = A^\dagger$. This condition has nothing to ...
tparker's user avatar
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3 votes
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Questions regarding measurement of a qubit

When you measure an observable $A$ in a state $|\psi\rangle$ the expectation value of the observable is $\langle\psi|A|\psi\rangle$. The possible measurement outcomes are the eigenvalues of the ...
alanf's user avatar
  • 7,131
3 votes

What does the state $a_k a_l^\dagger|0\rangle$ represent?

While the other answer is 100% mathematically and physically correct, I don’t think it satisfies the spirit of the question so I will actually answer a more “real” question: what happens if you have a ...
JohnA.'s user avatar
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3 votes

How to actually implement a global phase gate?

Quantum states are only defined up to a global phase. Thus, it is not meaningful to implement a "global phase gate", in fact, quantum gates are, for the same reason, also only defined up to ...
Norbert Schuch's user avatar
2 votes

Hermiticity of a radial momentum operator $\hat{p}_r$ and the spectral theorem

The operator $$\hat p_r \ (...) = -i \hbar \frac{1}{r}\partial_r \left[\ r \ (...)\right] $$ is obvoiusly hermitean/symmetric on $\mathit L^2(\mathbb R_+ , r^2 dr)$: $$\int_0^\infty \ r^2 \ dr \ \psi(...
Roland F's user avatar
2 votes

What does the state $a_k a_l^\dagger|0\rangle$ represent?

The state $\hat{a}_k\hat{a}_l^{\dagger}|0\rangle$ doesn't represent anything. In fact, since the result of this operation is the zero vector, the result isn't even a quantum state (since quantum ...
march's user avatar
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2 votes
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Deduction of Kinetic energy operator in quantum mechanics

In operatorial language for any operator $A$, $A^2$ means $A \circ A$, that is $A$ composed with itself. In the particular case of $A = \hat{p} = -i\hbar \frac{d}{dx}$, indeed $$\hat{p}^2 = -\hbar^2 \...
lucabtz's user avatar
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2 votes
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Difference between real operators and Hermitian operators in quantum mechanics

A real linear operator is one whose matrix elements are real. For example, given an orthonormal basis $\{|\psi_k\rangle\}$,and an operator $\hat O$, suitably defined on a Hilbert space $\mathcal H$; ...
Albertus Magnus's user avatar
1 vote

Parity operator action on quantized Dirac field

I don't have my copy of the book to hand, but I assume that $P$ is there a linear operator on the Hilbert space. It does not commute with the creation and annihilation operators: $$ P\hat a_{\bf k} ...
mike stone's user avatar
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1 vote

What will be wave function after application of operator?

No, the state vector after a position measurement is not given by $X |\psi\rangle$. Suppose your detector can decide whether the position of the particle is inside oder outside a certain (space) ...
Hyperon's user avatar
  • 4,751
1 vote

Deduction of Kinetic energy operator in quantum mechanics

Actually mathematical part of kinetic energy derivation is Ok : $$ \begin{align} \hat T &= \frac {\hat p^2}{2m}\\ &=\frac {1}{2m}\left(\hat p \cdot \hat p\right)\\ &=\frac {1}{2m}\left(-i\...
Agnius Vasiliauskas's user avatar
1 vote
Accepted

The time-derivative of the Hamiltonian for a 1D harmonic potential

In the Schrödinger picture (which seems like what you're working with), $\hat{x}$ and $\hat{p}$ don't have any time dependence. This means that the only time dependence of $\hat{H}$ comes from the ...
Rokas Veitas's user avatar
1 vote
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Non-perturbative matrix element calculation

First, some remarks : Wick's theorem only applies to free (and interaction picture) fields, hence it is fundamentally perturbative. the normal ordered product $:\phi(x)\phi(y):$ contains a term with ...
SolubleFish's user avatar
  • 5,508
1 vote
Accepted

Radial ordering in CFT

The main point is that the sum in OP's eq. (1) is only (absolutely) convergent if $|y|>|z|$, i.e. if OP's LHS $$\hat{T}(y)\hat{T}(z)~=~{\cal R}[\hat{T}(y)\hat{T}(z)]$$ is radially ordered. (...
Qmechanic's user avatar
  • 200k
1 vote

Unitary Operators

If $A$ is a unitary operator we have that: $$AA^{\dagger}=A^{\dagger}A=I,$$ where $I$ is the identity operator. So given that $B=A^{\dagger}aA$, we can apply $A$ to the left: $$AB=AA^{\dagger}aA=IaA=...
Albertus Magnus's user avatar
1 vote
Accepted

How does an operator in the denominator act on a state?

The comment by Filippo tells you everything you need to know, but I will explain a little. When we write a quantity such as $\exp(i \hat{\phi})$ we are giving ourselves the 'right' to evaluate a ...
Andrew Steane's user avatar
1 vote

Does every field correspond to a particle?

Particles in QFT are an approximate description of a field in regimes where the interaction terms between the particles are weak enough: http://philsci-archive.pitt.edu/15296/1/qm-continuum%20revised....
alanf's user avatar
  • 7,131
1 vote
Accepted

Schmidt decomposition of density operators

The density matrix can be written as $$ \rho = \sum_{ijkl} \rho_{ijkl} |i\rangle_A|j\rangle_B \langle k|_A\langle l |_B\;, $$ where $|i\rangle_A,|j\rangle_B, \langle k|_A$ and $\langle l |_B$ are ...
By Symmetry's user avatar
  • 9,091
1 vote

Hermiticity of a radial momentum operator $\hat{p}_r$ and the spectral theorem

Addressing a point which is not covered in the answer by @Roland F and the two pertinent comments to it: The OP is conflating spectral theory and the general theory of operators in Quantum Mechanics ...
DanielC's user avatar
  • 4,293
1 vote

What kind of physical process would correspond to an operator that doesn’t result in an eigenvalue equation: $ \hat{A}ψ=a ψ$?

The wavefunction, $\psi(x)$, is the position representation of the more abstract vector (or "ket"), $| \psi \rangle$, which is an element of the Hilbert space (a complex vector space with ...
Ben H's user avatar
  • 958
1 vote

How the supercharge operators act on superfields in quantum mechanics, and the adjoints of supercharges?

After applying an infinitesimal unbarred SUSY transformation generated by $\hat Q$, $$\delta_\epsilon X=[\epsilon\hat{Q},X]=\epsilon(\partial_\theta+i\bar\theta\partial_t)X(t,\theta,\bar\theta)$$ ...
Nihar Karve's user avatar
  • 8,415
1 vote
Accepted

On the Wigner symmetry representation theorem

I suppose that by the brackets you mean the equivalence class of the state, that is, $[\psi]\in\mathbb{PH}$ is the ray to which $\psi\in\mathbb{H}$ belongs. Now, if your question is whether every ...
Albert's user avatar
  • 307

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