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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
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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
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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. Operators like the momentum operator are real, for example, when we calculate its expectation value $\langle \psi|\hat p|\psi\rangle$ we ...
Albertus Magnus's user avatar
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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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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
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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
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Proof of boost generator $K_0$ commute with $S$-matix in Weinberg QFT 1

From equation (3.3.21), we see that there would be singularities in the matrix elements of $\mathbf{W}$ $$[\mathbf{K}_0, V] = - [\mathbf{W}, H]\tag{3.3.21}$$ The elements of $\mathbf{W}$, $$\left(\...
Ting-Kai Hsu's user avatar
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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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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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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
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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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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
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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
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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
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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
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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
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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
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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
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
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
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
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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
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How do we know the creation and annihilation operators for angular momentum give rise to a complete basis?

The irreducible representations of the Lie algebra of $\rm SU(2)$ (defined by $[T_k,T_\ell]= i \varepsilon_{k \ell m}T_m$) can be classified by the "weights" $j=0,\, 1/2,\, 1,\, 3/2, \,2, \...
Hyperon's user avatar
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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
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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
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How the $-2t\cos(k)$ term appears in the dispersion of the $1D$ tight binding model?

I guess it won't be important anymore but I still would like to clarify: I guess you are refering to the beginning of page 6. The step you are confused of is not using the commutator relations of ...
ChrisG's user avatar
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How to practically check if a Hermitian matrix is an observable?

What you've got from ChatGPT is amazingly wrong, operator representing an observable in quantum theory need not obey the equation $$ \langle \psi | A^\dagger A |\psi\rangle = 1 $$ and even writing it ...
Ján Lalinský's user avatar
8 votes

Why does $e^{-H}\partial_j e^{H} = \partial_j + \partial_jH$?

OP's identity$^1$ $$\begin{align} e^{-H}\partial e^H ~\equiv~&e^{-[H,\cdot]}\partial\cr ~\equiv~& \partial+[\partial, H] +\frac{1}{2}[[\partial, H],H] +\frac{1}{6}[[[\partial, H],H],H] +\ldots\...
Qmechanic's user avatar
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Why does $e^{-H}\partial_j e^{H} = \partial_j + \partial_jH$?

It's just Leibnitz rule. For any $\psi(x)$ we have $$ e^{-H(x)}\partial_x e^{H(x)} \psi(x) = e^{-H(x)} (\partial_x e^{H(x)})\psi(x) + e^{-H(x)} e^{H(x)}\partial_x \psi\\ = e^{-H(x)}e^{H(x)} \...
mike stone's user avatar
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Connection of a concrete Hamiltonian to the generator of time-translations

In principle, mathematically, yes. If you know $\frac{d}{dt}|\psi\rangle$ for every state $|\psi\rangle$ then you can calculate $\langle \phi|\frac{d}{dt}|\psi\rangle$ for every pair of states $\phi$ ...
Luke Pritchett's user avatar
2 votes

Connection of a concrete Hamiltonian to the generator of time-translations

Of course not - it is one property of the Hamiltonian, but it does not determine it. It leaves many options possible, like Hamiltonian of free particle, Hamiltonian of harmonic oscillator, Hamiltonian ...
Ján Lalinský's user avatar

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