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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 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 (...
adustduke's user avatar
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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 ...
Revo's user avatar
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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 ...
ziggurism's user avatar
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43 votes
2 answers
6k views

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
42 votes
3 answers
3k views

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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40 votes
5 answers
15k views

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 ...
vonjd's user avatar
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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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36 votes
5 answers
6k views

Why do we observe particles, not quantum fields?

My understanding is that, in the context of quantum field theory, particles arise as a computational tool. We perform an expansion in the path integral in some parameter, and the terms in these ...
Charles Hudgins's user avatar
35 votes
7 answers
7k views

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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34 votes
3 answers
4k views

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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33 votes
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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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33 votes
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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 \...
skywaddler's user avatar
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31 votes
4 answers
19k views

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
29 votes
3 answers
3k views

Do states with infinite average energy make sense?

Do states with infinite average energy make sense? For the sake of concreteness consider a harmonic oscillator with the Hamiltonian $H=a^\dagger a$ and eigenstates $H|n\rangle=n|n\rangle$, $\langle n|...
Weather Report's user avatar
28 votes
3 answers
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Heisenberg's uncertainty principle for mean deviation?

The Heisenberg uncertainty principle states that $$\sigma_x \sigma_p \ge \frac{\hbar}{2}$$ However, this is only for the standard deviation. What is the inequality if the mean deviation, defined as ...
Zach466920's user avatar
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25 votes
1 answer
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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 ...
user774025's user avatar
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25 votes
4 answers
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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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24 votes
2 answers
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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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24 votes
2 answers
3k views

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
23 votes
7 answers
3k views

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,...
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21 votes
5 answers
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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
20 votes
1 answer
1k views

Does a complete set of commuting observables always exist?

The Wikipedia article for complete set of commuting observables (CSCO) outlines a method for finding a CSCO. However, it does not suggest why such a set should exist. On one hand, I think it is very ...
awsomeguy's user avatar
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19 votes
2 answers
1k views

Are observables in QFT actually observable?

Consider some interacting QFT on a lattice (just to avoid infinitely large momentums). The size of the lattice is assumed to be much smaller than the size of the emergent particles (like in our world)....
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19 votes
3 answers
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In what sense (if any) is Action a physical observable?

Is there any sense in which we can consider Action a physical observable? What would experiments measuring it even look like? I am interested in answers both in classical and quantum mechanics. I ...
PPenguin's user avatar
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18 votes
2 answers
6k views

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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18 votes
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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.
346699's user avatar
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17 votes
5 answers
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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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17 votes
4 answers
3k views

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 ...
user avatar
17 votes
7 answers
4k views

Why is the measured value of some observable $A$, always an eigenvalue of the corresponding operator?

Explain why when we make a measurement of some observable $A$ in QM, the measured value is always an eigenvalue of the operator $A$.
alejandro123's user avatar
16 votes
7 answers
12k views

Heisenberg uncertainty principle in daily life

I need some examples of the Heisenberg uncertainty principle on a basic level, or if possible in daily life. Or maybe a simple explanation for validity of the principle in easier words. I cannot get ...
Akil's user avatar
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16 votes
4 answers
3k views

Why does the Heisenberg uncertainty principle apply to particles?

This might be a slightly naive question, and if so I apologize, but I am currently a little confused as to why the Heisenberg Uncertainty principle should apply to particles, i.e. our system (say an ...
Mason Giacchetti's user avatar
16 votes
4 answers
2k views

Is hermiticity a basis-dependent concept?

I have looked in wikipedia: Hermitian matrix and Self-adjoint operator, but I still am confused about this. Is the equation: $$ \langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x ...
OkThen's user avatar
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16 votes
3 answers
1k views

Paradox about the Stern-Gerlach experiment with $B=0$, $\nabla B\ne 0$

In a modern interpretation of the historical Stern-Gerlach experiment, a beam of neutral silver atoms, each with spin 1/2, was sent in the z direction through a nonuniform magnetic field having both a ...
user avatar
15 votes
6 answers
2k views

What does vector operator for angular momentum measure?

Consider the vector operator for angular momentum $\hat L=\hat L_x \vec i +\hat L_y \vec j + \hat L_z \vec k$. Does this mean that if we want to measure the angular momentum of a particle in state $\...
TaeNyFan's user avatar
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15 votes
2 answers
475 views

How do they measure quantum observables in general?

In most introductory quantum mechanics books they mention as an experimental example, which demonstrate the quantum nature of measurement, the Stern-Gerlach experiment and how they measure the spin. ...
Physor's user avatar
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14 votes
3 answers
1k views

Is a quantity calculated from observables, observable?

I am not a physicist, and not sure whether I want the adjective, or the noun, observable here. Example 1) If we view the mass and velocity of a classical particle as observables, we calculate the ...
Bob Terrell's user avatar
14 votes
5 answers
7k views

The Physical Meaning of Projectors in Quantum Mechanics

Let $O$ be a single-particle observable for a system, and $|L\rangle$ and $|R\rangle$ two orthonormal eigenstates of $O$. You may imagine that the system consists in two photons, and $|L\rangle$ and $|...
Lory's user avatar
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14 votes
2 answers
5k views

Can one define an acceleration operator in quantum mechanics?

It seems most books about QM only talk about position and momentum operators. But isn't it also possible to define a acceleration operator? I thought about doing it in the following way, starting ...
asmaier's user avatar
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14 votes
1 answer
956 views

Resolution of the identity of operator with mixed spectrum

In most quantum mechanics text books, the resolution of the identity or completeness relation is stated in the following (or similar) form $$ \mathbb I_\mathcal H = \sum\limits_n |\lambda_n\rangle \...
Tobias Fünke's user avatar
14 votes
0 answers
1k views

Where does a fermionic coherent state live (which Hilbert space)? [duplicate]

There have been a couple of questions on fermionic coherent states, but I didn't find any that covered the following question: If I define a coherent fermionic state in the 2-level-system spanned by $...
LFH's user avatar
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13 votes
3 answers
3k views

Why does the expectation value in quantum mechanics correspond to the classically measured value?

I understand that we can use the Heisenberg picture to show, for a Hamiltonian of the form $$ \hat{H}=\frac{\hat{P}^{2}}{2m}+\hat{V}(\hat{X}) $$ the Ehrenfest theorem: $$ m\partial_{t}\langle \hat{X}\...
Adrien Amour's user avatar
13 votes
1 answer
5k views

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
13 votes
2 answers
10k views

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 ...
Rajesh D's user avatar
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13 votes
1 answer
1k views

Sequential Stern-Gerlach devices - realizable experiment or teaching aid?

At least one textbook [1] uses sequential Stern-Gerlach devices to introduce to students that the components of angular momentum are incompatible observables. Viz., the $z$-up beam from a SG device ...
Robin Ekman's user avatar
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12 votes
5 answers
3k views

Why do eigenvalues correspond to observable quantities?

It makes sense to me that we can find some operator that gives us eigenfunctions that correspond to definite values for some desired observable. However, I do not see how the eigenvalues happen to ...
Jeff Bass's user avatar
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12 votes
6 answers
2k views

Why are expectation values of an observable important in QM?

I've been reading that expectation values of an observable is all what we can get and are the key quantities of the theory, but performing the same experiment many times would generate a distribution ...
user536450's user avatar
12 votes
2 answers
2k views

How to interpret quantum fields?

As an analogy of what I am looking for, suppose $f(x,t)$ represents a classical field. Then we may interpret this as saying at position $x$ and time $t$ the field takes on a value $f(x,t)$. In quantum ...
CBBAM's user avatar
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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
  • 2,558
12 votes
2 answers
2k views

Is every observable a function of position and momentum?

In the first answer of this question it is said that every quantum observable, let's say $\hat{A}$, can be represented as a function of position and momentum observables. In other words, as I ...
slaaidenn's user avatar
  • 584
11 votes
7 answers
2k views

Definition of four-velocity: why define it with proper time of the object?

The four-velocity(world-velocty) is defined by : $u^μ=\frac{dx^μ}{dτ}$ ,where $τ$ is the proper time of the object. I don't understand why it's defined with respect to the proper time but not the time ...
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