FREQUENTLY ASKED QUESTIONS

October 17, 2007

Content Questions

Why does an eigenfunction for the Hamiltonian have no spread in $E$?

An eigenfunction $u$ of the Hamiltonian satisfies $Hu=Eu$. So the expectation value $\langle H \rangle = \int u^* H u dx = \int u^* E u dx = E\int u^* u dx$, which is just $E$ since $u$ satisfies the normalization condition. Similarly $\langle H^2 \rangle = \int u^* H (Hu) dx = \int u^* H E u dx = E\int u^* Hu dx = E^2$. The spread of measured values of $E$ is the standard deviation, $\langle E^2 \rangle - \langle E \rangle^2 = E^2 - E^2 =0$, so the spread is $\Delta E = 0$. This means you'll always measure the same value of energy $E$, for any measurement of the eigenfunction $u$: energy is sharp.

What happens when something changes the energy?

If something changes the energy, that's equivalent to a force acting: so you need a new potential, a new Hamiltonian, and a solution to the new Schrödinger equation.

What is the line width and what does it tell us?

Suppose you have an excited state with a finite lifetime $\tau$. Then according to the energy-time uncertainty relation, the excited state $E_1$ can't have a precise energy: it will have a spread in energies given by $\Delta E_1 \sim \hbar/(2\tau)$. This means that transitions to the ground state will also be smeared out, with a range of frequencies $\Delta f = \Delta E_1/h \sim \hbar/(2 h \tau) = 1/(4 \pi \tau)$. So if you measure the spectrum of frequencies of light from the transition, instead of getting a sharp spectral line, you will see a finite width: the line will be broadened, even if your instruments are perfectly precise. The minimum possible value of the width, also known as the ``natural line width'', is $\Delta f = 1/(4 \pi \tau)$. This is an observable effect of the Heisenberg uncertainty principle.

How does $\Delta f \sim \frac{1}{4 \pi \tau}$ correspond to what we actually see?

When you observe transition frequencies (like in the spectroscope, for light from atomic transitions), instead of seeing an infinitely sharp spectral line, you will see a broadened, fuzzy line with a range of frequencies $\Delta f$. See also above question.

You said that for a particle that decays, we have $\int_{\rm all space} \vert \psi\vert^2 d({\rm space})=A(t)$, where the value of $A(t)$ changes with time. The information to describe $A(t)$ does not come from the particle's wave equation. Where does $A(t)$ come from? Why do particles decay?

Yes, if a particle could disappear with some probability, its wavefunction at time $t$ should be normalized according to the probability that the particle could exist at all (although this is not really a standard way to set up the description). For instance, if there is uniform rate of decay (equal probability of decay interaction happening per unit time interval), the probability of surviving after time $t$ is an exponential decay.

Particles decay for various specific reasons, but always due to some force acting, i.e. some kind of interaction potential. For the case of a radioactive nucleus, for instance, it could be the weak force. The theory of transitions, decays and interactions will be explored in future quantum courses; we don't yet have the tools to describe it fully.

How is it that energy is conserved over a long period of time if it isn't conserved on short time scales?

Energy always conserved in the long term, and violation of energy conservation is only allowed if it happens only over a really short interval. We can get away with short-term violation in the following way: the uncertainty principle says that $\Delta E \Delta t \ge \frac{\hbar}{2}$: for any wavefunction there will be a relation $\Delta E \Delta t \sim \rm {constant}$, where $\rm {constant}>\hbar/2$. Imagine you have a state that lasts for time $\Delta t$. Then it has an intrinsic uncertainty in its energy $\Delta E \ge \frac{\hbar}{2 \Delta t}$. That means that there's ``slop'' in the energy at least as big as $\Delta E=\frac{\hbar}{2 \Delta t}$. A particle with mass-energy less than this uncertainty can exist for the short time $\Delta t$. So a ``virtual'' particle with $\Delta m c^2 < \frac{\hbar}{2 \Delta t}$ can exist for a time $\Delta t < \frac{\hbar}{2 \Delta m c^2}$.

How can particles come out of nowhere?

Particles popping out of nowhere is a flagrant violation of conservation of energy. Since particles have rest-mass energy, they can't just appear out of nowhere, since then there would be more mass-energy before than after. But so long as particles appear only for a very brief time, with mass-energy less than the uncertainty allowed by the H. U. P., they're allowed to appear. The particles ``borrow some energy from the vacuum'' in order to exist, but they have to pay back that energy on a strict deadline, and they disappear again. If they are to remain in existence for the long term, there needs to be some real energy coming from somewhere.

How do we know that conservation of energy can be violated?

This is a consequence of the Heisenberg energy-time uncertainty principle: see above questions. Remember that you're only allowed to violate it on very short timescales.

The observable effects are all indirect: we can't observe virtual particles directly. But the presence of virtual particles does have measurable consequences. One measurable effect is the Casimir effect, which will cause a weak force between very closely spaced plates: here's a Wikipedia link. Another example is the precise measurement of certain quantities, such as the magnetic moment of the electron: the existence of virtual particles changes the expected value in a well-understood way.

But actually the idea of virtual particles runs very deep in our current theories of particle physics. The most successful model for describing the three strongest fundamental interactions- electromagnetic, strong and weak forces - is built on the idea that these interactions are ``mediated'' by virtual particles. For instance, electromagnetic forces between charged particles involves the exhange of virtual photons. You could say that our entire body of data in particle physics is consistent with the idea of virtual particles. BFG has a bit of material about this in section 16-3.

In the virtual particle example, does $\Delta p \Delta x \ge \hbar/2$ apply? How do we take relativistic effects into account? (e.g. mass of $e^+$ and $e^-$, speed of the particles, etc.)

The virtual particle uncertainty-principle argument works the same way for momentum and position. Imagine you have a state that exists within a region $\Delta x$. The ``slop'' in its momentum is $\Delta p \sim \frac{\hbar}{2 \Delta x}$. We can therefore violate conservation of momentum if this violation happens within a small enough space.

Relativistic effects are indeed relevant for this situation- in fact you have to take into account relativistic mass-energy in the total energy. The particles do not travel faster than $c$.

Do we know of many other processes which occur on a timescale on the order of magnitude of $\gamma \rightarrow e^ + e^- \rightarrow \gamma$? (i.e. do we know of many other processes which can not conserve energy for a small amount of time?)

Yes, pretty much any allowed process can happen in a virtual way: the electron-positron virtual pair was just a particular example. So long as it happens for a short enough time (and obeys certain known conservation laws, like conservation of charge), it can happen. But the larger the masses of the virtual particles, the shorter the time they are allowed to exist. For example, a proton-antiproton pair appearing out out nowhere is even more ephemeral than a positron-electron pair, because the mass-energy that must be created is larger.

What do the positron and the electron look like? Do they have their own $\psi$'s? (diagram showing wavefunctions combining).

Electrons and positrons in general can both be described by wavefunctions $\psi$.

If you are referring explicitly to the virtual case: there's actually a more advanced way of describing such situations (both virtual particles and interactions such as annihilation), which is beyond the scope of this class. Your text, BFG Chapter 16, gives a bit more information, but full treatment requires relativistic quantum mechanics formalism.

(Diagram of electron-positron pair at event horizon of black hole or some other huge potential.) Can this happen, where a photon splits into an electron/positron pair and one gets sucked away before they recombine? Could you create a stream of $e^-$ with a big potential and a stream of $\gamma$?

If you apply a potential to the situation, you are adding energy. If you add enough energy, you can indeed turn a virtual particle into a real one: and energy will be conserved in the long term. To make the particle real, you need to add at least its rest mass energy.

I think the situation you are talking about with the black hole is ``Hawking radiation'', which is sometimes described as a case for which one virtual particle is not allowed to reunite with its partner. In this case the separating force is gravity. Hawking radiation is created near the ``horizon'' of a black hole (the radius beyond which nothing can escape). The typical description is as follows: imagine virtual particle-antiparticle pairs being created at the horizon of a black hole (as they are created everywhere). One partner of the pair might escape before annihilation can happen, while the other falls into the black hole. In this way, the black hole converts some of its mass-energy to escaping particles. It's plausible that Hawking radiation exists, but it has never been observed.

What's an anti-turnip?

Antimatter is matter with very similar properties to regular matter (similar mass, spin, etc.) but opposite charge, and some other opposite quantum numbers. Every known particle has an antiparticle: the antiparticle of an electron is a positron, and the antiparticle of a proton is an antiproton, etc.. There is no physical reason that macroscopic objects, like turnips or people, could not be built out of antimatter particles. So an anti-turnip would be exactly like a turnip, with identical properties, except that all its particles would be antiparticles.

(In principle you can get virtual turnip-antiturnip pairs, but that would be vanishingly unlikely, and the pair would last such a vanishingly small time that they wouldn't have much chance to do anything, alas.)

So would anti-turnips be sort of radioactive? You said they would be dangerous to eat.

An anti-turnip would be almost exactly like a turnip, except its particles would all be antiparticles. Anti-turnips therefore wouldn't be any more radioactive than regular turnips, and they would be perfectly tasty and nutritious for anti-people. However, for us, anti-turnips are exceedingly dangerous objects to try to eat, because the antimatter in them would annihilate with the matter in our bodies as soon as the anti-turnip touched our lips. The resulting explosion would completely destroy you and maybe even take out nearby cities. Please stay away from them.

So is $p$ total energy? What exactly is $p$ in the equations with $p=\sqrt{2mE}$?

$p$ is momentum, not energy. The relation between (classical) energy and momentum is $E=p^2/2m$, from $E=\frac{1}{2} m v^2 = p v/2 = p^2/2m$.

When you say $E= p^2/2m + V_0$ in region 2, isn't $V_0$ the velocity, not the potential?

No, $V_0$ isn't anything to do with velocity; it's the height of the potential (remember potential means potential energy). When the particle is in region II, its total energy is kinetic ($p^2/2m$) plus potential ($V_0$) energy.

Why is there a finite repulsive force at the edges? How do we show this quantitatively?

It's a finite potential, not a finite force. Right exactly at the edge, in our idealization of a perfectly sharp-edged barrier, strictly the force $-dV/dx$ does have infinite slope. However, in the central part (and just inside the boundaries), the potential has a finite value of $V_0$ rather than an infinite value (as at the boundaries of the infinite well potential). So considering force as rate of change of momentum, $\Delta p/\Delta t$, we get a finite change of energy, and momentum, in an infinitely small time as the particle crosses. Note that the force is only really infinite in the true limit of perfect sharpness; in reality there will be a smoothed shape and finite force.

It's usually easier to think of the potential as a more ``fundamental thing'' than force (which is why it's the convention in quantum mechanices to deal with potentials rather than forces).

For the square potential, are we assuming that the edges are not perfectly horizontal so that the force is finite at each end?

In real life, the potential would be more smoothly shaped at the boundaries rather than an abrupt square shape, and indeed the force would not be infinite. We are considering the limit of squareness. See above question. But note that change of momentum and energy are finite in the perfectly sharp case, even for infinite force at the edges.

When you use the demon to detect the photon, you lose the wavelike nature, but if you believe that multiple universes exist and are created when you take a measurement, does the wavelike nature apply across universes?

Well, some people speculative that for every measurement, or interaction, the universe ``splits'' into many universes, each corresponding to a particular outcome of the measurement. I think if it did, in each ``multiverse'', or at least the vast, vast majority of them, the interference pattern would still be destroyed, because momenta imparted to the electron would still be random from measurement to measurement, in any of the universes. This ``many worlds'' scenario is intriguing, but a rather speculative interpretation of the probabilistic nature of quantum mechanics, and at present I'm not sure there's any way of testing experimentally whether it makes sense. In this course we'll stick to one universe.

What is the white hole? Is there anti-gravity?

A ``white hole'' refers to a time-reversed black hole, which emits rather than swallows matter: here's a Wikipedia link. The existence of such a thing is speculative (and perhaps not very likely) and none have ever been observed.

Anti-gravity (a repulsive gravitational force) has also never been observed, unless you count ``dark energy'', which is the name given to the phenomenon that is causing the accelerated expansion of the universe (see this link). The nature of dark energy is currently not at all understood, but it's not clear it could really be considered ``anti-gravity''.