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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64 views

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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217 views

Superposition and simultaneous observation

Trying to understand superposition. Ok, so double slit experiment. The multiple paths the particle simultaneously travels interfere with each other but as it is absorbed, it chooses one "actual" ...
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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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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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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
367 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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1answer
149 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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226 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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34 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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Commuting operators and Direct product spaces

Under what conditions is the common eigenspace of two commuting hermitian operators isomorphic to the direct product of their individual eigenspaces? When can an eigenket $|\lambda$1$\lambda$2$\rangle$...
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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
57 views

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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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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172 views

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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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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1answer
47 views

Gauge Bosons at Finite Temperature

I was reading a paper¹, 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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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
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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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3answers
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Are the authors saying that the observer effect plays no role in Bohr's thought experiment of the Heisenberg uncertainty principle?

Here is an excerpt from Eisberg & Resnick's Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. Here is introducing Bohr's though experiment to establish a physical origin for the ...
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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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4answers
406 views

Is commutation relation an equivalence relation?

I'm now learning quantum mechanics with Liboff. In the book it deals with "a compete set of mutually compatible observables" in order to make a state maximally informative. How can one find such set? ...
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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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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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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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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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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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2answers
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Susceptibilities and response functions

It is often confusing whether a susceptibility is the same as a response function, specially that often they are used interchangeably, in the context of statistical mechanics and thermodynamics. Very ...
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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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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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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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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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Is the expectation value always an eigenvalue?

Must the expectation value of an observable always be equal to an eigenvalue of the corresponding operator? I already know that 0 is not an eigenvalue, but are there any other examples?
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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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8answers
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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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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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Why does the measurement of some observable $A$, the measured value is always an eigenvalue of the operator?

Explain why when we make a measurement of some observable $A$ in QM, the measured value is always an eigenvalue of the operator $A$.
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2answers
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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?