Quantum-Mechanics-Notes
Notes for quantum mechanics of MIT-OCW 8.04.
’This is the goal of 8.04: we will step beyond the scale of daily experiences to develop an intuition for electrons, atoms, and superposition.’
1 Uncertainty Principle and Superposition
Think about a series of experiments.
The electron has two properties: white/black and hard/soft (Denoted by W/B & H/S).
We made boxes that can measure such properties. For instance, white electrons and black electrons will exit the box from different directions. From the direction, we know the white/black property of the electron. Denote ‘C-Box’ (color box) for the boxes that can measure the W/B property and, similarly, ‘H-Box’ (hardness box) for those that can measure the ‘H/S’ property.
We have mirrors and combinators that combine the two directions so we can continue more complicated experiments. Note that they only change the directions.
Below are the experiments from physicists.
First, there are some basic and naive experiments.
- Electrons —> C-Box —> W and B —> choose W => C-Box —> W (repeatability of color / hardness)
- Electrons —> H-Box —> H and S —> choose H =>C-Box —> W (50%) and B (50%) (lack of correlation of color with hardness)
Then comes the weird experiment.
- Electrons —> C-Box —> W and B —> choose W —> H-Box —> H and S —> choose H => C-Box —> Not (100%) W, but $50%$ W and $50%$ B
Apparently, the hardness box tampers with color. Therefore, experiment 3 tells us color and hardness properties cannot be measured simultaneously, which is the uncertainty principle.
Though we know it is white initially, as long as we measure the hardness, the white color property is not repeatable anymore.
Then, we use the combinator to dig deeper.
White electrons —> H-Box —> 50% H and 50% S —> combinator => H-Box —> 50% H and 50% S
White electrons —> H-Box —> 50% H and 50% S —> combinator => C-Box —> (100%) W
White electrons —> H-Box —> 50% H and 50% S —> put a wall in the S path —> combinator => C-Box —> 50% W and 50% B
Experiment 4 tells us combinator does not change the properties of electrons. Experiment 5 got a total W because the added combination part acts as if the H/S property isn’t measured; the W/B property will not be disturbed. Experiment 6 shows a disturbing one from which we can see the uncertainty principle plays a role.
After we understand the principle of the motion of electrons, we consider its being (or state): what is the state of an electron inside the apparatus in Experiment 5?
We know it is white, so it cannot be hard or soft. Otherwise, we will get the half-half result by the uncertainty principle. It cannot be both hard and soft since different properties will go in different directions. In one path, the electron cannot have both properties. It cannot be either since we can get some results.
Therefore, the only condition is the electron has a probability of being hard or soft (but we cannot know what exactly it is), and such a state is called superposition.
Being in a definite state of position means being in a superposition of having different values for momentum.
An inspiring example of this using polarization states:
2 Bell’s Inequality
In lecture 2, the professor demonstrated some exciting experiments that are waiting to be explained using quantum language:
Atom exists. But atomic orbits are unstable.
Atomic spectra are discrete. When excites/sparks the simple gas like $ H_2$, we get light sent out by it. If we shine the emitted light on a prism, we will find distinct patterns that indicate discrete spectra. Experimentalists found the wavelengths satisfy $1/\lambda=R\cdot(1/n^2-1/m^2)$.
Photoelectric effect. Stop voltage has no relation with intensity but is linear to frequency. It was found that the coefficient was just $h$, the Planck constant derived from Blackbody radiation.
Since the photon has zero mass, we can develop the momentum equation $p=h/\lambda$ with $E=pc$ and $E=h\nu$. This means the photon has the property of particles—momentum. Note that $E=h\nu$ is very unusual.
- By the way, I think a photon is massless because it is our world’s measurement datum.
Double-slit experiment with electron. The electrons came one in one and were found randomly positioned. But when the quantity is enough, the statistics tell us the electrons were interfered (with themselves).
Bragg scattering. We would find the interference if we send a beam of electrons to a crystal and watch out for the scattering from different angles. The intensity as a function of the angle of the out beam satisfies the function as if the electron is a wave with the wavelength $\lambda=2d\sin\theta/n$. The electron diffracts off the crystal.
Send out $C_{60}$ one by one through a diffraction grating, the result is also an interfered pattern, telling us that it interferes with itself.
Next comes the Bell’s inequality.
If the state of the stuff is definite, let $A,B,C$ be the set of condition $1,2,3$, and a bar means its complementary set. We have:
$$
\operatorname{Num}(A,\bar B)+\operatorname{Num}(B,\bar C)\ge \operatorname{Num}(A,\bar C)
$$
The proof is the following:
$$
\begin{aligned}
\operatorname{LHS}&=\operatorname{Num}(A,\bar B,C)+\operatorname{Num}(A,\bar B,\bar C)+\operatorname{Num}(A,B,\bar C)+\operatorname{Num}(\bar A,B,\bar C)\\
&\ge \operatorname{Num}(A,B,\bar C)+\operatorname{Num}(A,\bar B,\bar C)
=\operatorname{Num}(A,\bar C)
=\operatorname{RHS}
\end{aligned}
$$
However, Bell found by experiment that:
$$
\operatorname{Num}(\uparrow_0,\downarrow_\theta)+\operatorname{Num}(\uparrow_\theta,\downarrow_{2\theta})<\operatorname{Num}(\uparrow_0,\downarrow_{2\theta}).
\
$$
The up and down arrows indicate the electron’s spin angular momentum.
- I think this can be explained only if the three binary states are superimposed.
3 The Wave Function
Postulate of Wavefunction
An essential postulate: The configuration or state of a quantum object is completely specified by a wavefunction denoted as $\psi(x)$, $\psi\in\mathbb{C}$ .
$$
\text{Classical: } (\vec p,\vec x)\to \text{Quantum: }\psi(\vec x)
$$
Furthermore, $p(x)=|\psi(x)|^2$ determines the probability (density) that an object in the state $\psi(x)$ will be found at the position $x$. Note that the dimension of $\psi$ is the square root of 1/length since $\displaystyle \int|\psi|^2 \text{d}x=1$.
De Broglie Relations
From de Broglie’s generalization of the momentum-wavelength relation of photons, we have
$$
E=\hbar\omega, \
\vec p=\hbar \vec k
$$
where $\hbar:=h/2\pi$.
Mathematics of Superposition
Given to two possible states of a quantum system corresponding to two wavelengths $\psi_a$ and $\psi_b$, the system could also be in a superposition $\psi=\alpha\psi_a+\beta\psi_b$ with $\alpha$ and $\beta$ in $\mathbb{C}$ satisfying normalization.
Hence, for the superposition state, we have the probability density
$$
\begin{aligned}
\mathbb{p}(x)&=|\alpha\psi_a(x)+\beta\psi_b(x)|^2\\
&=|\alpha\psi_a(x)|^2+|\beta\psi_b(x)|^2+\alpha^\dagger\beta\psi_a^\dagger(x)\psi_b(x)+\alpha\beta^\dagger\psi_a(x)\psi_b^\dagger(x)
\end{aligned}
$$
which exhibits quantum interference aside from the usual addition probability. It’s the interference between the different momentum that leads to certainty in the position.
By using Fourier Transform, we see that any wavefunction $\psi(x)$ can be expressed as a superposition of states $e^{ikx}$ with definite momenta $p=k$ as
$$
\begin{aligned}
\psi(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \widetilde \psi(k)e^{ikx}\text{d}k\\
\widetilde\psi(k)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \psi(x)e^{-ikx}\text{d}x.
\end{aligned}
$$
- I think that means when state is expressed in wavefunction, position and momentum are tightly correlated, and this relationship is the root of the uncertainty.
4 Expectations
Uncertainty and Expectation
To characterize the uncertainty, we use the standard deviation
$$
\Delta a=\sqrt{\langle a^2\rangle-\langle a\rangle^2}
$$
The expectation value of a function of position in a given quantum state $\psi(x)$ is
$$
\langle f(x)\rangle =\int_{-\infty}^\infty \psi^\star(x) f(x)\psi(x);\text dx.
$$
However, when we consider the expectation of momentum $\langle p\rangle$, we get problem: $p$ and $x$ cannot be measured simultaneously.
(In fact we replace $f$ with $\hat f$ when compute the expectation value of an operator)
Operators
Notice some hints that indicate us to operator:
For wavefunction $\psi=e^{ikx}$,
$$
\begin{aligned}
\partial_x \psi=ik\psi\implies \frac{\hbar}{i}\partial_x \psi =\hbar k\psi=p\cdot\psi
\end{aligned}
$$Noether’s theorem states that to every symmetry is associated a conserved quantity .
Symmetry Conservation Translation : $\vec x\to \vec x+\Delta \vec x$ Momentum : $\dot{\vec p}=0$ Time : $t\to t+\Delta t$ Energy : $\dot E=0$ Rotation : $\vec x \to \overleftrightarrow R\cdot \vec x$ Angular Momentum : $\dot{\vec L}=0$ For translations behave for functions,
$$
\begin{align}
f(x)\to f(x+l)&=f(x)+l\partial_x f(x)+\frac{l^2}{2}\partial_x^2 f(x)+\dots\\
&=\sum_{n=0}^{\infty} \frac{(l\partial_x)^n}{n!}f(x)\\
&=e^{l\partial_x}f(x)
\end{align}
$$
Translations are generated by spatial derivatives $\partial_x$ and it should be associated with $p$.Similarly, $E\sim \partial_t,; L_z\sim\partial_\varphi$.
Finally, momentum in quantum mechanics is realized by an operator:
$$
\hat p=\frac{\hbar}{i}\frac{\partial}{\partial x}.
$$
- The hat above $p$ indicates it’s not a number, but a rule that operates on functions
And we use operator $\hat p $ to compute expectation values:
$$
\langle p\rangle=\int_{-\infty}^\infty \psi^\star(x)\hat p \psi(x);\text dx
$$
I think operator $\hat p$ is introduced more intuitively than mathematically. The key is that we want to calculate expectation of $p$ in a state associated with position $x$, but from uncertainty principle there is no meaning when $p$ and $x$ occured in one expression. Since momentum is related to position, we come out with operator. Then the $\langle p\rangle$ is the expectation of the momentum that can be observed. (Only the momentum that is able to be observed can be taken out from the operator)
The very interesting thing of momentum operator is it can get distribution of momentum when there is a superposition.
- BTW: At end of the lecture, the wishes from pirates are wonderful!
5 Operators and the Schrödinger Equation
Common operator
For each observable quantity we have an associate operator.
- Momentum : $\displaystyle\hat p:\hat pf(x)\to\frac{\hbar}{i} \partial_xf(x)$
- Position: $\hat x: \hat x f(x)\to xf(x)$
- Energy: $\displaystyle\hat E: \hat E f(x)\to\left[\frac{\hat p^2}{2m}+V(\hat x)\right]f(x)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]f(x)$
Measurement
Measuring the observable quantity $A$ associate with $\hat A$, is operating on the function again and again:
$$
\langle \hat A\rangle_\psi=\int\psi(x)^\dagger \hat A\psi(x) \ \mathrm{d}x.
$$Measuring gets you a value: this measured value has to be one of the real eigenvalues of $\hat A$
After measurement, system collapses into $\psi_a$ with eigenvalue $a$ measured
State can be expanded by $\hat A$‘s normalized eigenfunctions ${\psi_a(x)}$:
$$
\psi(x)=\int_a C_a\psi_a(x)\ \mathrm{d}x
$$
with $\vert C_a\vert^2$ being the probability of getting $a$ upon measurement.
Commutator and a deep equation
To measure the importance of operator (act) order, we define the commutator :
$$
[\hat A,\hat B]=\hat A\hat B-\hat B\hat A.
$$
Doing on position and momentum, we get the following lovely and deep equation:
$$
[\hat x,\hat p]=i\hbar \hat 1,
$$
which indicates the uncertainty principle. Assume the first thing we measure is accurate, then think about it, we cannot measure $x$ and $p$ both accurately, for the simple reason that their commutator is not zero: we cannot get the same result after the second one that is measured. Further more, when expressed in matrix, it is a very hard matrix equation and only infinite dimension matrix is can solve.
The Schrödinger Equation
Actually $e^{ikx}$ is an eigenfunction of operator $\hat p$ and every state $\psi(x)$ can be generated by it from Fourier’s Theorem.
From de Broglie, we know that any object is a wave, and consider a plane wave
$$
\psi(x,t;k)=e^{i(kx-\omega t)},
$$
which has momentum $p=\hbar k$ and energy $E=\hbar \omega$.
Similarly to hint of momentum, we notice
$$
i\hbar \frac{\partial\psi(x,t)}{\partial t}=\hbar\omega \psi(x,t)=E\psi(x,t),
$$
which depicts the time evolution of energy $E$. For waves it is trivial that $E=\hbar \omega$, but for particles maybe the energy $E$ is not trivial to get.
Changing $E\to \hat E$ complete the Schrödinger Equation:
$$
\begin{align}
i\hbar \frac{\partial}{\partial t}\psi(x,t)&=\hat E \psi(x,t)\\
i.e.\qquad i\hbar\frac{\partial \psi(x,t)}{\partial t}&=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x,t)}{\partial x^2}+V(x)\psi(x,t).
\end{align}
$$
The quantum mechanic model.
It generates from the observation of properties of the wave, and adding operators to make it ubiquitous for particles.
The equation describing the time evolution of a quantum state $\psi(x,t)$.
It is analogous to the equation of motion $\vec F=\dot {\vec p} $ and $\hat E$ is not defined as the operator $i\hbar\partial_t$ any more than $\vec F$ is defined to be $\hat{\vec p}$ .
6 Solving the Schrödinger Equation
To solve the Schrödinger Equation, we can first determine the energy eigenfunction $\phi_E(x)$, and then expand the wavefunction with the eigenfunctions, finally add the time evolution term by solving the time derivative in the Schrödinger Equation.
$$
\begin{aligned}
\hat E\psi(x)&=E\psi(x)\implies {\phi_E}\
\psi(x,0)&=\sum_E C_E\phi_E(x),\ C_E=\int_{-\infty}^{\infty}\psi^\dagger(x,0)\phi_E(x)\ \mathrm{d}x \
\implies \psi(x,t)&=\sum_E C_E e^{-i\omega t}\phi_E(x),\ \omega=E/\hbar.
\end{aligned}
$$
7 More on Energy Eigenstates
- A profound quality analysis on eiengy eigenstate function:
$$
-\frac{\hbar^2}{2m}\phi_E^{\prime\prime}(x)+U(x)\phi_E(x)=E\phi_E(x)\
\implies \frac{\phi_E^{\prime\prime}}{\phi_E}=-\frac{2m}{\hbar^2}\left[E-U(x)\right]
$$
8 Quantum Harmonic Oscillator II
brute force solving
9 Quantum Harmonic Oscillator II
adjoint operator
solving with
- lowering operator $\hat a$
- raising operator $\hat a^\dagger$
