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3

In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics.

1

I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace ...

0

If the Hamiltonian has $SU(2)$ symmetry then the Hilbert space can be spanned by eigenstates of Cartan subalgebra (one dimensional) operator and Casimir operator. As we know from elementary quantum mechanics \begin{align} \hat{L}^{2}\left|l,m \right\rangle &=l(l+1)\left|l,m \right\rangle,\\[3mm] \hat{L}_{z}\left|l,m \right\rangle &= m\left|l,m ...

0

Is the probability $|c_1|^2 + \ldots + |c_k|^2$? Yes or no. It happens $\frac{|c_1|^2 + \ldots + |c_k|^2}{|c_1|^2 + \ldots + |c_N|^2}$ fraction of the time in the long run. (Assuming the states you listed were all normalized.) What state does the system jump into after this measurement? There is no experimental evidence of jumps or anything discontinuous. ...

1

Unitarity of the time-evolution operator is exactly the point: Stone's theorem (see e.g. Reed, Simon: Theorems VIII.7, VIII.8) tells us If $A$ is self-adjoint, the spectral theorem holds. This gives us a functional calculus which makes it possible to define $U(t) = e^{itA}$ in the first place. A such defined $U(t)$ is a strongly continuous unitary group. ...

7

In QM, a real valued observable $A$ is mathematically represented by a projector valued measure over $\mathbb{R}$, $P^{(A)}$, i.e., if $E$ is a Borel subset of the real line, then $P^{(A)}(E)$ is a projector representing the proposition "the outcome of measuring $A$ falls in $E$". In principle, that's all you need for representing, mathematically, ...

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