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I think you are probably misinterpreting the context here. If you read the previous line carefully it says "there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. And later he just says due to the fact that photon can be scattered within the 2θ' angle ...


2

As everyone points out commutation is not a shorthand for equivalence, as, given your relations, the Jacobi identity, [A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0 dictates that when the first two terms vanish, the third must too, so that B must commute with [C,A], non vanishing in general, as remarked repeatedly. Lie algebra commutators do, nevertheless, parameterize ...


1

I'm sorry I first misinterpreted your braces as anticommutators and not as set inclusions! First absorb the constants into the normalizations of H and B so the (1,1) entry of each is now 1. The common eigenvectors of them both are then the transposes of (1,0,0), (0,1,1), and (0,1,-1); the eigenvalues of each are, under H: 1,-1, -1 ; and under B: 1, 1, -1, ...


1

Operators form a complete set if specifying eigenvalues of all of them already determines a state uniquely (up to normalization and phase). Since they commute, they will have simultaneous eigenspaces. Eigenstate is uniquely determined if all these eigenspaces are one dimensional.


1

You are free to choose which axis you make a measurement on and doing so will always yield and eigenvalue of the spin operator in that direction: you will always measure $\pm\hbar/2$ in whichever direction you choose to measure. The reason is that the state vector of the particle exists in a superposition of states with various probabilites $$ |\psi\rangle ...


0

Assuming $(X + zY)^*$ is the conjugate of $X + zY$, a counter-example is: $$z=0, Y=\begin{bmatrix} 0&0\\0&0 \end{bmatrix}, X=\begin{bmatrix} 0&-i\\i&0 \end{bmatrix}, \rho = \begin{bmatrix} 1/2&0\\0&1/2 \end{bmatrix}$$ Because: $tr[\rho(X+zY)^*(X+zY)]$ $= \text{Tr}\left[I/2\left(\begin{bmatrix} 0&-i\\i&0 ...


2

Yes it is. Use the fact that the operator in the trace different from $\rho $ is positive and compute the trace using a basis of eigenvectors of $\rho $, whose eigenvalues are also positive. (By positive I actually mean nonnegative). Here is the proof (assuming the first version of the question regarding $tr(\rho (X+zY)^\dagger(X+zY))$). As $\rho$ is ...


5

Your question has no answer. There is no "why". It is a postulate of quantum mechanics that a pure state vector contains all information you can possibly know about a physical state. Note that this is not the same as saying it contains all information you can possibly imagine having about a physical state, particularly since your (and my) imagination is ...


0

Technically one cannot say observables are probabilistic, since they are mathematically described by deterministic operators. Now when an observable has different eigenvalues, then the Born rule is used to predict which value the experiment will get, and this is where probabilities arise. The Born rule is a postulate of Quantum Mechanics, historically ...


7

This is a supplement to freude's correct answer: Hamiltonian is the infinitesimal generator of time translation defined as $$\mathrm{\hat{U}}(\mathrm dt)= 1- \frac{i}{\hbar} \mathrm{\hat{H}}(t)\mathrm dt\;.$$ Time-Evolution Operator: Let the system be at $|\phi\rangle\;.$ Now, let's wait for some time..... What is the probability amplitude of finding ...


6

This is a scalar value that is a projection of the state $H|\psi \rangle$ on the state $|\phi \rangle$. The state $H|\psi \rangle$ results from the action of the operator $H$ on the state $|\psi \rangle$. If the state $|\psi \rangle$ is an eigenstate of the operator $H$, the expression can be rewritten as $E \langle\phi|\psi \rangle$. If the state $|\phi ...


3

Actually, the outcome of the experimental apparatus is an interval $(x_0-\delta, x_0+\delta)$. $\delta>0$ stays for the accuracy of the instrument which can be made smaller and smaller but cannot be removed. It is therefore assumed (Luders-von Neumann's axiom) that, if the state immediately before the measurement was determined by the wavefunction ...



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