# Tag Info

## New answers tagged observables

1

Further to Anna V's answer, in the case of an electron there is an important physical meaning to the "lack" measurability of the phase of the electron's wavefunction. This is because the electron is coupled to the electromagnetic field. And, if one models this by the Minimal Coupling between the electron and the electromagnetic field, one gets the ...

3

What does measurable mean ? It means that one can do an experiment and get a value for a+ib , the complex number. A complex number to be measurable one should be able to measure a value at the same time for a and b and put a point on the complex plane. This means two independent variables, a and b can be measured and a point defined. In quantum mechanics ...

0

Ψ is supposedly a probability density amplitude. ΨΨ* is the probability density which in theory can be measured. For example in electron diffraction through a crystal a statistical measure of the electrons in, divided into the electrons out in a small region divided by the volume would allow ΨΨ* to be approximately measured. Phase information is lost when ...

2

$\newcommand{\ket}[1]{\lvert #1 \rangle}$You seem to be confused about what measuring an operator means. Let $A,B$ be two commuting self-adjoint operators as in your question, and let $\{u_n\}$ be a basis of simultaneous eigenvectors, that is $$A\ket{u_i} = a_i \ket{u_i} \ \vee \ B\ket{u_i} = b_i \ket{u_i}$$ Now, a generic state $\psi$ can be written as $$... 2 The algebraic approach gives the better idea of what the states and observables of a quantum theory are, and this holds in infinite dimensional systems as well. In the modern mathematical terminology, observables of quantum mechanics are the elements of a topological *-algebra, and states are objects of its topological dual that are positive and have norm ... 1 Any self-adjoint operator A on some Hilbert space H generates a one-parameter subgroup of the unitary group U(H) by$$t\mapsto U(t):= e^{itA}.$$Here we shall think that A is the Hamiltonian of some mechanical system, and that U(t) is the flow it generates. In the Heisenberg picture, the time evolution of an observable F(t) is given by ... 1 According to the Wikipedia article "Constant of Motion" A quantity A is conserved if it is not explicitly time-dependent and if its Poisson bracket with the Hamiltonian is zero That is to say, if both$$\frac{\partial A}{\partial t} = 0 \{A, H\} = 0 $$then$$\frac{dA}{dt} = \frac{\partial A}{\partial t} + \{A, H\} = 0, \, \Rightarrow A ...

1

Is the same idea of classical mechanics, but now this quantities can be undetermined if you apply some measurement of the complementary of this quantity. e.g. The total angular momentum is a constant of motion $\vec{L}$. You can measure some component $L_i$ of this angular momentum a lot of times and this yields to the same result (conservation), but if ...

7

Color charge in the sense of "being blue, red, green" is not a quantum mechanical observable because the $\mathrm{SU}(3)$ gauge transformations mix the colors. This means it is meaningless to say "We have a blue particle", because we can perform a gauge transformation and then we "have a red particle". Since physical descriptions related by gauge ...

1

In classical mechanics, each body has an exact position at all times (and hence an exact speed given by its derivative, and the equivalent quantities for rotary motion). To understand the difference to quantum mechanics, think of it as wave mechanics: Each body is described by a wave(function) and properties such as position or speed are only defined to the ...

1

Roughly speaking, and restricting to particles for now, a classical trajectory is a set of exact positions and corresponding velocities (or momenta) of the particle, which (usually) change over time. In quantum mechanics, the uncertainty principle says that it is not possible to know simultaneously the exact position and momentum of the particle, and so it ...

2

response function = susceptibility = (pure or mixed) second derivative of a (Helmholtz, Gibbs, etc.) free energy. Magnetization is not a response function as the free energy is not observable, so one cannot observe the response to a change of some variable.

0

I might be able to answer your question in the context of linear response theory: Response function: the power series expansion of the applied field generated by a weak external perturbation. Mathematically speaking, we can relate the average value of an observable $X$_i to the response function $\chi$ via \begin{align} \langle X_i(t)\rangle=\int_0^t dt'' ...

5

If you measure one particle to be in $\left|\uparrow\right\rangle$, then the other will be in $\left|\downarrow\right\rangle$, just as you said. So your two eigenvalues are $1$ and $-1$. Multiplying (not averaging) the results together: $$1\times (-1) = -1.$$ You'll get the same answer if you measure the particles to be down/up rather than up/down. I think ...

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