In physics, an operator is almost always either a square matrix or a linear mapping from one space of functions (often on $\mathbb{R}^N$ or $\mathbb{C}^N$) to the same or other like space of functions. Operators serve as *observables* and as *time evolution operators* in Quantum Mechanics. This tag ...

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Really how can an observable quantity be equal to an operator?

A wave-function can be written as $$\Psi = Ae^{-i(Et - px)/\hbar}$$ where $E$ & $p$ are the energy & momentum of the particle. Now, differentiating $\Psi$ w.r.t. $x$ and $t$ respectively, ...
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How are anyons possible?

If $|ψ\rangle$ is the state of a system of two indistinguishable particles, then we have an exchange operator $P$ which switches the states of the two particles. Since the two particles are ...
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Eigenvalues of a quantum field?

Fields in classical mechanics are observables. For example, I can measure the value of the electric field at some (x,t). In quantum field theory, the classical field is promoted to an operator-valued ...
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Evaluate $\langle \mathbf{p} | 1/\hat{r} | \mathbf{p}' \rangle$

In Sakurai's Problem 1.27 b), we use $\langle \mathbf{r} | \mathbf{p}\rangle = e^{i\mathbf{p}\cdot\mathbf{r}/\hbar}$ to show that $$ \langle \mathbf{p} | F(\hat{r}) | \mathbf{p}' \rangle = ...
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Why do we use operators in quantum mechanics?

In classical mechanics, physical quantities, such as, e.g. the coordinates of position, velocity, momentum, energy, etc, are real numbers, but in quantum mechanics they become operators. Why is this ...
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How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?

I know that $$\tag{1}\hat{p}~=~-i\hbar \frac{\partial}{\partial x}~.$$ How can I get $$\tag{2}\hat{x}~=~i\hbar \frac{\partial}{\partial p}~?$$ I think this simple and I'm just over thinking it, ...
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Is there a formalism for talking about diagonality/commutativity of operators with respect to an overcomplete basis?

Consider a density matrix of a free particle in non-relativistic quantum mechanics. Nice, quasi-classical particles will be well-approximated by a wavepacket or a mixture of wavepackets. The ...
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522 views

The cleverest way to calculate $\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]$ with $\left[\hat{a},\hat{a}^{\dagger}\right]=1$

Who can provide me some elegant solution for $$\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]\qquad\text{with} \qquad\left[\hat{a},\hat{a}^{\dagger}\right]~=~1$$ other than brute force calculation? ...
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Has anyone published the procedure to generalize ladder operators for any potential in Schrodinger's equation?

I know that the ladder operator for the quantum harmonic oscillator \begin{align} H\psi_m = \left(\dfrac{p^2}{2m}+\dfrac{1}{2}m\omega^2x^2\right)\psi_m=E_m\psi_m \end{align} is \begin{align} A = ...
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Is H=H* sloppy notation or really just incorrect, for Hermitian operators?

I saw it in this pdf, where they state that $P=P^\dagger$ and thus $P$ is hermitian. I find this notation confusing, because an operator A is Hermitian if $\langle \Psi | A \Psi \rangle=\langle A ...
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In what sense is the path integral an independent formulation of Quantum Mechanics/Field Theory?

We are all familiar with the version of Quantum Mechanics based on state space, operators, Schrodinger equation etc. This allows us to successfully compute relevant physical quantities such as ...
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Eigenstate of field operator in QFT

Why don't people discuss the eigenstate of the field operator? For example, the real scalar field the field operator is Hermitian, so its eigenstate is an observable quantity.
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Deriving the expectation of $[\hat X,\hat H]$

For a free particle of mass $m$, with Hamiltonian $$\hat{H} = \frac {\hat{P}^2} {2m},$$ where $$\hat{P} = -i \hbar \frac{\partial} {\partial x}.$$ The commutative relation is given by $$[\hat{X}, ...
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Action of Parity operator on Impulse representation

Is my derivation of the action of the parity operator $\mathbb{P}$ on the $|p\rangle$ representation correct? $$\left( \mathbb{P}\tilde\psi \right)(p)= - \tilde\psi (p).$$ Obtained from $$\left( ...
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Countable Matrix Representation

In my quantum mechanics class, my professor explained that the Hamiltonian along with position and momentum operators can be represented by matrices of countable dimension. This is especially usefull ...
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Why isn't the Heisenberg uncertainty principle stated in terms of spacetime?

As I understand it, there are two "versions" of the Heisenberg uncertainty principle: Position-Momentum uncertainty \begin{equation} \sigma_x \sigma_p \geq \frac{\hbar}{2} \end{equation} where ...
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682 views

What is the most general expression for the coordinate representation of momentum operator?

I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$. One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th ...
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Linearizing Quantum Operators

I was reading an article on harmonic generation and came across the following way of decomposing the photon field operator. $$ \hat{A}={\langle}\hat{A}{\rangle}I+ \Delta\hat{a}$$ The right hand side ...
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The Momentum Operator in QM

I've seen the 'derivation' as to why momentum is an operator, but I still don't buy it. Momentum has always been just a product $m{\bf v}$. Why should it now be an operator. Why can't we just multiply ...
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2answers
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How To Use Ladder Operators?

I'm studying for a test in quantum mechanics and I'm having a hard time understanding how to use ladder operators. There are no examples in my text book, only definitions that I can't understand how ...
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1answer
422 views

Wick Order and Radial Ordering in CFT

I am not so much familiar with the computations tools of conformal field theory, and I just run into an exercise asking to demonstrate the following formula (related to the bosonic field case): ...
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1answer
251 views

Help Simplifying a Commutator Equation

For the SHO, our teacher told us to scale $$p\rightarrow \sqrt{m\omega\hbar} ~p$$ $$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$ And then define the following $$K_1=\frac 14 (p^2-q^2)$$ $$K_2=\frac ...
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1answer
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State-operator map, and scalar fields

Up so far, i have been studied state-operator correspondence, $i.e$, i have been questioned State operator corrponding $i.e$ $S\times S$ to $R^2$ which was wrong question. By studing Ginsparg's ...
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Why is only one quantity of angular momentum i.e. $L_z$ quantized & not $L_x$ & $L_y$?

This is quoted from Arthur Beiser's Concepts of Modern Physics: Why is only one quantity of $\mathbf{L}$ quantized? The answer is related to the fact that $\mathbf{L}$ can never point in any ...
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Simple Quantum Mechanics Question about The Commutator of Translation Operators

Say there is $\hat{J} = \exp[-i \hat{p} l/ \hbar]$ and $\hat{U}= \exp[-i\hat{H}t/ \hbar]$, where $\hat{H}$ is time-independent. Can we say anything about $[\hat{J},\hat{U}]$? Is it zero? How do we ...
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1answer
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Rotation matrix is always a unitary operator

Can someone explain why the rotation matrix is a unitary, specifically orthogonal, operator?
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Expectation of momentum in the bound state

Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation ...
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1answer
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Non-Hermitian operator with real eigenvalues?

So we know that in Quantum Mechanics we require the operators to be Hermitian, so that their eigenvalues are real ($\in \mathbb{R}$) because they correspond to observables. What about a non-Hermitian ...
2
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1answer
209 views

Why are Hermitian operators linked to observables?

In Quantum Mechanics, why is it that a self-adjoint operator is linked to an observable? What makes it measureable? And why isn't a non-Hermitian operator linked to an observable? Also, what type of ...
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2answers
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Why is quantum mechancis is not content with symmetric operators, but wants self-adjoint operators?

A symmetric operator has only real eigenvalues and different eigenvectors corresponding to different eigenvalues are orthogonal. These are exactly what we want for a physical observable. I think ...
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1answer
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Eigenvalues of Angular Momentum in Quantum Mechanics

The eigenvalue equation of the $L^2$ operator is given by $$L^2f_l^m = \hbar ^2l(l+1)f_l^m$$ Side: So a determinate state for some observable $Q$ is a state where every measurement of $Q$ returns ...
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Density matrix formalism

The density matrix $\hat{\rho}$ is often introduced in textbooks as a mathematical convenience that allows us to describe quantum systems in which there is some level of missing information. ...
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Self-adjoint and unbounded operators in QM

An operator $A$ is said to be self-adjoint if $(\chi,A\psi)=(A\chi,\psi)$ for $\psi, \chi \in D_A$ and $D_A=D_{A^\dagger}$. But for the free particle momentum operator $\hat{p}$ these inner products ...
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Applying an operator to a function vs. a (ket) vector

I have a question regarding the effect of quantum mechanical operators. The definition that I'm familiar with says that an operator $A$ acts on a vector from a Hilbert space, $|\psi\rangle$, and the ...
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Differences between symmetric, Hermitian, self-adjoint, and essentially self-adjoint operators

I am a physicist. I always heard physicists used the terminology "symmetric", "Hermitian", "self-adjoint", and "essentially self-adjoint" operators interchangeably. Actually what is the difference ...
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Regularisation of infinite-dimensional determinants

Can a regularisation of the determinant be used to find the eigenvalues of the Hamiltonian in the normal infinite dimensional setting of QM? Edit: I failed to make myself clear. In finite ...
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Recovering QM from QFT

Reading through David Tong lecture notes on QFT. On pages 43-44, he recovers QM from QFT. See below link: QFT notes by Tong First the momentum and position operators are defined in terms of ...
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Does the commutator of anything with itself not vanish?

In a quantum mechanics exam one question was to write the commutator of a couple of operators. Everybody got points taken away since they did not write $[Q_i, Q_i] = 0$ for all the operators $Q_i$ in ...
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Non-associative operators in Physics

Are non-associative operators (or other kind of elements) used in Physics? For example, in QM I'm looking for something like this: $A(BC)|\psi\rangle \ne (AB)C|\psi\rangle$ NOTE: I think that this ...
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Does Heisenberg equation of motion imply the Schrodinger equation for evolution operator?

Let us choose to postulate (e.g. considering the analogy of the Hamiltonian being a generator of time evolution in classical mechanics) $$ i\hbar \frac{d\hat{U}}{dt}=\hat{H}\hat{U}\tag{1} $$ where ...
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Implications of unbounded operators in quantum mechanics

Quantum mechanical observables of a system are represented by self - adjoint operators in a separable complex Hilbert space $\mathcal{H}$. Now I understand a lot of operators ...
5
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1answer
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Directional derivatives in the multivariable Taylor expansion of the translation operator

Let $T_\epsilon=e^{i \mathbf{\epsilon} P/ \hbar}$ an operator. Show that $T_\epsilon\Psi(\mathbf r)=\Psi(\mathbf r + \mathbf \epsilon)$. Where $P=-i\hbar \nabla$. Here's what I've gotten: ...
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1answer
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The dual role of (anti-)Hermitian operators in quantum mechanics

Hermitian (or anti-Hermitian) operators are of central importance in quantum mechanics in at least two different incarnations: Observables are represented by Hermitian operators on the quantum ...
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Unitary spacetime translation operator

Srednicki writes: We can make this a little fancier by defining the unitary spacetime translation operator $$ T(a) \equiv \exp(-iP^\mu a_\mu/ \hbar) $$ Then we have $$ T(a)^{-1} \phi(x) T(a) = ...
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Canonical Commutation Relations in arbitrary Canonical Coordinates

If one were to formulate quantum mechanics in an arbitrary canonical coordinate system, does he impose canonical commutation relations using Dirac's recipe? ...
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What is the value of a quantum field?

As far as I'm aware (please correct me if I'm wrong) quantum fields are simply operators, constructed from a linear combination of creation and annihilation operators, which are defined at every point ...
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Commutator with exponential $[A, \exp(B)]$

How can I tell if $A$ and $\exp(B)$ commute? For $[A, B]$ it's simply $AB-BA$ and for $[\exp(A), \exp(B)]$ I think it'd be $\exp(A)\exp(B) - \exp(B)\exp(A) = \exp(A+B) - \exp(B+A) = 0$. Update: it's ...
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Associating a Unitary operator to proper Lorentz transformations?

If one reads eg page 32 of Srednicki where he says: In quantum theory, symmetries are represented by unitary (or antiunitary) operators. This means that we associate a unitary operator U(Λ) ...
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2answers
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What is the Momentum Operator?

I know the equation for the momentum operator, but what exactly is the momentum operator? It's bizarre to me that taking the derivative of the wave function, which is an operator, should return ...
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Energy is actually the momentum in the direction of time?

By comparatively examining the operators a student concludes that `Energy is actually the momentum in the direction of time.' Is this student right? Could he be wrong?