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.

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 (self-adjoint operator) in quantum mechanics? E.g.…
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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 of a particle is an observable because it can be…
Revo
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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 observable of time? in 2012. Answer by Arnold…
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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 by their level of detail and by how easy (not…
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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.[17][18] The first footnote is to the appendix in…
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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 than the other you get a predictably different…
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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-Hermitian operators (by observables, I obviously…
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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 expansions correspond to Feynman diagrams which can be…
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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 it is a hermitian operator (operator-valued…
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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 simultaneous basis of eigenvectors of each of them.…
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Time as a Hermitian operator in quantum mechanics

In non-relativistic QM, on one hand we have the following relations: $$\langle x | P | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial x} \psi(x),$$ $$\langle p | X | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial p} \psi(p).$$ On the other…
skywaddler
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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') operators. The eigenvalues of the operator are…
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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 $\hat A$ as the result. To me, this result seemed to…
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Why doesn't the phase operator exist?

In many articles about quantum optics, the phase-number uncertainty relation $$\Delta \phi \Delta n \ge 1$$ has been mentioned and used as a heuristic argument, but they say that the phase-number uncertainty relation does not exist in a strict…
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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 $H|n\rangle=n|n\rangle$, $\langle n|m\rangle=\delta_{n,m}$. Then prepare a…
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