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2

I believe in the last line, the plane-wave functions $u_k(x)$ should carry different coordinates and momenta, e.g $$[a(k)^\dagger,a(k')]u_k(x)u_{k'}(x')$$ You may note that the commutator $[\phi(x),\pi(x')]=i\hbar\delta(x-x')$ holds if one choses $[a_k,a_{k'}^\dagger]=\delta_{kk'}$. However, this indirect reasoning is no proof that this choice is unique. ...

1

I didn't read your answer, but let's think about just computing the operator $\partial_x^2 f$. First we need to compute the operator $\partial_x f$. Now I am saying "the operator" because we are viewing $\partial_x f$ as a composition of first multiplying by $f$ and then taking the derivative. By the product rule, we know $\partial_x f = (\partial_x f) + f ... 0 I think I got answer myself. The vanish commutator and EPR paradox are not correlated. The vanishing commutator simply says, once a measurement at$x$was done, the obtained state will not bring uncertainty for measurement of$y$, by the simutaneous eigenstate property. Not like one measures momentum, the state becomes$|p \rangle$, then measure position ... 1$[q_r q_s p_r , q_s p_r q_s]=q_r [q_s p_r , q_s p_r q_s ] + [q_r q_s , q_s p_r q_s]p_r=q_r q_s [p_r , q_s p_r q_s ] + q_r[q_s , q_s p_r q_s]p_r + q_r [q_s , q_s p_r q_s ]p_r + [q_r , q_s p_r q_s]q_s p_r$Now only the last term is non-vanish because the others three have inside the commutators only operators that commutes one to each other. So:$[q_r q_s ...

1

Don't forget the bosonic creation/annihilation operators (harmonic oscillator operators) $[a,a^{\dagger}] = 1$ and the fermion counterparts $\{ c, c^{\dagger} \} \equiv cc^{\dagger}+c^{\dagger} c = 1$

3

I also know that L and S commute, but I am unsure why. I've heard that it is simply because they act on difference variables, but I don't understand exactly what this means. Is there a way to show this explicitly? Suppose we have two Hilbert spaces $H_1$ and $H_2$, an operator $A_1$ acting on $H_1$, and an operator $A_2$ acting on $H_2$. Let $H = H_1 ... -2 When two qm operators do not commute, it means that we are missing stuff in Nature. That is quantum mechanics is a theory of measurement but not of Nature because of non-commutation. Hence this means that the stuff we miss cannot be described by quantum mechanics, and this leads to the conclusion that qm is not a complete description of Nature. 1 To summarize, assume hypothetically we managed to find a way in the future where we can have a look at an electron without disturbing it by measurement or causing its wave function to collapse, would the uncertainty principle still hold in such a case?? Why/Why not? To start with any measurement when looked at the quantum mechanical level involves ... 1 But in practice, is there any measurement that will NOT disturb the system at all? To prove that uncertainty is beyond measurement, we must design a measurement process that does not disturb the system. If such a process cannot be designed then the statement that "uncertainty is beyond measurement" cannot be experimentally tested. Isn't it? I don't know ... 6 I write below the statement of the mentioned theorem which assumes, as hypotheses, the validity of so-called "Wightman axioms" in the four dimensional Minkowski spacetime. You see that there is nothing imposed by hand. It is actually a no-go theorem. Quantizing free fields, it establishes in particular that the standard choice is the only possible. ... 4 Hint:$pq$-order$^1$your last expression $$2(p^2q^2+q^2p^2)-(pq+qp)^2.$$$pq$-ordering means commuting all$p$'s to the left and all the$q$'s to the right by using$^2$the CCR formula$qp = pq +i\hbar{\bf 1}$, possibly repeatedly. (There are shorter ways, but$pq$-ordering is at least a systematic approach.) What remains will be a$c\$-number. In fact, the ...

0

It seems to me that the approach formulated by the OP is perfectly sound! The OP demonstrates that he is familiar with the commutator between q^n and p. In order to use this formula, he proposes to apply Taylor series expansion to the target function. That is okay. The only problem is that the OP got cold feet and stopped his calculation at this point. ...

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