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Questions tagged [observables]

A quantum observable is a measurable operator whose corresponding property of the state can be determined by some sequence of physical operations ("observation"), such as submitting the system to various electromagnetic fields and eventually reading a value. In systems governed by classical mechanics, any experimentally observable value can be shown to be given by a real-valued function on the set of all possible system states.

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What's the deal with momentum in the infinite square well?

Every now and then a question comes up about the status of the momentum operator in the infinite square well, and while we have two good answers on the topic here and here, I'm generally not satisfied ...
Emilio Pisanty's user avatar
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3 answers
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Time as a Hermitian operator in quantum mechanics

In non-relativistic QM, on one hand we have the following relations: $$\langle x | P | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial x} \psi(x),$$ $$\langle p | X | \psi \rangle ~=~ i \hbar \...
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31 votes
4 answers
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Why do we use Hermitian operators in QM?

Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') ...
Benjamin Hodgson's user avatar
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4 answers
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Heisenberg Uncertainty Principle scientific proof

Heisenberg's uncertainty principle states that: $$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$ What is the scientific proof of this principle? Operators Uncertainty
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Is there a time operator in quantum mechanics?

The question in the title has been asked many times on this site before, of course. Here's what I found: Time as a Hermitian operator in QM? in 2011. Answer states time is a parameter. Is there an ...
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Hermiticity of Momentum Operator (matrix) Represented in Position Basis

I read that the momentum operator, $\hat P$ should be Hermitian (some would say by QM postulate). That makes perfect sense when it is represented in the momentum (p) basis: $\hat P =\int_{-\infty}^{\...
David's user avatar
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What Hermitian operators can be observables?

We can construct a Hermitian operator $O$ in the following general way: find a complete set of projectors $P_\lambda$ which commute, assign to each projector a unique real number $\lambda\in\mathbb R$...
Cristi Stoica's user avatar
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5 answers
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How does non-commutativity lead to uncertainty?

I read that the non-commutativity of the quantum operators leads to the uncertainty principle. What I don't understand is how both things hang together. Is it that when you measure one thing first and ...
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Does every hermitian operator represent a measurable quantity?

In Quantum mechanics, observables are represented by hermitian operator. But does every hermitian operator represent a observable? If not , how do we know that whether a hermitian operator represent ...
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43 votes
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Not all self-adjoint operators are observables?

The WP article on the density matrix has this remark: It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable.[...
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Is mass an observable in Quantum Mechanics?

One of the postulates of QM mechanics is that any observable is described mathematically by a hermitian linear operator. I suppose that an observable means a quantity that can be measured. The mass ...
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Which derivative with respect to time is which in the Heisenberg picture of quantum mechanics?

For an observable $A$ and a Hamiltonian $H$, Wikipedia gives the time evolution equation for $A(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$ in the Heisenberg picture as $$\frac{d}{dt} A(t) = \frac{i}{\hbar} ...
Qiaochu Yuan's user avatar
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What is the Physical Meaning of Commutation of Two Operators?

I understand the mathematics of commutation relations and anti-commutation relations, but what does it physically mean for an observable (self-adjoint operator) to commute with another observable (...
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Conservation of momentum in infinite square well

This is inspired by Griffiths QM section 2.2, on the infinite square well, which is about how far I've gotten (so, sorry if this is addressed later in the book). For any given starting wavefunction, ...
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Why isn't the time-derivative considered an operator in quantum mechanics? [duplicate]

Based on my understanding when doing quantum mechanics we deal with a small set of mathematical objects: namely scalars, kets, bras, and operators. But then in the Schrodinger equation we have this ...
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In what sense is a scalar field observable in QFT?

Consider a QFT consisting of a single, hermitian scalar field $\Phi$ on spacetime (say $\mathbb R^{3,1}$ for simplicity). At each point $x$ in spacetime, $\Phi(x)$ is an observable in the sense that ...
joshphysics's user avatar
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Why hermitian, after all? [duplicate]

This question is going to look a lot like a duplicate, but I've read dozens of related posts and they don't touch the subject. Here we go. Why are observables represented by hermitian operators? ...
QuantumBrick's user avatar
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35 votes
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Is there something behind non-commuting observables?

Consider a quantum system described by the Hilbert space $\mathcal{H}$ and consider $A,B\in \mathcal{L}(\mathcal{H},\mathcal{H})$ to be observables. If those observables do not commute there's no ...
Gold's user avatar
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10 votes
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How are local observables encoded in this formulation of quantum field theory?

I've recently begun trying to understand a formulation of quantum field theory as a functor from a category of spacetimes-with-boundaries (bordisms) to a category of Hilbert spaces, as reviewed in ...
Chiral Anomaly's user avatar
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Why do we use Eigenvalues to represent Observed Values in Quantum Mechanics?

One of the postulates of quantum mechanics is that for every observable $A$, there corresponds a linear Hermitian operator $\hat A$, and when we measure the observable $A$, we get an eigenvalue of $\...
Ishan Deo's user avatar
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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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Is color charge a quantum mechanical observable?

If I had 2 pions that were identical, except one was comprised of a red and anti-red, and the other was comprised of a green and anti-green, would I be able to perform an experiment that distinguishes ...
Izzhov's user avatar
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1 vote
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Heisenberg choice of the observables and different outcomes [duplicate]

Somebody could help me to clarify how its possible that different choices of the observable to measure can lead to different outcomes of the observed system state?
Andrea Scaglioni's user avatar
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Intuitive meaning of Hilbert Space formalism

I am totally confused about the Hilbert Space formalism of Quantum Mechanics. Can somebody please elaborate on the following points: The observables are given by self-adjoint operators on the ...
Koushik's user avatar
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13 votes
1 answer
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Are eigenstates of the position operator continuous?

The way I've understood it is that eigenfunction of an operator are the different states which the actual wavefunction can take when the property/observable corresponding to the given operator is ...
Marek Zakrzewski's user avatar
7 votes
1 answer
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The state space is somehow defined by the observables?

In Quantum Mechanics states of a system are described by vectors in a Hilbert space called the state space while the physical quantities associated to the system are described by hermitian operators ...
Gold's user avatar
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5 votes
5 answers
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Velocity definition in quantum mechanics?

What is the definition of velocity in quantum mechanics? is it an operator?
Mohamed Daoud's user avatar
2 votes
2 answers
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How to derive Uncertainty Principle relation?

How to derive Heisenberg Uncertainty Principle relation?
Pushkar Soni's user avatar
37 votes
6 answers
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Must observables be Hermitian only because we want real eigenvalues, or is more to that?

Because (after long university absence) I recently came across field operators again in my QFT lectures (which are not necessarily Hermitian): What problem is there with observables represented by non-...
Quantumwhisp's user avatar
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17 votes
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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, we ...
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Particle in a 1-D box and the correspondence principle

Consider the particle in a 1-d box, we know very well the solutions of it. I'd like to see how the correspondence principle will work out in this case, if we consider position probability density ...
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11 votes
1 answer
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"An operator is hermitian". Implications?

Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter. The first part of his second postulate can be stated as: Every dynamical variable may be ...
ZAC's user avatar
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10 votes
3 answers
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Reason for Uncertainty principle

$$\Delta x \Delta p_x \geq \frac{\hbar}{2} $$ I understand what does Heisenberg's uncertainty principle states i.e. it's definition and it has been proven experimentally. But, can anyone please ...
23rduser's user avatar
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8 answers
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What exactly is the 'observer' in physics and/or quantum mechanics? [duplicate]

Possible Duplicate: nature of an observer For instance, in the double slit experiment, what is exactly defined as an observer? I remember from somewhere, light is also an observer?
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Are two states with the same measurement probabilities necessarily equal up to unitary equivalence?

Let $\rho$ and $\rho'$ be $n\times n$ density matrices, and suppose that for every observable $A$ and every $\lambda$ in the spectrum of $A$ we have $$ \text{tr}(\rho P_{\lambda})=\text{tr}(\rho' P_{\...
dezign's user avatar
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9 votes
3 answers
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Are angles ($\theta$ and $\phi$) in spherical coordinates treated as operators in quantum mechanics?

Position is specifically considered as an operator in quantum mechanics. I want to know if $\theta$ and $\phi$ are explicitly considered as operators in quantum mechanics for solutions to 3D ...
prateek's user avatar
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8 votes
3 answers
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Proof for $\langle q| p \rangle = e^{iqp}$

What would be the proof for $\langle q| p \rangle = e^{iqp}$? Is it derived from canonical commutation relation? ($|q \rangle $ represents the position eigenstate, while $|p \rangle$ represents the ...
RRRR's user avatar
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2 votes
3 answers
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Can we measure "wavefunction" of quantum particles?

We know that there is uncertainty principle, so question: can we ever measure wavefunction of particles? I do not think this is possible, but I am not sure. I guess that everything is probabilistic. (...
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1 answer
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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 ...
turnip's user avatar
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2 votes
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What are the Time Operators in Quantum Mechanics? [duplicate]

I don't understand at all what the time operators are in quantum mechanics. I thought that given a wave function, because it's a function of time, we could simple put in any time in the future to find ...
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1 vote
3 answers
39k views

Deriving a QM expectation value for a square of momentum $\langle p^2 \rangle$

I already derived a QM expectation value for ordinary momentum which is: $$ \langle p \rangle= \int\limits_{-\infty}^{\infty} \overline{\Psi} \left(- i\hbar\frac{d}{dx}\right) \Psi \, d x $$ And I ...
71GA's user avatar
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25 votes
4 answers
950 views

Is every quantum measurement reducible to measurements of position and time?

I am currently studying Path Integrals and was unable to resolve the following problem. In the famous book Quantum Mechanics and Path Integrals, written by Feynman and Hibbs, it says (at the beginning ...
Jay's user avatar
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23 votes
7 answers
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Why is a Hermitian operator a "quantum random variable"?

To me, as a stupid mathematician, a random variable is a measurable function from some probability space $(\Omega, \sigma, \mu)$ to $(\Bbb{R}, B(\Bbb{R}))$. This makes sense. You have outcomes, events,...
user avatar
21 votes
5 answers
1k views

What is the physical meaning of the sum of two non-commuting observables?

Scenario: ${\mathcal A}$ and ${\mathcal B}$ are two observables. Mathematically we model them by two Hermitian operators $A\colon H \to H$ and $B\colon H \to H$ on a separable Hilbert space. ...
Nobody-Knows-I-am-a-Dog's user avatar
12 votes
2 answers
3k views

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 ...
hwlin's user avatar
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12 votes
5 answers
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Why is Spin Less Classical than Position?

It is often repeated that "the spin observable is purely quantum and has no classical counterpart". What is actually meant by that? I see no principle difference between the set of spin observables ...
Lior's user avatar
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10 votes
3 answers
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Susceptibilities and response functions

It is often confusing whether a susceptibility is the same as a response function, specially that often they are used interchangeably, in the context of statistical mechanics and thermodynamics. Very ...
user929304's user avatar
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8 votes
1 answer
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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 ...
doetoe's user avatar
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8 votes
1 answer
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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 ...
SuperCiocia's user avatar
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7 votes
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"Commuting observables share common eigenstates"

I am struggling to find a precise definition of this line from my quantum mechanics textbook: If $[A,B] = 0$, then the operators commute, and "commuting operators share common eigenstates". This ...
Sophia Tevosyan's user avatar