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Notice that $\psi(x)$ is defined on a circle of circumference $a$. Multiplying $x$ on this circle is really multiplying a periodic extension of $x$, i.e., the sawtooth function $x - a\lfloor x/a\rfloor$, where $\lfloor y\rfloor$ means the largest integer not greater than $y$. So, the commutator of the position and momentum operators involves the derivative ...

5

This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land ... 3 It's not "multiplied by$r^2$to get the probility density". The issue is that the volume element in spherical coordinates is $$\mathrm{d}V = r^2\sin(\theta)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$ and since the probability to find a particle in a subspace$X\subset \mathbb{R}^3$is $$P(X) = \int_X\lvert \psi(r)\rvert^2\mathrm{d}V$$ by definition of a ... 3 No. The$p_x$and$p_y$orbitals are always real-valued, and the complex-valued$|m|=1$orbitals are always denoted$p_{1}$and$p_{-1}$. Depending on what scheme you're using, the third$l=1$orbital can be denoted either$p_z$or$p_0$, with both notations completely equivalent. Just because the$p_x$and$p_y$orbitals are linear combinations of$p_1$... 2 Yes, you multiply each of the super-positioned states (energy eigenstates) with their time evolution as $$\Phi(x,t) = \phi_n(x)e^{i E_n t/\hbar} + \phi_{n+1}(x)e^{i E_{n+1} t/\hbar}$$ here$\phi_n(x)$is the first state and$\phi_{n+1}(x)$is the second. As for the frequency of this state, try to write it as $$\Phi(x,t) = e^{i \omega t} ... 2 Note: All products between \chi's are to be understood as tensor products. Assuming the fermions are both spin 1/2 particles, one recalls that the spin-part of the total wavefunction in the spin 1 triplet is either one or a linear combination of$$ \chi(j=1,m=1) = \chi(1/2,1/2)\chi(1/2,1/2)  \chi(j=1,0) = \frac{1}{\sqrt{2}}\big( ... 2 You have stumbled upon a difference between how chemists and physicist denote orbitals. The$p_x, p_y, p_z$notation is common in chemistry because the resulting orbitals are real. Physicist use$p_{-1}, p_0, p_1$where the subscripts are the values of m and embrace the complexity of the resulting wave function. It is purely a matter of choice since all ... 1 In general, what we use for periodic boundary conditions is defined by$Y(0)=Y(L)$and$\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition$Y(0) = Y(L)$I believe is not sufficient to impose conditions on$k\$. This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the ...

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Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2.

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So to your first question, when you calculate the time-evolution of a wavepacket, you first transform your wavepacket representation from a positional basis to a momentum basis. $$|\psi\rangle = \sum_k c_{x,p}|p\rangle$$ This moment basis is useful as they are eigenstates to your hamiltonian and satisfy $$H |p\rangle = E_p |p\rangle$$ This in turn allows ...

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