Physics 401 - Spring 2015



The HET Group







George Siopsis - April 21, 2015


It is possible to have a system governed by laws that possess a certain symmetry, without the system itself being symmetric. As an example, consider the orbit of the earth around the sun. The earth moves because of the gravitational potential which is spherically symmetric, yet its orbit is an ellipse with a well defined direction that breaks the spherical symmetry. Any other axis of the ellipse would be a perfectly sensible physical state. It would be related to the axis chosen by the earth through a rotation. Out of all possible directions, the earth made a choice for historical reasons. Any other choice, though, would have been just as good.

To ponder upon this observation, consider a ball moving in a one-dimensional harmonic oscillator potential \[ U(x) = \frac{1}{2} K x^2 \] It oscillates back and forth rolling up and down in the potential well as determined by its total (conserved) energy \[ E = \frac{p^2}{2m} + U(x) \] Both the Hamiltonian ($H=E$) and the trajectory are symmetric under reflection, \[ x \to -x \] The ground state has $E=0$ (minimum energy) and a simple trajectory ($x=0$), which is also symmetric under reflection.

Now modify this harmonic oscillator, choosing \[ U(x) = \frac{1}{2} K x^2 + \frac{1}{4} \xi x^4 \ \ , \ \ \ \ \xi \gt 0 \] None of the above conclusions will change, although we can no longer easily compute the trajectories.

However, if $K\lt 0$, the potential has minima at \[ x_{min} = \pm v \ \ , \ \ \ \ v = \sqrt{\frac{-K}{\xi}} \] The ground state (of minimum energy) may correspond to two different trajectories, $x = +v$, or $x= -v$. Neither is invariant under reflection; one is obtained from the other by reflection. For energies slightly above the minimum, we obtain trajectories around either $x=+v$, or $x=-v$. Thus, the symmetry (reflection) is broken. It should be noted that the underlying physical laws are still invariant under reflection, however this symmetry is not at all evident to the ball rolling up and down the hill, unless it has sufficiently high energy to go past the middle point $x=0$.

A low-energy ball rolling around $x=+v$ is oblivious to the existence of the other minimum of the potential (at $x=-v$), and will naturally describe its motion in terms of the coordinate \[ x' = x-v \] which is the distance from its equilibrium point, and write the potential as \[ U = \xi v^2 {x'}^2 + \xi v {x'}^3 + \frac{1}{4} \xi {x'}^4 \] ignoring an irrelevant additive constant. The potential has no apparent symmetry. An astute ball will realize that the potential is symmetric under \[ x' \to -x' \ \ , \ \ \ \ v \to -v \] but there is really no motivation for the ball to study this symmetry because $v$ is a (God given) parameter of its world.

Now place the ball at $x=0$. This is a point of unstable equilibrium. Suppose it corresponds to energy $E=0$ (same as the potential the ball wrote). For $E\ge 0$, the trajectories are symmetric under reflection.

Perturb the system by kicking the ball to the left, applying a small constant force, thus changing the potential to \[ U(x) \to U(x) + hx \ \ , \ \ \ \ h \gt 0 \] Observe the response of the system and then let $h\to 0$. Since $V'(0) = h\gt 0$, the ball sitting at $x=0$ will roll down to the left and perform oscillatory motion determined by its energy $E=0$. As $h\to 0$, the original shape of $U$ is restored together with its symmetry, but for the trajectory of the ball, the average position $\langle x\rangle \to -v$ as $h\to 0$, and the symmetry is spontaneously broken.

This is similar to a ferromagnet: if we switch on a uniform magnetic field $H$, we break rotational symmetry and the material gets magnetized. When we turn $H$ off, the magnetization does not go to zero - the symmetry has been (spontaneously) broken.

The above classical picture changes drastically if one includes quantum effects. The system can be approximated by a harmonic oscillator around each minimum, however, the states corresponding to the two minima are not eigenstates due to quantum tunneling. There is a unique ground state which is symmetric under reflection (even function). Therefore, there is no symmetry breaking, because $\langle x\rangle = 0$.

This conclusion does not change even when you perturb the potential by adding a kick. The effect of the perturbation is to shift the relative energies of the energy levels of the two wells, but the states can still mix! So if we call $|\pm\rangle$ to corresponding ground states of the wells, then $\langle + |-\rangle \ne 0$ and tunneling is possible. In the limit $h\to 0$, we thus recover $\langle x\rangle \to 0$, unlike in the classical system.

So how can magnets exist? Are they not quantum mechanical systems? A magnet has not one but $\sim 10^{26}$ oscillators. So even though after applying the kick (external magnetic field), the energy difference of the two wells is small for each harmonic oscillator, the total energy difference is enormous. Thus $\langle + |-\rangle$ is minute and tunneling is next to impossible. In the limit $H\to 0$, one needs an enormous amount of energy to restore the symmetry, which is not readily available, so the material remains magnetized.

How can we apply these ideas to weak interactions? Recall the equation for a massive spin 0 particle. Suppose it has no charge. Furthermore, suppose $\Psi$ is uniform. Then the wave equation reduces to \[ \ddot\Psi = - \frac{c^2}{\lambda^2} \Psi \] so $\Psi$ is like a harmonic oscillator with $K = c^2/\lambda^2$ and $m=1$ (compare with $m\ddot x = -Kx$). By analogy, we define the potential energy \[ U = \frac{1}{2} \frac{c^2}{\lambda^2} |\Psi|^2 \] which has a minimum at $\Psi =0$. Notice that $\Psi$ is complex, so $\Psi = \Psi_1 + i\Psi_2$, and $|\Psi|^2 = \Psi_1^2 + \Psi_2^2$.

Now modify this potential to \[ U = \frac{1}{2} K|\Psi|^2 + \frac{1}{4} \xi |\Psi|^4 \ \ , \ \ \ \ K \lt 0 \] The wave equation changes to (think of $\Psi_1$ and $\Psi_2$ as two coupled harmonic oscillators) \[ \ddot\Psi = - U' = - K\Psi - \xi |\Psi|^2 \Psi \] and if we restore the space dependence of $\Psi$, we obtain \[ \frac{\partial^2\Psi}{\partial t^2} - c^2 \nabla^2 \Psi = - U' = - K\Psi - \xi |\Psi|^2 \Psi \] The potential has minima along the circle \[ |\Psi| = v = \sqrt{\frac{-K}{\xi}} \] Let \[ \Psi = \varrho e^{i\sigma} \] Then in the ground state \[ \varrho \approx v \] Notice that this is possible quantum mechanically, because $\Psi$ is equivalent to not 1, not $10^{26}$, but an infinite number of oscillators (one for each point in space).

We can express the wave equation in terms of $\sigma$. After some algebra, we obtain from the imaginary part of the wave equation \[ \frac{1}{c^2} \frac{\partial^2\sigma}{\partial t^2} = \nabla^2 \sigma \] which shows that $\sigma$ is a spin-0 massless particle - the Goldstone boson.

Now give the charge back to $\Psi$. The wave equation is \[ \left( \frac{\partial}{\partial t} - \frac{q}{i\hbar} V \right)^2 \Psi - c^2 \left( \vec\nabla + \frac{q}{i\hbar} \vec A \right)^2 \Psi = - K\Psi - \xi |\Psi|^2 \Psi \] Check that it is invariant under a gauge transformation.

But now, if I perform a gauge transformation with \[ \chi = \frac{\hbar}{q} \sigma \] I get \[ \Psi \approx v e^{i\sigma} \to v \] Where did the Goldstone boson $\sigma$ go?

Look at the charge and current densities \[ \rho = - \frac{i}{2} \left[ \Psi^* \left( \frac{\partial}{\partial t} - \frac{q}{i\hbar} V \right) \Psi - \mathrm{c.c.} \right] \ , \ \ \vec J = - \frac{i}{2} \left[ \Psi^* \left( \vec\nabla + \frac{q}{i\hbar} \vec A \right) \Psi - \mathrm{c.c.} \right] \] Show that the current is conserved \[ \frac{1}{c} \frac{\partial\rho}{\partial t} + \vec\nabla\cdot\vec J = 0 \] For $\Psi \approx v$, the charge and current densities reduce to \[ \rho = - qv^2 V \ \ , \ \ \ \ \vec J = - \rho v^2 \vec A \] This is precisely the form we used before to modify the Maxwell equations and give mass to the photon! But now we are not breaking the symmetry (gauge invariance). It looks broken, but it is only because of the choice of the ground state (spontaneous breaking). The laws of Nature are still gauge invariant.

Since the photon has mass, it has 3 degrees of freedom. The massless photon, which had 2 degrees of freedom, ate the Goldstone boson (1 degree of freedom) and became massive. This is the Higgs mechanism.

Around the minimum, $\varrho$ fluctuates, so in our world, we would define \[ \varrho = v + \varrho' \] Check that $\varrho'$ is a massive particle. Unfortunately, it has only a residual effect on our world, after giving mass to the weak photon (as well as everything else in our world that has mass). Even its mass cannot be predicted. $\varrho'$ is the Higgs particle was found at the Large Hadron Collider in Geneva, Switzerland in 2011. Its mass was found to be \[ m_{\varrho'} = 126~\mathrm{GeV} \]



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