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20
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2answers
4k views

Time as a Hermitian operator in QM?

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 ...
12
votes
3answers
5k views

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') ...
23
votes
3answers
3k views

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 ...
18
votes
7answers
25k 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 ...
7
votes
8answers
3k 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?
4
votes
2answers
354 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 ...
5
votes
1answer
410 views

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 ...
2
votes
1answer
190 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 ...
2
votes
1answer
276 views

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 ...
16
votes
3answers
1k views

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 ...
32
votes
5answers
3k views

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 ...
5
votes
5answers
839 views

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 ...
10
votes
1answer
1k views

“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 ...
1
vote
3answers
636 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. ...
2
votes
3answers
496 views

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 ...
18
votes
2answers
2k views

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 ...
22
votes
6answers
2k views

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 ...
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, ...
7
votes
2answers
1k views

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 ...
10
votes
2answers
463 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.
4
votes
2answers
526 views

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) ...
5
votes
2answers
299 views

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 ...
3
votes
1answer
142 views

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 ...
2
votes
4answers
167 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 ...
5
votes
3answers
406 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 ...
3
votes
2answers
462 views

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 ...
1
vote
3answers
282 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
1answer
1k views

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 ...
2
votes
1answer
216 views

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 ...
1
vote
2answers
222 views

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 ...
21
votes
4answers
545 views

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 ...
5
votes
1answer
360 views

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)? ...
11
votes
3answers
543 views

Is it possible to define a “it went through two slits” observable?

This concerns the famous two-slit experiment. Electrons or photons or your favorite particle, doesn't matter. As we all know, the attempt to detect which slit the quanta pass through leads to loss ...
7
votes
2answers
2k views

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 ...
6
votes
3answers
415 views

Some questions on observables in QM

1-In QM every observable is described mathematically by a linear Hermitian operator. Does that mean every Hermitian linear operator can represent an observable? 2-What are the criteria to say whether ...
4
votes
1answer
310 views

Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?

Is the Hermitian operator $\hat{\mathcal{O}}=\hat{\phi}^{-}(x)\hat{\phi}^{+}(x)$, where $\hat{\phi}^{+}(x)$ is positive frequency part of the scalar field operator, a well defined observable in QFT? ...
3
votes
1answer
415 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 ...
6
votes
4answers
2k views

What does observation mean in two-slit electron diffraction experiment? [duplicate]

My question is clear, that I ask: What do we mean by "observation" in 2-slit experiment for electrons (or any other wave-particle)? You know, we say that :"if we observe the electron, it shows a ...
2
votes
2answers
6k views

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 ...
3
votes
1answer
316 views

The Physical Meaning behind a Commutator [duplicate]

I've just been introduced to the idea of commutators and I'm aware that it's not a trivial thing if two operators $A$ and $B$ commute, i.e. if two Hermitian operators commute then the eigenvalues of ...
3
votes
2answers
302 views

Understanding Well Defined States

I am self-studying from a text in QM. Well defined states are mentioned several times. By and large these are consistent and seem to be readily apparent: states of well defined energy are basis ...
0
votes
2answers
64 views

What is the correct way to treat operators that has “time” in QM? [duplicate]

I don't know if this question has already been resolved but considering that $i\hbar\partial_t$ is the energy operator, and $\partial^2_t$ is the waves operator (or helmholtz), I can't accept that $t$ ...
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?
6
votes
2answers
139 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 ...
4
votes
2answers
276 views

Uniqueness of eigenvector representation in a complete set of compatible observables [duplicate]

Possible Duplicate: Uniqueness of eigenvector representation in a complete set of compatible observables Sakurai states that if we have a complete, maximal set of compatible observables, ...
2
votes
1answer
87 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 ...
2
votes
1answer
180 views

About states, observables and the wave functional interpretation in QFT with gauge fields

First of all, I'm a mathematician, so forgive me for my possible trivial mistakes and poor knowledge of physics. In a QFT, we just start with a field (scalar, vectorial, spinorial, gauge etc), so I ...
2
votes
2answers
817 views

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 ...
1
vote
2answers
181 views

Is there a deterministic observable that has only single eigenvalue?

Is there an observable in quantum mechanics which has only one eigenvalue and an eigenspace associated with that single eigenvalue? This observable is deterministic in the sense that it gives same ...
1
vote
2answers
184 views

Is there any non-hermitian operator on Hilbert Space with all real eigenvalues?

The property of hermitian is the sufficient condition for eigenvalue being real. Is there any non-hermitian operator on Hilbert Space with all real eigenvalues? If there exist, then can all ...