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2
votes
1answer
61 views

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

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 ...
2
votes
4answers
146 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 ...
2
votes
1answer
68 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, ...
2
votes
2answers
50 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 ...
7
votes
2answers
1k views

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?
1
vote
0answers
29 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 ...
3
votes
0answers
48 views

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 ...
1
vote
1answer
157 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" ...
3
votes
2answers
276 views

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 ...
7
votes
8answers
2k views

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?
9
votes
2answers
416 views

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.
7
votes
5answers
282 views

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 $A$.
3
votes
2answers
109 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- ...
4
votes
0answers
62 views

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?
1
vote
0answers
43 views

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 ...
0
votes
0answers
26 views

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 ...
5
votes
2answers
161 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 ...
4
votes
2answers
351 views

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 ...
1
vote
0answers
59 views

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

What does an experiment look like, in which both quantities are measured simultanously?
0
votes
1answer
71 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 ...
1
vote
2answers
44 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 $x_0$. ...
18
votes
7answers
22k views

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 ...
1
vote
3answers
575 views

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. ...
0
votes
1answer
56 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 ...
3
votes
0answers
81 views

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 ...
1
vote
0answers
52 views

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 ...
3
votes
1answer
185 views

On the Equivalence of Schrodinger and Heisenberg Descriptions of Quantum Mechanics and Observability

I'm not a physicist, but rather a control (feedback) systems engineer eager to understand more than just a cursory explanation of quantum mechanics. The StackExchange has been an excellent forum for ...
0
votes
1answer
43 views

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 ...
0
votes
2answers
101 views

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 ...
2
votes
0answers
81 views

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 ...
5
votes
3answers
7k views

Proof of Canonical Commutation Relation (CCR)

I am not sure how $QP-PQ =i\hbar$ where $P$ represent momentum and $Q$ represent position. $Q$ and $P$ are matrices. The question would be, how can $Q$ and $P$ be formulated as a matrix? Also, what is ...
1
vote
3answers
245 views

Why is only one quantity of angular momentum i.e. $L_z$ quantized & not $L_x$ & $L_y$?

This is quoted from Arthur Beiser's Concepts of Modern Physics: Why is only one quantity of $\mathbf{L}$ quantized? The answer is related to the fact that $\mathbf{L}$ can never point in any ...
2
votes
0answers
41 views

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 ...
4
votes
2answers
282 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 ...
4
votes
4answers
158 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 $A : U\subset \mathcal{H}\to \mathcal{H}$ ...
1
vote
1answer
75 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 ...
0
votes
1answer
96 views

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, ...
1
vote
3answers
75 views

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 ...
0
votes
2answers
118 views

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: ...
0
votes
3answers
117 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.. " ...
6
votes
2answers
132 views

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 ...
2
votes
1answer
151 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 ...
0
votes
1answer
43 views

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 ...
3
votes
1answer
359 views

The dual role of (anti-)Hermitian operators in quantum mechanics

Hermitian (or anti-Hermitian) operators are of central importance in quantum mechanics in at least two different incarnations: Observables are represented by Hermitian operators on the quantum ...
1
vote
1answer
43 views

“Independent simultaneous eigenbras” in Dirac's book 'Principles of Quantum Mechanics'

I've been puzzling through this book off and on and can usually work out what is going on via other external references on the Intertubes. But, this paragraph from pages 55 and 56 has me a bit ...
1
vote
1answer
102 views

The Eigenstate Existence Problem in Dirac's book 'Principles of Quantum Mechanics'

In Chapter II of Dirac's book Principles of Quantum Mechanics, Dirac explains that in general it is very difficult to know whether, for a given real linear operator, that any eigenvalues/eigenvectors ...
0
votes
3answers
87 views

Do we get the same answer at any time if we measure a system's energy?

Schrödinger's equation says that the only allowed energy states of a system are the eigenvalues of the energy operator $H$. This means that if we measure the energy of the system at any time we ...
13
votes
4answers
1k views

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, ...
0
votes
1answer
58 views

Particle in a box - speed probability distribution

Consider a particle in a box with infinite barriers. By solving the Schrödinger we can find the probability of finding the particle at some points in the box. How can we find the probability of ...
0
votes
2answers
108 views

Would $[\hat{Q},\hat{H}]$ correspond to an observable? [closed]

Would $[\hat{Q},\hat{H}]$ correspond to an observable? Where $\hat{Q}$ is an observable and $\hat{H}$ is the Hamiltonian. Surely that would just mean that $[\hat{Q},\hat{H}]$ would commute i.e. = 0?: ...