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1

How to prove it? EDIT: Consider a state $|\psi_n\rangle$, which is an eigenstate of the Hamiltonian $H$ having eigenvalue $E_n$. Then consider the action of $H$ on these two other states: $$ |\alpha\rangle \equiv AB|\psi_n\rangle $$ and $$ |\beta\rangle \equiv BA|\psi_n\rangle $$ EDIT: Ask yourself: Are these eigenstate of the Hamiltonian? If so, ...


0

Agreed, I don't see what is so troubling. If two operators don't commute, so be it, there's nothing you can do about that. It may make the problem at hand much more difficult however. For example, in path integral formulations of QM we encounter operators of the form $e^{\hat{T}+\hat{V}}$, which is not equal to $e^{\hat{T}}e^{\hat{V}}$ if ...


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your result is correct $$ [a_k, a_q] = -2 a_k a_q $$ which is consistent with $$ [a_k, a_k ]= - 2 a_k a_k = 0 $$ because $$ a_k a_k = \frac{1}{2}\{a_k, a_k \} = 0 $$ And in general you can use $$ [A,B] = 2AB - \{A,B\}$$ which would also give $$[a_k^\dagger, a_q] = 2a_k^\dagger a_q - \delta_{kq}$$


4

Suppose the space-time group includes dilatations which expand or contract space. Points in space $x^{i}\in V_{3}$ transform under a small dilatation $\epsilon$ near the identity as, \begin{equation} x'^{i}=x^{i}+\epsilon x^{i} \ . \end{equation} The change in the coords is, \begin{equation} \frac{d x^{i}}{d\epsilon}=x^{i} \end{equation} In the Hamiltonian ...


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Does this mean that the operator $\hat O$ (an observable) is special in some way? I believe it means there is no such $\hat O$. If $\hat O$ corresponds to an observable, we require the eigenvalues to be real. Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$: $$\hat O |o\rangle = o |o\rangle$$ Now consider the following $$\hat O ...


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$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 | #2 \right>}$ $\newcommand{bok}[3]{\left< #1| #2 |#3\right>}$ It means basically that all of the energy eigenstates has zero energy eigenvalue. Ups... Let $\left| \psi \right>$ be a normalized energy eigenstate with energy ...


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\begin{align*} -i\gamma^5 \gamma^i \gamma^j \partial_j &= \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^i \gamma^j \partial_j \\ &= \tfrac{1}{3!}\gamma^0 \epsilon_{klm}\gamma^k\gamma^l\gamma^m\gamma^i\gamma^j\partial_j\\ &= \tfrac{1}{2}\gamma^0 \epsilon_{kli}\gamma^k\gamma^l\gamma^j\partial_j\\ &= \gamma^0 \epsilon_{kji}\gamma^k\partial_j\\ ...


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I think I'm essentially supposed to show... Why do you think this? Is this a homework question? If so then it should be tagged as such. If this is not a homework question and you are interested to read an explanation involving no explicit equations then consider the following: The operator $\vec L$ is the generator of rotations. Therefore, any ...


4

In this case $p_0$ is just a number. It's the amount of phase that you add to the wave function. $\text{e}^{ip_0 x/\hbar}$ is not a translation operator. It's just multiplying the wave function by a complex number of norm $1$. In you previous question you had $\hat{p}$ which is an operator. It acts on the wave function by differentiating it.


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Yes, that's what a commutator is. Remember that the momentum operator is $\vec p = -i\hbar\vec\nabla$, so $\nabla^2 = -p^2/\hbar^2$; also that $\nabla^2$ is a scalar, not a vector. Usually when you find a commutator you need a "test function" for the operators to operate on; that's how you find for instance that $[\hat x, \hat p_x]=+i\hbar$.


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You made a mistake somewhere. My approach was to write $xp=i\hbar +px$ and $px=xp-i\hbar$ on the side of my paper and just substitute away. I first wrote $$\{x^2,p^2\}-\frac{(\{x,p\})^2}{2}=x^2p^2+p^2x^2-\frac{xpxp+xp^2x+px^2p+pxpx}{2}$$ The terms in the fraction are $$xpxp=x^2p^2-i\hbar xp$$ $$xp^2x=x^2p^2-2i\hbar xp$$ $$px^2p=x^2p^2-2i\hbar xp$$ ...


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I'm digging this thread out just to clarify some things for those who might have a similar question. Summary We cannot use $\mathcal T$. Space-like four-vectors are essentially like $(0,x,y,z)$, so we can ignore the time and do three-dimensional rotations to get $(0,-x,-y,-z)=-(0,x,y,z)$. A la Valter Moretti As Valter Moretti already pointed out, you ...



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