*Ex nihil, nihil fit*, is the Latin phrase, that from nothing, nothing comes. If there is something, then why? How did it get here?

It is then popular in certain circles that place a high value on scientific understanding --- people who perhaps don't understand math well enough to study it for real, but who nonetheless appreciate human efforts to understand the natural world in terms of rational processes and read as much of it as they can understand --- to make the rebuttal claim that, according to the physical understanding of quantum mechanics, something

*can*come from nothing.

You can see an example of this conversation in the below video:

The idea is that in quantum field theory, study has shown that even in the state representing a vacuum, i.e. a system with zero particles, there is still the constant process of random particle-antiparticle pair creation and annihilation going on all the time. You start with zero particles, and for brief instances you have two particles. Or, in higher order interactions, four, or one hundred and twenty four. Therefore, something -- particle-antiparticle pairs -- can come from nothing -- the quantum vacuum.

This idea is right, and it's wrong. I think both people are talking past each other, and in this post, I would like to try to clarify.

I'm not a field theorist. I've had some grad classes in it, but it's not anything in which I'm an expert (in fact, there probably isn't

*anything*in which I'm an expert, but it's a helpful caveat). Still, what I'm about to say is very basic to field theory (if anything in field theory can be called "basic"), and I'm more or less directly citing the text

*Field Quantization*by Greiner and Reinhardt (available on Amazon for only $\$20$!). What follows is a very, very brief outline of how quantum field theory leads to the understanding of the quantum vacuum, but also how the results therein do not mean what many people think it means. I have some wikipedia links throughout, so that hopefully people who do not understand math can at least follow along with what I'm trying to say -- the math isn't important, but the physics is.

Classical field theory is, more or less, the replacement of a collection of discrete variables $\{\psi_i\}$ with a single field, $\psi(x)$, which is a function of a parameter $x$, so that $\psi(x_i)\sim \psi_i$. The typical example is to consider a chain of beads linked together by springs. We hold it horizontal to the ground and ignore effects due to gravity or anything besides the spring stretchiness. In equilibrium, with no forces acting, we could write down the position of each bead as $y_i(0)$. If we displace one of the beads, it will start wobbling around, and set all the others wobbling around, causing the effect to propagate through the chain. We are no longer in equilibrium, and we might describe the position of each bead at a given time by $\eta(t) = y_i(t) - y_i(0)$, its displacement from equilibrium, as it is this displacement that goes in to the spring force, $F_i = -k_i \eta_i(t)$. This is a set of discrete variables $\{\eta_i\}$. If rather than a string of beads, we want to talk about, say, a continuous rubber band, then we let the number of beads go to infinity, and we end up with $\eta(x)$. Here $x$ gives a position along the rubber band, and at that position $\eta(x)$ specifies the displacement of the itty-bitty bit of rubber band right there at $x$. So we go from a discrete set $\{\eta_i\}$ to a continuous set describable by the function $\eta(x)$.

The next part of classical field theory is a field equation. This is the part with the really difficult math, but basically we write an equation describing how the field $\eta(x)$ changes in time and in space --- so we include derivatives $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial t}$. The field equation will involve $\eta$ and its derivatives and the solution to the field equation will give a description (in this case) of the deformation of the rubber band at each point in time and space.

One example of a classical field theory is quantum mechanics. We usually use "classical physics" to mean

*non-quantum physics*, but funnily enough, standard quantum mechanics is a classical field theory. So the wavefunction $\psi$ in QM is a classical field.

In going from classical mechanics to quantum mechanics, the process taken is sometimes called canonical quantization and is an extremely elegant theory. Basically, take your favorite equation from classical physics and write it in terms of the position, $x$, and the momentum, $p$. Now all we do is make a replacement of $x$ and $p$, so that they are no longer mere variables, but instead become what are called operators $\hat{x},\hat{p}$ that act on wavefunctions. The standard replacement is that $\hat{x}$ is the operator defined by $\hat{x}\psi=x\psi$, that is, $\hat{x}$ changes the function $\psi$ to now be the function $x\psi$, or $x$ times $\psi$. So if $\psi = x^2$, then $\hat{x}\psi = x\cdot x^2 = x^3$. Likewise $\hat{p}$ will get the more difficult replacement $\hat{p}=-i\hbar \frac{\partial}{\partial x}$. So if $\psi=x^2$, then $\hat{p} = -i\hbar \frac{\partial x^2}{\partial x} = -i\hbar 2x.$

(If you don't know that $\partial$ means, it means a partial derivative, treating everything besides $x$ as fixed, and if you don't know what a derivative is, then it doesn't particularly matter, so long as you understand that a derivative acting on a function changes it to another function; that's what all operators do -- they change one function in to another.)

In going from the

*classical*field theory of quantum mechanics to the

*quantum*field theory of quantum mechanics (those are different things) we perform what is sometimes called "second quantization". Before we only quantized the dynamic variables by replacing them with operators. Now we further quantize

*the fields themselves*by replacing them with operators. So we replace our Schrodinger wavefunction $\psi$ with the field operator $\hat{\psi}$.

The basic definition of $\hat{\psi}$ is that when its Hermitian conjugate $\hat{\psi}^\dagger$ (a generalization of complex conjugate) operates on a quantum state, it has the effect of creating a particle in the state $\psi$. So consider some state of two particles, in the states $\psi_1, \psi_2$. We will describe this with the funny symbol $|\psi_1\psi_2\rangle$ called a "ket", which we use because it simplifies the theory by a bunch (by the way, the LaTeX command to get $|\psi\rangle$ instead of $|\psi>$ is to use "\rangle"). The whole ket describes the quantum system, and each $\psi_i$ in the ket stands for a particle with the wavefunction $\psi_i$ in that system. Then if we have the quantized field operator $\hat{\psi}$, we define $\hat{\psi}^\dagger|\psi_1\psi_2\rangle = |\psi\psi_1\psi_2\rangle$, meaning now there is a third particle with state $\psi$. That isn't technically precise because there's shenanigans with Fermions and Bosons, but we don't need that complication to make the point we're trying to make.

Because the $\hat{\psi}$ describes a general wavefunction, we would like to express it more particularly. A common way to do this in physics is to give it a planewave expansion --- so we interpret $\psi(x,t)$, the wavefunction, as consisting of a bunch of planewaves $e^{ikx-i\omega t}$ added together, to varying extents. In equations,

$$\psi(x,t) = \int\int dk \,d\omega \;a(k,\omega) e^{ikx-i\omega t}$$

where $a(k,\omega)$ is now a Fourier amplitude telling to what extent the particular planewave $e^{ikx-i\omega t}$ contributes to $\psi(x,t)$. For the field operator, $\hat{\psi}$, this is the same, except now the Fourier amplitudes $a(k,\omega)$ also become quantized to operators, to $\hat{a}(k,\omega)$. It is the operators $\hat{a}$ and their Hermitian conjugates $\hat{a}^{\dagger}$ that are important for this discussion; everything before this was just to show where they come from.

We can now express any field operator $\hat{\psi}$ by a sum of operators $\hat{a}, \hat{a}^\dagger$. These operators are called by the evocative title creation and annihilation operators, and are crucial to QFT.

*If you haven't followed anything so far*,*here is all that you need to know to understand what I'm trying to say.*In QFT, we work with the creation operator $\hat{a}^\dagger(k,\omega)$ and the annihilation operator $\hat{a}(k,\omega)$. The effect of these is to produce or eliminate a particle with definite momentum $k$ and energy $\hbar\omega$. To keep things simple, let us look only at one particular momentum and one particular frequency. Say there are $n$ particles in our system with this $k$ and $\omega$; we describe this with the ket $|n\rangle$. Then

$$\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$$

and

$$\hat{a}^\dagger |n\rangle = \sqrt{n+1}|n+1\rangle.$$

In other words, the creation and annihilation operators raise or lower the number of particles in a particular state.

What happens if we only have a single particle with $k$ and $\omega$, and we act on it with $\hat{a}$? Then we get the vacuum state,

$$\hat{a}|1\rangle = |0\rangle = |\rangle$$

sometimes written either way. This vacuum state, the state with zero particles, is what is called the quantum vacuum, and it is this vacuum state that is unstable and has all of the association with something coming from nothing.

However, the vacuum state is

*not*nothing! The vacuum state is the ground state field of quantum field theory. It is a system of zero particles, but it is a system with a very complicated structure.

To make this clear, if we are in the vacuum state, and we act on it with the creation operator $\hat{a}^\dagger$, then we arrive at

$$\hat{a}^\dagger |0\rangle = |1\rangle,$$

the single particle state. We can then keep acting on it, as much as we want, to get to

$$|n\rangle = (\hat{a}^\dagger)^n|0\rangle,$$

the $n$ particle state. So the vacuum has properties, and even if it has no particles, it still behaves just like a system with particles.

Furthermore, the vacuum has energy. For the single frequency $\omega$ we are considering, the energy of the $n$th state can be written as $E_n = \hbar\omega(n+\frac{1}{2})$. That means the vacuum has energy $E_0=\frac{1}{2}\hbar\omega$. If you consider that the true QFT vacuum contains contributions from

*every*frequency $\omega$, and that there are an infinite number of such frequencies, then we arrive at the conclusion that

*the quantum mechanical vacuum has infinite energy.*We typically ignore this infinity in our calculations, since this infinite energy will appear in every other field configuration without contributing anything, but it is still there. The quantum vacuum is a structured field state with infinite energy. It isn't nothing.

To make this more emphatic, there

*is*a thing in QFT that can be rightfully called nothing. If you operator $\hat{a}$ on the $|1\rangle$, you get the vacuum, $|0\rangle.$ But from $|0\rangle$, you are still in a quantum state, and you can get back to any other state through application of $\hat{a}^\dagger$. But if you operate $\hat{a}$ on the vacuum, it's done.

$$\hat{a}|0\rangle = 0.$$

It is not a quantum state. It is nothing. It is destroyed. You can act $\hat{a}^\dagger$ on it all you want, nothing happens:

$$\hat{a}^\dagger \hat{a}|0\rangle = \hat{a}^\dagger 0 = 0.$$

There is no more state. There is no more system. There is nothing.

*And*

*from nothing, nothing comes.*

This property, that the annihilation operator acting on the vacuum destroys the system entirely, is extremely useful in calculation. If you are dealing with Fermionic or Bosonic systems such as $|n_1n_2\cdots n_j\cdots\rangle$ and you need to figure out what the normalization factor $C$ needs to be to make sure probabilities sum to 1, then the only way to figure this out is to turn it in to the product of creation operators,

$$\begin{align}

|C|^2&=\langle n_1n_2\cdots n_j\cdots|n_1n_2\cdots n_j\cdots\rangle\\

&= \langle 0|\cdots (\hat{a}_j)^{n_j}\cdots(\hat{a}_2)^{n_2}(\hat{a}_1)^{n_1}(\hat{a}_1^\dagger)^{n_1}(\hat{a}_2^\dagger)^{n_2} \cdots(\hat{a}_j^\dagger)^{n_j}\cdots |0\rangle,

\end{align}$$

and use commutation relations between the operators to express it as a ginormous sum of terms, and eliminate all of the terms that give an annihilation operator on the vacuum. It is so tedious and painful, but thankfully made much simpler by the fact that annihilation on the vacuum gives

*nothing.*

It is true that in this vacuum state, in QED considerations, the vacuum will undergo a very, very, very complicated process of pair creation and annihilation, a single term of which may take months to solve explicitly. So people who appreciate science are correct in stating that at the quantum mechanical level, yes, the quantum vacuum is unstable and goes through processes in which particles are randomly created and destroyed. They are merely mistaken in calling the quantum vacuum nothing.

So the real question should be, then,

*why is there a quantum vacuum rather than nothing?*

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