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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Our choice of base 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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28 views

The question about an observer and an observed environment [on hold]

Imagine that Bob is observing everything that is not him (environment). The experiment: the environment changes somewhere far away from Bob. The conclusion: The change propagates in the ...
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45 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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26 views

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: ...
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42 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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52 views

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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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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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 ...
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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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73 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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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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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 ...
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68 views

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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59 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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107 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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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
94 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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172 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
94 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
62 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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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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129 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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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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85 views

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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54 views

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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81 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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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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0answers
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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 ...
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405 views

Why don't non-Hermitian operators with all real-eigenvalues correspond to observables? [duplicate]

Suppose you could construct an operator that was non-Hermitian but had all real eigenvalues or could at least be restricted in a way to create only real eigenvalues, why would this operator not ...
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4answers
183 views

How to determine the observables rigorously?

In Quantum Mechanics, as I know, if a system is described by a Hilbert space $\mathcal{H}$, each physical quantity is associated with some hermitian operator $A : U\subset \mathcal{H}\to \mathcal{H}$ ...
3
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1answer
135 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 ...
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Is the perturbation Hamiltonian an observable?

In fine structure calculation we use the perturbation theory. The basic Hamiltonian $H_0$ is perturbed as: $H = H_0 + W$ First, the basic problem assume that $H_0$ is an observable. That allows to ...
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spin independent observable [closed]

Let's consider a spin independent observable $O$ (the terms of the operator don't depend upon the spin operator). If we are interested to find an eigenfunctions' basis of the wave-functions' space, ...
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Function of observables in mathematical words

In mathematical words, an observable is an operator that a set of linearly independent eigenfunctions constitutes a complete basis of the wave-functions' space. Now, let's consider some observables: ...
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428 views

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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3answers
135 views

Both Eigenvalues and Operators are “Observables”? [duplicate]

I am having a bit of difficulty wading through the what seems to be multiple usages for Observables in Quantum Mechanics. " Mathematically observables are postulated to be Hermitian operators.. " ...
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1answer
204 views

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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What does it mean to observe? [duplicate]

This is a layman's question. The only thing I know about quantum physics is from casual reading and documentaries. I can imagine electrons being probabilistic waves. Their position is an infinite ...
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Lorentz Symmetry

Quick question about Lorentz symmetry. From the wiki page the feature of nature that says experimental results are independent of the orientation or the boost velocity of the laboratory through ...