# Tag Info

7

I will try to make it as simple and intuitive as possible. In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$): $$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$ Which ...

6

If $|\phi⟩$ and $|\psi⟩$ are linearly independent, then it is always possible to assign them to the column vectors $$|\phi⟩\mapsto\begin{pmatrix}1\\0\end{pmatrix} \text{ and } |\psi⟩\mapsto\begin{pmatrix}0\\1\end{pmatrix},$$ but if they're not orthogonal you're obviously going to need to work harder on the representation of the inner product in this basis. ...

5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...

4

The equation you phrase as $$|l,m\rangle=\int_\text{all space}\psi_{lm}(r,\theta,\phi)\,\left|r,\theta,\phi\right\rangle r^2\,\mathrm dr\,\mathrm d\Omega$$ is, and must be, wrong. The reason is that $|l,m⟩$ inhabits the orbital part of Hilbert space, $\mathcal H_\Omega$, and the right-hand side is a vector in the full Hilbert space $\mathcal H$, which is the ...

4

I believe the difficulty stems from the archaic notation. What he is trying to do is show that $-\mathrm{i}\mathrm{D}$ is Hermitian, where $\mathrm{D}=\mathrm{d}/\mathrm{d}q$. Note the first equation you wrote. It is just saying that $\mathrm{D}$ acting "backwards" on the bra is equivalent to $\mathrm{D}$ acting "forwards" on the ket. He is trying to derive ...

3


1

Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?

1

Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...

1

It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...

1

$\newcommand{\ket}[1]{\lvert #1 \rangle}$Recall that the vacuum is annihilated by all annihilation operators: $$a_i\ket{\Omega} = 0$$ and that all the occupied states are created from the vacuum as $$\ket{\chi_i} = a^\dagger_i \ket{\Omega}$$ Now, if you apply an annihilation operator to a state which doesn't have the corresponding electron in it, the ...

1

You chose the $\lvert \pm \rangle$ to be an eigenvector of $S_z$ with eigenvalue $\pm\frac{1}{2}$ - that's what the $m_s$ is: The eigenvalue of the state w.r.t. the $z$-spin. Since $S_x$ and $S_y$ do not commute with $S_z$, $\lvert \pm \rangle$ is not an eigenvector of them, hence the state cannot stay the same after they are applied to it. That the spin ...

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