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 ...

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

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 $...
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2answers
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What happens to Pauli's argument (that says that there is no time operator) when applied to $X$ operator for some simple systems?

An argument by Pauli is usually referred to in the literature when it is stated that there cannot be a time operator in quantum mechanics. This argument can be found as a footnote to P63 of W. Pauli, ...
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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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1answer
113 views

Why are general wave functions expressed in terms of energy eigenfunctions?

I have read that the eigenfunctions of any hermitian operator can be used as a basis to express any function, but I have only ever really seen the eigenfunctions of the Hamiltonian used. Why is this? ...
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Why isn't there a Time Operator in Quantum Mechanics? [duplicate]

I was wondering about a scenario where you subject a quantum particle to an intense gravitation field. Why can't we apply a sort of time operator to the particle to see how time changes for the ...
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2answers
58 views

How do I get observables to calculate uncertainty?

Given an infinite potential square well with $0<x<L$, I need to calculate the uncertainties of position and momentum. The eigenstates in the position basis are $$\lvert E_n\rangle\to \psi_n(x)=\...
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1answer
368 views

In quantum mechanics, how exactly do we associate Hermitian operators to classical observables? [duplicate]

In a first course on quantum mechanics, everybody learns some version of the following statement: Postulate: To every classical observable $A$ of a physical system, there corresponds a Hermitian ...
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Expectation value of an Observable and Eigenstates

I am learning about Quantum Mechanics at the moment and I was wondering about Eigenfunctions and Observables. The question I would like to ask is, If a wavefunction is not an eigenstate of an ...
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2answers
231 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? ...
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1answer
35 views

Bohr frequency of an expectation value?

Consider a two-state system with a Hamiltonian defined as \begin{bmatrix} E_1 &0 \\ 0 & E_2 \end{bmatrix} Another observable, $A$, is given (in the same basis) by \begin{bmatrix} 0 &a \...
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1answer
58 views

Interpretation of two different observables, both with the same resolution of the identity

Suppose you have a resolution of the identity $\hat{\mathbb{1}}=\sum_i\hat{p_i}$ (pairwise othogonal), and construct two (non-degenerate) pvm observables, $\hat{B}=\sum_ib_i\hat{p_i}$ and $\hat{C}=\...
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7answers
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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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Our choice of basis surely cannot effect possible outcomes of a measurement?

Common sense says that, of course, the outcome of a measurement on a quantum system cannot be affected by what base we choose to represent it in. However, while studying QM text, it seems like they ...
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1answer
47 views

Gauge Bosons at Finite Temperature

I was reading a paperĀ¹, and it states: " Therefore, the gauge fields themselves cannot be entities of the physical reality, as any observations should be independent of the chosen gauge" I'm trying ...
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Can linear combinations of expectation values be considered a valid expectation value?

I ask this question with Bell's paper in mind. I certainly don't hope that is actually necessary for anyone to look at but here is the link anyway: http://journals.aps.org/rmp/abstract/10.1103/...
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1answer
44 views

Does Heisenberg's uncertainty hold for any two quantum measurements?

Heisenberg's uncertainty principle is most commonly expressed in terms of the uncertainty in measurement of position and momentum of a particle, $$\Delta x\Delta p \geq \hbar$$and uncertainty in ...
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Uncertainty Principle with the corresponding operators

Why does the corresponding operator do not commute if there is uncertainty related to two observables A and B that states $\Delta A\,\Delta B > 0 $ ?
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1answer
81 views

Heisenberg's uncertainty principle derivation in a ring [duplicate]

The standard derivation But now suppose the space is a ring of length $L$, it seems the derivation could work out exactly the same and we get $$\Delta p \Delta x \geq \hbar/2.$$ But since $\Delta x$ ...
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2answers
71 views

How to recognize a Complete Set of Commuting Operators (CSCO)

A question about 'completeness'. These two operators are commuting, but I want to know more about their completeness. How do you know if {H}, {B}, {H,B} and/or {$H^2$,B} are forming (a) Complete Set(...
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1answer
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Taking Measurements of Quantities in QM

I have a quick question relating to Annihilation and Creation operators, and in taking observables in general. Let's say, for instance, that I prepare a particle so that I consider the projection of ...
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2answers
79 views

Trace of an observable [closed]

If $X$ and $Y$ are two observables and $\rho$ is a density operator, is it true that for every complex number $z$ the quantity $$ \mathrm{tr}[\rho (X+zY)^*(X+zY)] $$ is non-negative?
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2answers
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If a quantum state is pure why are its observables still probabilistic?

As I understand it, a pure quantum state is one that can be represented as a ket $\lvert\psi\rangle$ in a Hilbert space, and it contains all the information about the state of the system. As such, we ...
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What is $\langle \phi | H | \psi \rangle$ in QM?

I know that $\langle \phi | \psi \rangle$ is the probability of going from the $\psi$-state to the $\phi$-state, and that $\langle \phi | H | \phi \rangle$ is the expectation value of the energy for ...
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1answer
123 views

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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State after measurement, 'Observable operator on ket' still a physical state?

If we do Sx measurement on initially z direction spin of + state, say |z> state, we will get either |+x> or |-x> state. But if we represent $|z>=\frac{1}{\sqrt2}(|x>+|-x>) $,Then $S_x |z&...
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Is the probability current an observable?

Is the probability current in Quantum Mechanics an observable? If so, how can it me measured (directly or indirectly)?
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1answer
61 views

Reason behind the uncertainty principle [duplicate]

I know that Heisenberg Uncertainty principles states that the momentum and position of a quantum object can not be determined at the same time. This is very strange to me. I want the basic reason ...
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1answer
113 views

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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6answers
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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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1answer
105 views

Which of the properties of particles are intrinsic properties and why? [closed]

For macroscopic objects it's clear that - once observed - the observed property does exist for a while, even if we are no longer observing it. That has to do with the complexity and stability of such ...
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4answers
187 views

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 ...
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1answer
100 views

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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2answers
66 views

Why do $\hat{X}$ and $\hat{P}$ have to correspond to position and momentum?

As far as I understand, in QM we treat observables as operators, and the eigenvalues of these operators are the possible values we can measure of the observables. It is usually simpler to work in the ...
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Simultaneous measurement of non-commuting observables without uncertainty

A pair of non-commuting Observables $\hat{X}$ and $\hat{P}$ does not have a common set of eigenfunctions, i.e., it can not be measured simultaneously. Let us for the sake of simplicity assume that $[\...
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1answer
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Volume Operator / volume phase-space-function in thermodynamics

In Thermodynamics, one often encounters the derivation of pressure as the generalised force that belongs to the extensive state-variable of the volume. Postulates: One looks just at a system of many ...
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131 views

Interpretation of $\langle \phi | A | \psi \rangle$ [duplicate]

If the current state of some quantum system is $| \psi \rangle$, what is the physical interpretation of $$ \langle \phi | A | \psi \rangle $$ where $|\phi\rangle$ is some other -maybe the same- ...
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Measuring the Dirac field

If the Dirac field $\psi(x)$ is to the electron as the Electromagnetic field is to the photon, why is it that we can measure the Electromagnetic field, whereas the Dirac field we cannot?
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Measuring expectation value in quantum field theory and in quantum mechanics

There is a way of calculating the vacuum expectation value $\langle 0|\hat\phi|0\rangle$ theoretically in a quantum field theory like there is a rule to compute expectation value of any operator A (...
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What are physical observables that are connected to orbital angular momentum?

We considered a system that is confined to a curved surface. In the quantization process, we have obtained an additional orbital angular momentum that are from the surface geometrical deformation. Now ...
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297 views

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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How to measure $\mathbb{L}^2$ and $L_z $ simultaneously

What does an experiment look like, in which both quantities are measured simultanously?
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1answer
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Is the collapsed wavefunction a solution of Time-dependent Schrodinger equation?

For measurement of any observable associated with the particle, should the wavefunction after collapse be a solution of the time-dependent Schrodinger equation? A general solution of the time ...
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2answers
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Measurement of position after collapse of a wavefunction

Suppose I have a wavefunction which collapses to a certain eigenstate after a measurement of energy. In that state, I perform a calculation of position and obtain a certain position value, say $x_0$. ...
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1answer
85 views

Relativistic Commutation relation for momentum and position

We all know that the canonical commutation relation give you $$[x_i,p_j]=i\hbar\delta_ij,$$ is there a relativistic version such as $$[x^a,p_b]=i\hbar\delta_a^b?$$ If so what is the time ...
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Is there a physical significance to non-normal states of the algebra of observables?

Quantum theory may be formalized in several different ways. Generally, the physical discussion of different states of a quantum system distinguishes pure and mixed states, and then subsumes both in a ...
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Why Hamiltonian is Hermitian? [duplicate]

Everyone knows that this is needed to make eigenvalues real, but still why we enforcing such a structure at first place? An arbitrary operator can have as complex as real eigenvalues, we can simply ...
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1answer
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Distinguishing degenerate states physically

Suppose there is a free particle on a circle with radius r. The energy spectrum is then $$E_n = \frac{n^2\hbar^2}{2mr^2} \,.$$ Thus, when $n \neq 0$, then the spectrum of energies is degenerate ...
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Physical quantities have definite values?

I don't really know if this question has an anwser but I thought it was worth to try asking. My point here is the following: in Quantum Mechanics, to describe the states of a system we use a Hilbert ...
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Is there any additional complexity in the physical laws that seems unnecessary for us to exist? [closed]

I am wondering if the universe is as simple as possible, at least given the constraint that humans exist on Earth. This is my second attempt at this question, which was marked as too broad, since I ...
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Have Witten-type TQFT's nonconservation of energy and momentum in interactions?

Witten-type topological quantum field theories are based on cohomology theories. Every observable must lie in a cohomology class. May be $G$ a geometric field. Then every observable expectation value ...