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We usually say that if two operators, $\hat{A}$ and $\hat{B}$ commute, then they have a simultaneous set of eigenstates. Saying that the eigenstates are the same isn't really correct. For example, let operator $\hat{A}$ be hermitian and act on elements of the Hilbert Space $\mathcal{H}_A$ and let operator $\hat{B}$ also be hermitian and act on elements ...

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Assumptions: I will be talking about Hermitian (more generally self-adjoint) operators only. This means that I will assume that the operators in question have a set of eigenvectors that span the Hilbert space. As mentioned by tomasz in a comment, this is not exactly necessary, since more general statements can be made, but since we are dealing with basic QM, ...

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I almost voted to close your question as a duplicate of How do you rotate spin of an electron?. This would be controversial because the question looks related at a first glance, but ACuriousMind's answer to that question also (indirectly) answers this question. When we talk about rotating an electron, or any fermion, we are not talking about a physical ...

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If you are sending some light to a surface there are two ways to send more energy. First, send more photons, each with the same energy. This increases the intensity, but keeps the frequency (and thus the energy per photon) fixed. Second, increase the energy per photon. This requires increasing the frequency. You could do neither (and thus not increase the ...

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Essentially this question boils down to "why does the energy of light depend on frequency? The answer is:This is what has been observed, measured, for photons. Light is composed by zillions of photons and the energy of the light wave is dependent on the amplitude of the classical electromagnetic wave. This means many more photons are needed for low ...

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The energy $E$ of a photon is directly related to its frequency $f$ via $E = h f$. This relation is a fact of Nature and is described by the Schrodinger equation for the time evolution of a quantum thing $|\Psi\rangle$: $$H |\Psi(t)\rangle = i \hbar \frac{d}{dt}|\Psi(t)\rangle \, .$$ If $|\Psi\rangle$ happens to have a definite energy, then $H|\Psi(t)\rangle ... 1 It has been shown experimentally that the formation of H2 can happen in the presence of free electrons: so a photon is emitted in this two step process, taking energy away . Note that this is for low densities. The three body process in the answer by John dominates with increasing density, as discussed in the link. 1 Your orbitals are not$sp^2$hybridized orbital. They are just real spherical harmonics$s, p_x, p_y, p_z$. Their expectation value of$ \vec{r}$is zero, because they are symmetrical function. The three$sp^2$hybridized orbitals are$ \phi_0 = \frac{1}{\sqrt 3 }s - \frac{1}{\sqrt 6 }p_x + \frac{1}{\sqrt 2 }p_y\phi_1 =\frac{1}{\sqrt 3 }s - ...

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Two isolated hydrogen atoms cannot form an $H_2$ molecule for the simple reason that they have too much energy. Any system formed from the two atoms will have an energy greater than the dissociation energy of $H_2$ so no bound state will be formed. Observation tells us that the process must happen because there is a lot of $H_2$ around. It happens when ...

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