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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27
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7k views

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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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') ...
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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 ...
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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 ...
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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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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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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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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 $\lambda\in\mathbb R$...
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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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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 ...
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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}^{\...
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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?
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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 ...
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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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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 ...
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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 ...
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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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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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“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 ...
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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 ...
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proof for $\langle q| p \rangle = e^{ipq}$

What would be the proof for $\langle q| p \rangle = e^{ipq}$? Is it derived from canonical commutation relation? ($|q \rangle $ represents the position eigenstate, while $|p \rangle$ represents the ...
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1answer
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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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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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1answer
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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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How to derive Uncertainty Principle relation?

How to derive Heisenberg Uncertainty Principle relation?
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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,...
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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 ...
2
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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 ...
2
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1answer
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Lagrangian gauge theory with physically observable local degrees of freedom

In my answer at What, in simplest terms, is gauge invariance?, I mentioned that in certain contexts there can be a "gauge theory" with a local symmetry that leave the Lagrangian/Hamiltonian invariant ...
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3answers
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What happens with a tunneling particle when its momentum is imaginary in QM?

In classical mechanics the motion of a particle is bounded if it is trapped in a potential well. In quantum mechanics this is no longer the case and there is a non zero probability of the particle to ...
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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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1answer
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Does spontanous symmetry breaking affect Noethers theorem?

Does spontanous symmetry breaking affect the existence of a conserved charge? And how does depend on whether we look at a classical or a quantum field theory (e.g. the weak interacting theory)? (In ...
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Is the wave function objective or subjective?

Here is a question I am curious about. Is the wave function objective or subjective, or is such a question meaningless? Conventionally, subjectivity is as follows: if a quantity is subjective then ...
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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? ...
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2answers
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Inexact measurement and wavefunction collapse

As is usually said, measurement of an observable $q$ leads to collapse of wavefunction to an eigenstate of the corresponding operator $\hat q$. That is, now the wavefunction in $q$ representation is $\...
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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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2answers
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Diff(M) and requirements on GR observables

This question is kind of inspired in this one: Diff(M) as a gauge group and local observables in theories with gravity The conundrum i'm trying to understand is how is derived the (quite) ...
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4answers
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Is what statisticians call a “random variable” what physicists call an “observable” in QM? [duplicate]

I read at http://www.statlect.com/fundamentals-of-probability/random-variables that A random variable is a variable whose value depends on the outcome of a probabilistic experiment. Its value is a ...
3
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1answer
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Observables - what are they?

I often read in books that an observable is represented by an Hermitean operator. But it is deceiving as operator isn't the observable. As far as I've read the observable is denoted like $\langle \...
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1answer
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Quantum mechanics - measuring position

I am watching Susskind's Stanford Lectures on quantum mechanics. The eigenvectors (eigenfunctions) of the position operator are of the form $\delta(x-k)$. But $$\int\delta^{*}(x-k)\delta(x-k)\, \...
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2answers
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What exactly implies the need of quantum mechanics for self-adjoint and not only symmetric operators? [duplicate]

We know that quantum mechanics requires self-adjoint operators, not only symmetric. Can we say that this follows ONLY from the two following axioms of quantum mechanics, namely that each observable $...
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1answer
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Why we dont have “direct” velocity operator just as $p$? ( as use $p$ space not $v$? ) in quantum mechanics?

why there is no direct velocity operator on quantum mechanic while there is for mumentum ( $p_{x}=d/dx$ ) Also why use mumentum space not velocity?
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Uniqueness of eigenvector representation in a complete set of compatible observables

Sakurai states that if we have a complete, maximal set of compatible observables, say $A,B,C...$ Then, an eigenvector represented by $|a,b,c....>$, where $a,b,c...$ are respective eigenvalues, is ...
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1answer
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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 ...
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Deriving a QM expectation value for a square of momentum $\langle p^2 \rangle$

I alredy 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 ...
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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 ...
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1answer
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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 ...
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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 ...
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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 ...