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2021-03-04
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To get the most out of this week's colloquium by Eva Andrei, I need to introduce a bit of basic background in quantum mechanics (QM). I already alluded to a few aspects of this physics in my talk last week. If desired, you can read more about these topics in Knight chapters 38, 39, and 40.
At small scales, the world is not continuous like in classical physics, it is granular. Example: at long length scales compared to the size of a water molecule, water looks like a perfectly continuous material, like when we pour ourselves a cup of water out. But at length scales of nanometres, if we have a powerful microscope, we can see that water is actually made up of individual lumps known as molecules. So at molecular/atomic/subatomic scales, we must use quantum physics to describe how the world behaves.
Newton thought of light as being made of lightweight corpuscles
, or particles. Later, physicists switched to thinking of light as a wave. In your PHY152 course you have learned about how electric and magnetic fields behave from Gauss's Law; there are also Faraday's Law, and Ampère's Law. When we put together all of the EM equations that were discovered through experiment, we find Maxwell's equations, which unify $\vec{E}$ and $\vec{B}$ together to describe light as a wave with wave speed $c$. This is the classical view of EM fields. Quantum mechanically, important experiments including the Photoelectric Effect
have shown that at small scales, EM fields are actually made up of lots of indivisible quantum lumps or quanta
known as photons
.
Fun fact: Albert Einstein didn't actually win the Nobel Prize in Physics for discovering relativity -- he actually won for explaining the PhotoElectric Effect! You will learn more about that interesting phenomenon before the end of PHY152, as well as others. I also have some brief notes about it here.
Quanta have both particle-like and wave-like properties; this is known as particle/wave duality
. Subatomic particles (e.g. photons, electrons, quarks, the Higgs) do it, composite particles (e.g. neutrons) do it, atoms do it, and even molecules do it! I like to call quanta wavicles
to remind us that they have particle/wave duality. Physicists more broadly tend to use the more boring and somewhat imprecise word wavepacket
for this. Which behaviour you see -- particle or wave -- depends on what kind of experiment you do.
Fun fact: in 2016 I made a little video discussing an analogy between particle/wave duality and my nonbinary gender identity; I hope you all enjoy and/or learn from it.
How do you know when the wavelike behaviour of a quantum is important? This happens when your experimental apparatus has sufficiently small features that they are in a similar size ballpark as the wavelength of the quantum. Then you end up being able to discern wavelike behaviours like interference and diffraction. If your apparatus is a lot bigger than the quantum wavelength, you will basically see particle-like behaviour.
For photons, which make up EM fields, we characterize them by a wavelength (or equivalently a frequency $f=c/\lambda$) which is related to the energy $E$ and the momentum $\vec{p}$ via $$ E=c|\vec{p}| = {\frac{c\,h}{\lambda}} \qquad {\rm (for\ photons)} \,. $$ Here, $h$ is Planck's constant, and $c$ is the speed of light. For a CBC Radio One photon with frequency , this amounts to about $7 \times 10^{-26}$ Joules of energy, a piffling amount.
How about for massive particles? In 1924, a young Louis de Broglie proposed for his Ph.D. thesis, with very little evidence, that if a massive particle is moving with spatial momentum $\vec{p}$ then its quantum wavelength is given by
$$
\lambda_{\rm deB} = {\frac{h}{|\vec{p}|}} \qquad {\rm (for\ all\ quanta)} \,.
$$
This was later spectacularly confirmed in a number of famous experiments. (Note: if we think of the quantum as a particle, it moves at a velocity known as the group velocity
of the wave.)
The next question that comes up is -- how do matter waves evolve? Can we write down a differential equation describing how they change over space and through time? The Schrödinger equation shows us how to do this. It postulates that a field called the wavefunction
$\Psi({\vec{x}},t)$ describes the physics, and this is generally a complex function of both space and time coordinates.
For a non-relativistic particle, the equation describing how the wavefunction changes over space and time is known as the time-dependent Schrödinger equation
,
$$
i\hbar {\frac{\partial}{\partial t}} \Psi({\vec{x}},t) = H \Psi({\vec{x}},t)
= \left( -{\frac{\hbar^2}{2m}} {\vec{\nabla}}^2 + U({\vec{x}},t) \right) \Psi({\vec{x}},t)\,,
$$
where $H$ is known as the Hamiltonian. The first piece in $H$ represents the kinetic energy coming from wavefunction gradients, and the second piece represents external force(s), if any, through a potential energy function.
If the Hamiltonian of the system does not depend on time, then the total energy is conserved (this is an example of Noether's Theorem). Then the quantum states described by the wavefunction are referred to as stationary states. In that case, we can factor off the time dependence as $\Psi({\vec{x}},t)=\exp(-iEt/\hbar) \psi({\vec{x}})$, and we obtain the time-independent Schrödinger equation
,
$$
H \psi({\vec{x}}) = E \psi({\vec{x}}) \,.
$$
This has the form of an eigenvalue equation, where the eigenvalue is the energy and the eigenvector is the spatially dependent wavefunction. The job of a physicist is then to find the quantized energy levels and the wavefunctions that go along with them.
If we take the absolute magnitude squared of the wavefunction (say, of an electron), that gives the probability $\mathscr{P}$ of finding it, $$ \mathscr{P} ({\vec{x}}) \propto \left| \psi({\vec{x}}) \right|^2 \,. $$ This is why when we draw pictures of atomic electrons, we represent them as probability clouds, not as tiny planets whizzing around the central nucleus, held in orbit by electric attraction. If we integrate the probability over all space, we should get 1, which teaches us how to fix the proportionality constant above. Wait -- what do we mean by probability? Isn't physics supposed to be deterministic?!
No. In quantum mechanics, all we can actually predict with certainty is how the wavefunction evolves over space and time, which tells us the probability of finding the particle at a specific place in space at a specific point in time. But what actually ends up happening is purely up to random chance! This essential randomness of quantum mechanics kills the idea of classical determinism dead. The universe does not run like a clockwork machine like was believed in Newton's day. (BTW, Albert Einstein disliked this indeterminacy of quantum mechanics, famously saying God does not play dice with the universe!
. History proved him wrong about that particular thing.) Quantum determinism took over from classical determinism.
The wavelike nature of quanta, and this randomness of quantum mechanics, introduces some fuzziness to physical observables like momentum or position. The Heisenberg Uncertainty Principle is a mathematically precise statement of this fuzziness. It governs only particular pairs of physical observables, so it does not apply across the board (contrary to popular belief). After certain reasonable assumptions about the physical states being measured, you can express the Heisenberg Uncertainty relation as an equation which says that the uncertainty about momentum in a particular direction times the uncertainty about position in the same direction has to be bigger than $h/2$,
$$ \Delta p \, \Delta x\geq {\frac{h}{2}} \,. $$ Energy and time are also governed by an uncertainty relation $$ \Delta E \,\Delta t \geq {\frac{h}{2}} \,. $$
Intrinsic spin $s$ is one of the properties of an elementary particle that fully characterizes it, along with its mass $m$ and any force charges such as the electric charge $q$. Intrinsic spin cannot be removed via any physical process; it is very literally intrinsic to the elementary particle. It is quantized in units of $\hbar/2$, where $\hbar = h/(2\pi)$ and $h$ is Planck's constant,
$$
s = n {\frac{\hbar}{2}} \,, \qquad n=0,1,2,3,\ldots \,.
$$
This condition is needed in order to make sense of the space of possible wavefunctions for the elementary particle known as Hilbert space
.
For massive particles, intrinsic spin is a genuine angular momentum. If you measure its component along some axis, say the $z$-axis, then the allowed values of $s_z$ are $-s, -s+ 1, \ldots, s-1, s$, i.e., $(2s+1)$ options, where $s=|{\vec{s}}|$. For a massless particle, things work a bit differently: you only end up with two options: $h = \pm s$, where $h$ characterizes the helicity
. For example: an electron with $s=1/2$ can have $s_z=-1/2$ or $+1/2$, while a photon with $s=1$ can have $h=\pm 1$, corresponding to left and right circular polarizations.
If you have even $n$, you are a boson
, named for Bengali physicist and mathematician Satyendra Nath Bose. If you have odd $N$, you are a fermion
, named for Italian physicist Enrico Fermi. Bosons don't mind company: you can put an in principle arbitrary number of them into the same quantum state. Fermions, by contrast, need some personal space / have elbows -- they obey the Pauli Exclusion Principle
, which states that no two fermions can be in the same quantum state at the same time. The PEP is why atoms with higher atomic number are physically larger: you can only fit two electrons (one with spin-up along some axis and one with spin-down) into any one electronic energy level in an atom; if a level is full up, you have to put the next ones into higher energy levels which are less strongly bound to the atomic nucleus and the electron clouds are centred further away from the nucleus.
When we cool things down to low temperature, boson and fermion behaviours are very different. For bosons, if you arrange things just right experimentally (this is hard!), you can get a group of identical bosons to go into a Bose-Einstein condensate
, where all of them are in the same quantum state! By contrast, fermions insist on having some personal space from each other because of Pauli exclusion, and this gives rise to a pressure known as the Fermi degeneracy pressure
. In white dwarf stars, electron degeneracy pressure counteracts gravitational attraction to produce a stable equilibrium for the star; in neutron stars, neutron degeneracy pressure does the job of holding up the star. Neutrons are composite particles made out of three quarks (up, down, down), and when you combine an odd number of fermions into a composite particle you get a fermion. Combining even numbers of fermions gives a composite particle that is a boson.
Let us now try to solve the time-independent Schrödinger equation for the case of a free particle all alone in one-dimensional space. We need to solve $$ -{\frac{\hbar^2}{2m}} {\frac{d^2}{dx^2}} \psi(x) = E \psi(x) \,. $$ Let us try an oscillatory wavefunction: $$ \psi = A_\pm \exp\left( \pm i kx \right) \,, $$ where $k$ and $A_\pm$ are constants. Plugging it in produces $$ E = {\frac{\hbar^2 k^2}{2m}} = {\frac{p^2}{2m}}\,, $$ where $p=\hbar k$. What we have is an oscillatory wavefunction which can have any value of the wavevector $k$, and the higher its magnitude, the greater energy it carries. The general solution is a sum of these two types of oscillatory exponentials, or equivalently a sum of sines and cosines.
Now imagine that our particle is restricted to live inside a box of side length $L$, because of an infinite potential well keeping it in there. In this case, the wavefunction has to go to zero at the walls: it is not allowed to leak outside of the interval $0 \leq x \leq L$. These boundary conditions select out sines rather than cosines, $$ \psi(x) = A \sin(kx) \quad {\rm for}\ \ 0\leq x \leq L\ \ {\rm or }\ \ 0\ \ {\rm otherwise}\,. $$ What are the allowed values of the wavevector $k$? To find out, let us check the boundary conditions. We need our wavefunction to be zero at both $x=0$ and $x=L$. This is obviously true at $x=0$. At $x=L$ we require $\sin(kL)=0$, which selects out $$ k_n = n {\frac{\pi}{L}}\,,\qquad n=1,2,3,\ldots \,. $$ Putting this back into the energy formula gives $$ E_n = n^2 {\frac{\pi^2\hbar^2}{2mL^2}} \,, $$ which shows that the energy levels are quantized. The last thing we need to do to nail down our solution is to figure out the constant $A$. This is fixed by insisting that at any given time we have to find the particle somewhere along the $x$-axis: $$ \int_{-\infty}^{+\infty} dx\, \left| \psi_n(x)\right|^2 = 1 \,, $$ which if you do the algebra you will find gives $$ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left( {\frac{n\pi x}{L}} \right) \quad {\rm for}\ \ 0\leq x \leq L\ \ {\rm or }\ \ 0\ \ {\rm otherwise} \,. $$ A very rough plot of the probability of finding the particle at various positions is given below for $n=1$.
![[potential well with infinitely high walls]](198s/figs/infinite-square-well.jpg)
What happens if we allow our potential energy well to have walls of finite height, say $U_0$, instead of infinite height? In this case, the equation we are solving becomes $$ -{\frac{\hbar^2}{2m}}{\frac{d^2}{dx^2}}\psi(x) = (E-U_0) \psi(x) \,. $$ If $E>U_0$, i.e. the kinetic part of the energy is positive, then we obtain oscillatory exponentials of the form $\exp(\pm i kx)$ (or sines and cosines) again, with real $k$ given by $$ k^2 = {\frac{2m(E-U_0)}{\hbar^2}} \,. $$ This is morally like our previous case and is referred to as the classically allowed region in the potential well.
What if instead the kinetic energy is below the height of the potential energy barrier? In this classically forbidden region, the two linearly independent real solutions of the Schrodinger equation turn out to be not oscillatory functions, but rather positive and negative exponentials $\exp(\pm \kappa x)$ with real $\kappa$ given by $$ \kappa^2 = {\frac{2m(U_0-E)}{\hbar^2}} \,. $$ The kind of exponential that grows away from the finite-height square well can be ruled out by applying the physical boundary condition that the particle cannot penetrate infinitely far into the forbidden region if the potential barrier persists all the way out to $x\rightarrow \pm\infty$. That leaves us with the kind of exponential that dies away from the square well. The particle has a finite probability of being found inside the energetically forbidden region, which is exponentially smaller the further you penetrate into it.
To solve for the full wavefunctions, you need to match oscillatory behaviour in the classically allowed region onto dying exponential behaviour in the classically forbidden regions. The mathematical details of this are not very illuminating, so we have skipped them. A very rough plot of the probability of finding the particle at various positions is given below for $n=1$.
![[square well potential with finite height]](198s/figs/finite-square-well.jpg)
Let us end with one final question: what if the potential well of finite height $U_0$ only had a finite thickness $W$ in the $x$ direction? Then in the region $x\in[0,L]$ (classically allowed) the wavefunction would be oscillatory, in the regions $x\in[-W,0]$ and $x\in[L,L+W]$ (classically forbidden) it would be a dying exponential, and for $x\lt -W$ and $x\gt L+W$ (classically allowed) it would be oscillatory again -- but with a smaller amplitude, such that the wavefunction was continuous across all of space.
This story is referred to as quantum tunnelling. It means that quanta can literally 'walk' through walls, as long as they are not infinitely tall and thick!
2021-03-04