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1

You can prove this by induction. I'll drop the operator hats as they're a pain to write. First step: Suppose $\hat H = \hat q_k$. Then $[ H, p_i] = [q_k, p_i] = i\hbar \delta_{ik} = -\frac{\hbar}{i} \frac{\partial H}{\partial q_i}$ So in the special case that the Hamiltonian has this form, the claim is true! Now suppose that the claim is true for $H = ... 3 The commutators in the above expressions are sued to change the order of the Hamiltonian and annihilation or creation operators. I'll show you the first one in some detail, the second one should not give you problems afterwards. We start from$\hat{H}\hat{a}\psi_n$. Using the commutator$[\hat{H},\hat{a}] = \hat{H}\hat{a}-\hat{a}\hat{H} = ...

3

Start with your $\hat{H} = \hbar \omega \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right)$. I will omit hat notation from this point. The commutator then reads as \left[ H, a \right] = \hbar \omega \left[ \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) a - a \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) \right] = \hbar \omega ...

3

On the Wikipedia page you link to there is a derivation of the commutation relation between $\hat{a}$ and $\hat{a}^{\dagger}$, $$[\hat{a},\hat{a}^{\dagger}] = 1.$$ This directly leads to (use the relation $[AB,C]=[A,C]B+A[B,C]$) $$[\hat{a}^{\dagger}\hat{a},\hat{a}] = -\hat{a} , \qquad [\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}] = +\hat{a}^{\dagger}.$$ Up ...

1

The derivation by Sakurai is by no means mathematicaly rigorous, so you should expect something like your argument about the scalar product. Indeed, we have everything more or less fine until $$[x,\mathcal{T}(\epsilon)]|z\rangle=\epsilon|z+\epsilon\rangle$$ where we want to replace $|z+\epsilon\rangle$ by $|z\rangle$ and claim that it is ok in the first ...

2

I) The associative non-commutative Moyal/Groenewold/star product $f\star g$ is explained on Wikipedia. The corresponding $\star$-commutator is defined as $$\tag{1} [f\stackrel{\star}{,} g]~:=~f\star g-g\star f.$$ In particular, the Jacobi identity for the $\star$-commutator is a consequence of the associativity of the $\star$-product. II) On one hand ...

1

Here's the most logical way to proceed if you ask me. Given any $a\in\mathbb R$, we define the translation operator $T_a$ by its action on position basis vectors $$T_a|x\rangle = |x + a\rangle$$ One can prove the following properties: $T_a$ is unitary for each $a\in\mathbb R$. $T_aT_b = T_{a+b}$ for all $a,b\in\mathbb R$. It follows (by Stone's ...

1

Main point: You should allow the possibility of sign factors appearing into the definition of the Hilbert space representation of fermionic operators, cf. fermionic Fock space. In more detail, consider the CAR algebra $$\tag{1} \{c_{\sigma}, c_{\tau}\}~=~0, \qquad \{c_{\sigma}, c^{\dagger}_{\tau}\}~=~\hbar {\bf 1}, \qquad\{c^{\dagger}_{\sigma}, ... 2 I think you are right. Using really simple commutator math. All you need is this:$$ [AB,C] = A[B,C] + [A,C]B $$Then in your case:$$ A=PB=XPC=P [PXP,P] = PX [P,P] + [P,P]XP = PX[P,P] + P[X,P]P + [P,P] XP  As you said, [P,P] is antisymmetric to itself, and therefore we can remove all the [p,p] terms. We then have left only one term: ...

2

You teacher seems to have made a mistake. I imagine that he/she did something like this: \begin{align} [PXP, P] &= P[XP,P]+[PX,P]P \\ &= P(X[P,P]+[X,P]P)+(P[X,P]+[P,P]X)P \\ &= P[X,P]P+P[X,P]P \\ &= 2i\hbar P^2 \end{align} Notice that the first equality is wrong. You can't peel operators off to the left and right if there are three ...

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