# Tag Info

7

Groenewold is working in the framework of deformation quantization, where the (reduced) Planck constant $\hbar$ is treated as a formal parameter that doesn't have to be the actual physical value $\sim 10^{-34}{\rm Js}$. Eq. (1.30) is explained by an unconventional normalization of the commutator $$[{\bf a},{\bf b}]~:=~\frac{i}{\hbar}({\bf ab}-{\bf ba}). \... 7 How did he assume that \frac{\hbar}{i}=1? He didn't. Check the definition he gives of the commutator in equation (1.02). And if \hbar (as we have learned it) is a constant how can we say that it goes to zero? I think the point here is to say: if \hbar \rightarrow 0 we recover the classical mechanics (CS), therefore if in nature \hbar = 0 we ... 5 \omega is presumably a bilinear form on \mathbb{R}^n, cf. the Heisenberg group. In other words, the comma separates the two arguments x,y\in\mathbb{R}^n of \omega. 4 As it has been already pointed by others: The Hamiltonian \hat{H} = k\hat{x}\hat{p} is not Hermitian, i.e. it does not correspond to anything measurable. One can obtain Hermitian Hamiltonians by using either symmetrizing or anti-symmetrizing the operator product \hat{x}\hat{p}:$$\hat{H}_+ = \frac{k}{2}(\hat{x}\hat{p} + \hat{p}\hat{x}),\\ \hat{H}_- = \...

3

A good guess to start would be to identify classical quantities that are conserved. This is because, for most cases, \begin{align} [\hat H,\hat {\cal O}]=i\hbar\{H,{\cal O}\}_{PB} \end{align} where the operation on the right hand side is the classical Poisson bracket. The Poisson bracket is $0$ when the quantity is conserved, so in a central field, (in ...

2

Yes. For example, when calculating angular momentum of a hydrogen atom (ignoring fine structure or hyperfine structure), the eigenbases (which are described by spherical harmonics) are simultaneously eigenbases of the Hamiltonian operator $H$, the total angular momentum operator $L^2$, and the z-component angular momentum operator $L_z$. The proof can be ...

2

The meaning of conjugate variables is usually that they are related by the Fourier transform (other transforms are possible, but I am not sure they are interesting). In quantum mechanics, this is simply a change of basis in Hilbert space. The relation between energy and time is similar, but time is a parameter in quantum mechanics, not an observable (this is ...

2

I agree with the answer by Dvij D.C. Let me add, however, that the answer to your question is trivially yes from a purely mathematical point of view. Consider, for example, the degenerate operator $A$: $$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$$ (i.e....

2

My question is, can you always find an operator that commutes with a degenerate operator to form of a complete set and 'lift the degeneracy'. I think this is exactly the opposite of the right way to think about this. How would you know that the spectrum of an observable is degenerate? Let's say we want to see if the spectrum of an operator $\hat{A}$ is ...

2

I don't think Sakurai is talking about the domains of the operators, because claims about their commutator wouldn't help us determine if they share the same domain or not. However, assuming they share the same domain, the eigenstates of each operator would span the domain because observables are Hermitian operators and the eigenstates of a Hermitian ...

2

I'm not sure this is the right answer, but one problem that I can immediately see is that if you had a Hamiltonian $$\hat{H} = k \hat{p}\hat{x},$$ then it wouldn't be Hermitian, since you could show that $$\hat{H}^\dagger = k^* \hat{x}\hat{p},$$ which, since $\hat{x}$ and $\hat{p}$ don't commute, is not equal to $\hat{H}$. Thus, $\hat{H} \neq \hat{H}^\dagger$...

1

For a slightly different perspective, in natural units one can set $\hbar = 1$. That is, in natural units we agree to measure action in units of $\hbar$ (instead of, say, $\rm J\cdot s$). Seen this way, it makes no more sense to send $\hbar$ to $0$ than it does to send $1 \, \rm J \cdot s$ to $0$. Put differently, sending $\hbar$ to $0$ is like sending $1 \... 1 Let me answer the first question which is presently the title, and maybe experienced users will direct you to existing resources on Stackexchange to help you sharpen the other questions. In general, the brute force way to know if two operators commute is to identify a suitable basis for the Hilbert space: a set of states that satisfy completeness. Then ... 1 I assume you're referring to the line(s) which say $$\phi_1^+N(\phi_2\cdots\phi_m) = N(\phi_2\cdots\phi_m)\phi_1^+ + [\phi_1^+,N(\phi_2\cdots\phi_m)]$$ $$=N(\phi_1^+\phi_2\cdots\phi_m) + N\big([\phi_1^+,\phi_2^-]\phi_3\cdots \phi_m + \phi _2[\phi_1^+,\phi_3^-]\phi_4\cdots\phi_m + \cdots\big)$$ The implication here is that$N(\phi_2\cdots\phi_m)\phi_1^+=N(\...

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