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8

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 ...


1

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.


1

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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