Title: ETH_Sem3 / Physik_III / Woche_10

Finite square well:

We solve for the three regions seperately.
By considering symmetry and normalizability we can reduce the complexity.
We find two types of solutions:
Bound states, which are localized near the well
Unbound states, which are just a combination of free particle wavefunctions.
We see that the bound states have discrete energy eigenvalues and unbound states have continuous energy values

Example potential step:

Again we solve the two regions seperately:
Region I:
\(\psi_E(x) = A_I e^{ikx} + B_Ie^{-ikx}\)  (\(A_1\) traveling from left to right and \(B_1\) the opposite)
Region II:
\(\psi_E(x) = A_{II} e^{\sqrt{k_0^2-k^2}x}+ B_{II}e^{-\sqrt{k^2_0-k^2}x}\)
with
\(k_0 = \sqrt{\frac{2mV_0}{\hbar^2} }, k = \sqrt{\frac{2mE}{\hbar^2} }\)
For \(E>V_0\) we get \(k_0^2-k^2 < 0\).
Which results in a complex exponential in Region II and thus a moving particle.
By comparing \(A_I,B_I,A_{II},B_{II}\) we can define transmission and reflection coefficients.

Note that no matter the heigth of the step there will always be some reflection

For \(E < V_0 \) we get a real exponential in Region II, which causes an exponentialy decaying probability of finding the particle on top of the potential.

Trick 1 to solving the harmonic oscilator in QM:
We start with \(\hat{H} = \frac{\hbar \omega}{2}(\hat{X}^2 + \hat{P}^2)\)
and "quadratically expand" but we have to consider the commutator.
\(\hat{H} = \frac{\hbar\omega}{2}((\hat{X}-i\hat{P})(\hat{X}+i\hat{P}) - i[\hat{X},\hat{P}])\)
\(= \hbar\omega(\hat{a^\dagger}\hat{a} + \frac{1}{2})\)
with \(\hat{a} = \frac{1}{\sqrt{2} }(\hat{X}+i\hat{P}), \hat{a}^\dagger = \frac{1}{\sqrt{2} }(\hat{X}-i\hat{P})\)

Trick 2 to solve the harmonic oscillator in QM:
Starting from \(\hat{H}= \hbar\omega(\hat{a}^\dagger\hat{a} + \frac{1}{2})\)
Assuming we have an eigenstate \(\ket{n}\) with
\(\hat{a}^\dagger\hat{a} \ket{n} = n\ket{n}\) and \(E_n = \hbar\omega(n + \frac{1}{2})\)
Note that \(n\) can be any number (even complex)
consider
\(\hat{a}\hat{a}^\dagger\hat{a}\ket{n} = \hat{a}n\ket{n} =(\hat{a}^\dagger\hat{a} + 1)\hat{a}\ket{n}\)
Plugging in our assumption
\(\implies \hat{a}^\dagger\hat{a}(\hat{a}\ket{n}) = (n-1)(\hat{a}\ket{n})\)
The two solution to this are
- \(n=0\) or
- \(\hat{a}\ket{n}\) is proportional to an eigenstate of \(\hat{a}^\dagger\hat{a}\) with eigenvalue \(n-1\) \(\ket{n-1} = C_{n-}\hat{a}\ket{n}\)
We now know that \(n\) are actually \(\mathbb{N}_{\geq0}\) and not any eigenvalues.

Ladder operator
in position representation
\(\hat{a} = \frac{1}{\sqrt{2\hbar m\omega} }(\hbar\dede{}{x} + m\omega x)\)
with dimensionless quantities
\(\hat{a} = \frac{1}{\sqrt{2} }(\hat{X} + i\hat{P})\)

Q: [...]
[...]
A: Ground state of the QM harmonic oscilator in position representation:; \(\psi_0(x) = (\frac{m\omega}{\pi\hbar})^\frac{1}{4} e^{-\frac{m\omega}{2\hbar}x^2}\)

Higher energy states of the QM harmonic oscilator:
Recursion:
\(\ket{n} = \frac{(\hat{a}^\dagger)^n\ket{0} }{\sqrt{n!} }\)
In general:
\(\psi_n(X) = (\frac{m\omega}{\pi\hbar})^{1/4}\frac{1}{\sqrt{2^nn!} }H_n(x)e^{-x^2/2}\) with \(X =\)\(\sqrt{\frac{m\omega}{\hbar} }x\) and \(H_n(X) \) the Hermite polynomials.
Visually:

Q: Position and momentum operators in terms of the ladder operators:
[...]
A: \(\hat{X} = \sqrt{\frac{\hbar}{2m\omega} }(\hat{a}^\dagger+\hat{a}), \hat{p} = \sqrt{\frac{\hbar m\omega}{2} }(\hat{a}^\dagger-\hat{a})\)

Expectation values and uncertainties of p,x in the QM harmonic oscilator:
\(\braket{n|\hat{p}|n}=\) \(0\)
\(\braket{n|\hat{X}|n} =\) \(0\)
\(\braket{n|\hat{X}^2|n} = \)\(\frac{\hbar}{2m\omega}(2n+1)\)
\(\braket{n|\hat{p}^2|n} = \)\(\frac{\hbar m\omega}{2}(2n+1)\)
\(\Delta x = \) \(\sqrt{\braket{\hat{x}^2} - \braket{\hat{x} }^2} = \)\(\sqrt{\frac{\hbar}{2m\omega} }\sqrt{2n+1}\)
\(\Delta p = \)\(\sqrt{\braket{\hat{p}^2} - \braket{\hat{p} }^2} = \) \(\sqrt{\frac{\hbar m\omega}{2} }\sqrt{2n+1} \)
Note that the ground state has non-zero energy and non-zero uncertainty in p and x:
We call these zero-point energy and zero-point position/momentum
Back Extra: We also note that \(\Delta x \Delta p = \frac{\hbar}{2}(2n+1)\), which satisfies the Heisenberg uncertainty

Quantum states that exhibit classial behaviour are called coherent states and are a superposition of the energy eigenstates.
For the classical harmonic oscilator those states are:
\(\ket{\alpha} = e^{-|a|^2/2}\sum_n^\infty\frac{\alpha^n}{\sqrt{n!} }\ket{n}\)
We observe that the expected value for \(\hat{x}\) follows:
\(\braket{\alpha|\hat{x}|\alpha} = \sqrt{\frac{2\hbar}{m\omega} }|\alpha|cos(\omega t + \phi)\)
Which is a classial harmonic oscilator.
Furthermore this state minimizes Heisenberg uncertainty in p and x for all time

Q: Schrödinger Equation in 3D:
[...]
A: \(i\hbar \dede{\Psi}{t} = -\frac{\hbar^2}{2M}\nabla^2\Psi + V\Psi\)

Q: All (simple) operators  for different Carthesian coordinates [...]
A: commute

Q: The trick to solving the SE in 3D is [...] and [...]
A: writing the SE in spherical coordinates; separating the different angle and radial components and solving seperately.

Q: [...]
A: Spherical harmonics \(Y_l^m(\theta,\phi)\)