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It’s commonly said that a Quantum Field is a bunch of Harmonic Oscillators. To see what’s really going on, let’s start with a particle in a box:

$V(x)=\left\{\begin{array}{ll}{0,} & {0<x<L} \\ {\infty,} & {\text { otherwise }}\end{array}\right.$Inside the box, the Schrodinger equation is:

$i \hbar \frac{\partial}{\partial t} \psi(x, t)=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} \psi(x, t)+V(x) \psi(x, t)$For a box of size $L$, we have the following boundary conditions:

$\psi(0)=\psi(L)=0$To obtain $\psi(x)$, just guess an exponential. The solution is:

$\begin{aligned} \psi_{n}(x)&=\sqrt{\frac{2}{L}} \sin \left(k_n x\right)\\ &=\sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right) \end{aligned}$where

$k_n=\frac{n \pi}{L}$Using the De Broglie relation, we note that momentum is now quantized:

$p_n=\hbar k_n = \frac{n \pi \hbar}{L}$the energy is, for $n=0, \pm 1, \pm 2, \ldots$

$E_{n}=\frac{\hbar^{2} \pi^{2} n^{2}}{2 m L^{2}} = \frac{p_n^{2}}{2 m}$However, the momentum is not a good quantum number since it does not commute with the Hamiltonian. due to the $V(x)$ term. This causes some technical issues since our momentum states aren’t eigenstates. We can simply just get rid of the potential $V(x)$ and impose periodic boundary conditions. The new Schrodinger equation is:

$i \hbar \frac{\partial}{\partial t} \psi(x, t)=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} \psi(x, t)$With periodic boundary conditions:

$\psi(0)=\psi(L), \quad \psi^{\prime}(0)=\psi^{\prime}(L)$which imply

$\psi(x+L)=\psi(x)$And so the solution is

$\psi_{n}(x)=\frac{e^{i k_{n} x}}{\sqrt{L}}$where

$k_{n}=\frac{2 \pi n}{L}, \quad n=0, \pm 1, \pm 2, \ldots$And the energy is, for $n=0, \pm 1, \pm 2, \ldots$

$E_{n}=\frac{\hbar^{2} \pi^{2} n^{2}}{2 m L^{2}} = \frac{p_n^{2}}{2 m}$Now we can see that momentum is now an eigenstate:

$\begin{aligned} \hat{p} \psi_{n}(x)&=-i \frac{\partial}{\partial x} \psi_{n}(x)=p_{n} \psi_{n}(x) \\ p_{n}&=\hbar k_{n}=\frac{2 \pi \hbar n}{L} \end{aligned}$The results are the same as before, except now our **new momentum** $p_n$ **is double of the old momentum**, and the new momentum states are now eigenstates.

Let’s now investigate a non interacting multi particle system. Let’s also rename the letter $n \to p_m$, so we can free up the letter $n$ for something new.

Suppose the first particle is in state $|p_1\rangle$, and the second particle is in state $|p_2 \rangle$. Then as usual, we describe the two particle system with:

$|p_1 p_2 \rangle$Apply the momentum and energy operators we get:

$\hat{p}\left|p_{1} p_{2}\right\rangle=\left(p_{1}+p_{2}\right)\left|p_{1} p_{2}\right\rangle$$\hat{H}\left|p_{1} p_{2}\right\rangle=\left(E_{p_{1}}+E_{p_{2}}\right)\left|p_{1} p_{2}\right\rangle$and we can write the total energy is given by:

$\sum_m n_{p_m} E_{p_m}$where $n_{p_m}$ is the total number of particles in the state $|p_m\rangle$.

In single particle quantum mechanics we might write:

$|ABC\rangle$where

- $A$ is the state of the first particle,
- $B$ is the state of the second particle, and
- $C$ is the state of the third particle.

So if

- $A = p_1$
- $B = p_2$
- $C = p_1$

Then we would write

$|{ABC}\rangle=\left|p_{1} p_{2} p_{1}\right\rangle$This would make sense if particles were distinguishable, but they are not. The laws of indistinguishable particles are:

There are only two types of particles in the Universe:

**fermions**and**bosons**.Looking at the wave functions in the position basis:

We can define the exchange operator:

$\hat{P}^2 |AB\rangle =\hat{P}|BA\rangle = |AB\rangle$Since $\hat{P}^2 = 1$, this leaves:

$\hat{P}\left|AB\right\rangle=\lambda\left|BA\right\rangle$We see that $\lambda^2 = 1$ and so $\lambda = \pm 1$.

- If you swap two identical bosons nothing happens: $|AB\rangle = |BA\rangle$. This is $\lambda = 1$.
- If you swap two identical fermions you get a minus sign: $|AB\rangle = -|BA\rangle$. This is $\lambda = -1$ and is known as the
**Pauli exclusion principle**.

We can make a change of notation to accommodate this. This new notation is called **Second quantization**, and the old notation is called **First quantization**.

Before in first quantization, we would have said: ** Which particle is in which state?** But according to the rules above, this is nonsense. In second quantization, we ask:

We now list the number of identical particles in each quantum state:

Particles | Old notation | New notation |
---|---|---|

0 | $\|0\rangle$ | $\|00\rangle$ |

1 | $\|q_1\rangle$ | $\|10\rangle$ |

1 | $\|q_2\rangle$ | $\|01\rangle$ |

2 | $\|q_1q_1\rangle$ | $\|20\rangle$ |

2 | $\|q_1q_2\rangle$ | $\|11\rangle$ |

2 | $\|q_2q_2\rangle$ | $\|02\rangle$ |

3 | $\|q_1q_1q_1\rangle$ | $\|30\rangle$ |

So, now when we apply the Hamiltonian, we get:

$\hat{H}\left|n_{1} n_{2} n_{3} \ldots\right\rangle=\left[\sum_{m} n_{p_{m}} E_{p_{m}}\right]\left|n_{1} n_{2} n_{3} \ldots\right\rangle$The computation is now very natural in second quantization, we just count the number of particles and multiply by the energies:

$\sum_{m} n_{p_{m}} E_{p_{m}}=n_1 E_{p_1}+n_2 E_{p_2}+\ldots$The quantum harmonic oscillator has energy

$E = \left(n + \frac{1}{2}\right)\hbar \omega$If we throw away the zero point energy, we can write this as:

$E = n\hbar \omega$For $N$ independent oscillators, we have the total energy is:

$\sum_{k=1}^{N} n_{k} \hbar \omega_{k}$The total energy of the oscillator looks exactly the same as the many body system. These two systems are special in that their ** energy eigenvalues are linearly spaced**. This is why we can associate the energy of

So having $2 \hbar \omega$ of quanta can mean **either** we have two photons with energy $\hbar \omega$, or a single photon with energy $2 \hbar \omega$.

Harmonic Oscillator | Identical Particles |
---|---|

Quanta | Particles in momentum states |

$k$th Oscillator | $m$th momentum mode |

$\sum_{k=1}^{N} n_{k} \hbar \omega_{k}$ | $\sum_{m=1}^{N} n_{p_{m}} E_{p_{m}}$ |

Note that we’ve done this **mathematically**: we haven’t physically replaced identical particles with harmonic oscillators.

States defined by the quantum harmonic oscillator come with the raising $\hat{a}^{\dagger}$ and lowering $\hat{a}$ operators, allowing us to raise and lower states in the **harmonic oscillator** system. Here $|n\rangle$ is the state of the single harmonic oscillator:

The wavefunctions $\psi(x)$ of the Harmonic Oscillator are quite complicated, but if you are curious about what the wave functions of the Harmonic Oscillator look like they are here:

The **raising operator** allows us to raise states:

The **lowering operator** allows us to lower states:

This is just like climbing a ladder! For this, these are also called **ladder operators**.

*Image: David J. Griffths, Introduction to Quantum Mechanics pg.61, Figure 2.5*

Since the quantum harmonic oscillator is the same as the identical particles system, we can use the raising $\hat{a}^{\dagger}$ and lowering $\hat{a}$

$\left|n_{1} n_{2} \cdots\right\rangle=\prod_{k} \frac{1}{\sqrt{n_{k}!}} \left(\hat{a}_{k}^{\dagger}\right)^{n_{k}}|0\rangle$And say that $\left|n_{1} n_{2} \cdots\right\rangle$ is the state of the many body system! To create a particle with momentum $k$, just use the raising operator $\hat{a}^\dagger_k$!

Some physicists will tell you that particles *are* operators. This can be confusing, but just remember what they act on: the vacuum state $|0\rangle$.

- In second quantization, we can create and destroy particles in a state using the raising and lowering operators of the harmonic oscillator.
- Right now our box is finite, so we still have to take the infinite limit.
- We’re still in momentum space so we have to Fourier transform this later.

Suppose that we had two states as such:

$\hat{a}_{p_{1}}^{\dagger}|0\rangle=|10\rangle, \quad \hat{a}_{p_{2}}^{\dagger}|0\rangle=|01\rangle$Let’s add one more particle:

$\hat{a}_{p_{2}}^{\dagger} \hat{a}_{p_{1}}^{\dagger}|0\rangle \propto|11\rangle, \quad \hat{a}_{p_{1}}^{\dagger} \hat{a}_{p_{2}}^{\dagger}|0\rangle \propto|11\rangle$Both expressions are proportional to the same state, so they must be proportional:

$\hat{a}_{p_{1}}^{\dagger} \hat{a}_{p_{2}}^{\dagger}=\lambda \hat{a}_{p_{2}}^{\dagger} \hat{a}_{p_{1}}^{\dagger}$We said before that $\lambda = \pm 1$ depending on whether the particles are fermions or bosons, so let’s take a look at each case:

When $\lambda = 1$, we have the particles are bosons. Therefore, we have that the operators commute

$\hat{b}_{p_{2}}^{\dagger} \hat{b}_{p_{1}}^{\dagger}=\hat{b}_{p_{1}}^{\dagger} \hat{b}_{p_{2}}^{\dagger}$and so the commutator, defined as:

$\left[\hat{A}, \hat{B}\right]:=\hat{A}\hat{B}-\hat{B}\hat{A}$is zero, since the operators commute.

$\left[\hat{b}_{i}^{\dagger}, \hat{b}_{j}^{\dagger}\right]=\hat{b}_{i}^{\dagger} \hat{b}_{j}^{\dagger}-\hat{b}_{j}^{\dagger} \hat{b}_{i}^{\dagger}=0$The same is true for the lowering operators:

$\left[\hat{b}_{i}, \hat{b}_{j}\right] =\hat{b}_{i} \hat{b}_{j}-\hat{b}_{j}\hat{b}_{i}=0$To make it exactly the same as the harmonic oscillator, we define:

$\left[\hat{b}_{i}, \hat{b}_{j}^{\dagger}\right]=\delta^i_{j}$That is, if you raise and then lower you it is equal to the Kronecker delta:

$\delta^i_{ j}=\left\{\begin{array}{ll}{0} & {\text { if } i \neq j} \\ {1} & {\text { if } i=j}\end{array}\right.$And so if the particles are bosons it doesn’t matter which order you put particles in the states.

$\hat{b}_{p_{1}}^{\dagger} \hat{b}_{p_{2}}^{\dagger}|0\rangle=\hat{b}_{p_{2}}^{\dagger} \hat{b}_{p_{1}}^{\dagger}|0\rangle=\left|1 1\right\rangle$When $\lambda = -1$, the particles are fermions. The operators then anti-commute:

$\hat{c}_{p_{2}}^{\dagger} \hat{c}_{p_{1}}^{\dagger}=-\hat{c}_{p_{1}}^{\dagger} \hat{c}_{p_{2}}^{\dagger}$And so the anti-commutator, defined as

$\left\{\hat{A}, \hat{B}\right\}:=\hat{A}\hat{B}+\hat{B}\hat{A}$is zero, since the operators anti-commute.

$\left\{\hat{c}_{i}^{\dagger}, \hat{c}_{j}^{\dagger}\right\}=\hat{c}_{i}^{\dagger} \hat{c}_{j}^{\dagger}+\hat{c}_{j}^{\dagger} \hat{c}_{i}^{\dagger}=0$The same is true for the lowering operators:

$\Big\{\hat{c}_{i}, \hat{c}_{j}\Big\} =\hat{c}_{i} \hat{c}_{j}+\hat{c}_{j}\hat{c}_{i}=0$And similarly, we define:

$\left\{\hat{c}_{i}, \hat{c}_{j}^{\dagger}\right\}=\delta^i_{j}$The anti-commutation relations are the exact same as the quantum harmonic oscillator, except that we have anti-commutators instead of commutators.

Note that setting $i = j$ we get that

$\hat{c}_{i}^{\dagger} \hat{c}_{i}^{\dagger}+\hat{c}_{i}^{\dagger} \hat{c}_{i}^{\dagger}=0$and we get the **Pauli Exclusion principle**, which says that each state can only contain a single fermion.

Lastly, we have that

$\hat{c}_{i}^{\dagger} \hat{c}_{j}^{\dagger}|0\rangle=-\hat{c}_{j}^{\dagger} \hat{c}_{i}^{\dagger}|0\rangle$So it matters which order you put the particles into the states!

Our particles are in confined in a box, but a quantum field theory usually needs to work outside a finite box. Just take the length of the box to be infinity, and our Kronecker deltas become Dirac deltas in 1D:

$\delta^i_{j} \rightarrow \delta(x-x_0)$Where the Dirac delta is the continuous version of the Kronecker delta:

$\delta(x)=\left\{\begin{array}{ll}{+\infty,} & {x=0} \\ {0,} & {x \neq 0}\end{array}\right.$The Kronecker delta has the property that:

$\sum_{i=-\infty}^{\infty} v_{i} \delta^i_{j}=v_{j}$The Dirac delta behaves similarly:

$\int_{-\infty}^{\infty} \delta(x-y) f(x) d x=f(y)$And so for 3D space, we have:

$\delta^{(3)}(\mathbf{x} - \mathbf{x_0}) := \delta(x-x_0)\delta(y-y_0)\delta(z-z_0)$There’s one last technique we need which is the **change of basis** in second quantization. So far, our operators work only in momentum space. Here we will give a proof that will work for changing to any basis.

In first quantization we have:

$\psi_{\alpha}(x) \equiv\langle x | \alpha\rangle$For both fermions and bosons we have:

$|\alpha\rangle= \hat{a}_{\alpha}^{\dagger}|0\rangle$where $\hat{a}_{\alpha}^{\dagger}$ is the respective raising operator.

Suppose we want to change to the basis $|\tilde{\alpha}\rangle$.

$\tilde{\psi}_{\tilde{\alpha}}(x) \equiv\langle x | \tilde{\alpha}\rangle$with

$|\tilde{\alpha}\rangle= \hat{a}_{\tilde{\alpha}}^{\dagger}|0\rangle$We use the identity operator:

$I=\sum_{\alpha}|\alpha\rangle\langle\alpha|=\sum_{\tilde{\alpha}}|\tilde{\alpha}\rangle\langle\tilde{\alpha}|$hence

$\hat{a}_{\alpha}^{\dagger}|0\rangle=|\alpha\rangle=\sum_{\tilde{\alpha}}|\tilde{\alpha}\rangle\langle\tilde{\alpha} | \alpha\rangle \\= \sum_{\tilde{\alpha}}\langle\tilde{\alpha} | \alpha\rangle|\tilde{\alpha}\rangle=\sum_{\tilde{\alpha}}\langle\tilde{\alpha} | \alpha\rangle \hat{a}_{\tilde{\alpha}}^{\dagger}|0\rangle$therefore:

$\hat{a}_{\alpha}^{\dagger}=\sum_{\tilde{\alpha}}\langle\tilde{\alpha} | \alpha\rangle \hat{a}_{\tilde{\alpha}}^{\dagger}$Taking the complex transpose/dagger of both sides:

$\hat{a}_{\alpha}=\sum_{\tilde{\alpha}}\langle\alpha | \tilde{\alpha}\rangle \hat{a}_{\tilde{\alpha}}$The inverse transformations are derived very similarly:

$\begin{aligned} \hat{a}_{\tilde{\alpha}}^{\dagger} &=\sum_{\alpha}\langle\alpha | \tilde{\alpha}\rangle \hat{a}_{\alpha}^{\dagger} \\ \hat{a}_{\tilde{\alpha}} &=\sum_{\alpha}\langle\tilde{\alpha} | \alpha\rangle\hat{a}_{\alpha} \end{aligned}$The wave functions follow a very similar relation, the same as in first quantization:

$\psi_{\alpha}(x)=\langle x | \alpha\rangle = \langle x|\left(\sum_{\tilde{\alpha}}\langle\tilde{\alpha} | \alpha\rangle|\tilde{\alpha}\rangle\right)\\=\sum_{\tilde{\alpha}}\langle\tilde{\alpha} | \alpha\rangle\langle x | \tilde{\alpha}\rangle=\sum_{\tilde{\alpha}}\langle\tilde{\alpha} | \alpha\rangle \tilde{\psi}_{\tilde{\alpha}}(x)$and

$\tilde{\psi}_{\tilde{\alpha}}(x)=\sum_{\alpha}\langle\alpha | \tilde{\alpha}\rangle \psi_{\alpha}(x)$Finally, we have the **field operators.** We put a hat on them so we don’t confuse them with wavefunctions. These are the operators that create and destroy particles at a position $x$. Technically physicists use “creation and destruction” interchangeably with “raising and lowering” since they are behave the same mathematically, but I think it sounds cooler and is less confusing if we give different names depending on whether we are in the position or momentum space.

We can get them for free from the equations

$\begin{aligned} \hat{a}_{\tilde{\alpha}}^{\dagger} &=\sum_{\alpha}\langle\alpha | \tilde{\alpha}\rangle \hat{a}_{\alpha}^{\dagger} \\ \hat{a}_{\tilde{\alpha}} &=\sum_{\alpha}\langle\tilde{\alpha} | \alpha\rangle\hat{a}_{\alpha} \end{aligned}$The field operators are the operators that create and destroy particles in the **position basis**. Just like in first quantization, position and momentum space are related by the Fourier transform!

For bosons, we have:

$\begin{aligned}\left[\hat{\psi}(x), \hat{\psi}\left(x^{\prime}\right)\right] &= 0 \\ \left[\hat{\psi}^{\dagger}(x), \hat{\psi}^{\dagger}\left(x^{\prime}\right)\right]&=0 \\ \left[\hat{\psi}(x), \hat{\psi}^{\dagger}\left(x^{\prime}\right)\right] &=\delta\left(x-x^{\prime}\right) \end{aligned}$And for fermions, we have:

$\begin{aligned}\left\{\hat{\psi}(x), \hat{\psi}\left(x^{\prime}\right)\right\} &= 0 \\ \left\{\hat{\psi}^{\dagger}(x), \hat{\psi}^{\dagger}\left(x^{\prime}\right)\right\}&=0 \\ \left\{\hat{\psi}(x), \hat{\psi}^{\dagger}\left(x^{\prime}\right)\right\} &=\delta\left(x-x^{\prime}\right) \end{aligned}$Second quantization differs from first quantization by listing the number of identical particles in each quantum state. It is mainly used to describe many-body physics.

We do this by raising the vacuum state $|0 \rangle$.

There are three commutation rules for Bosons:

$\left[b_{i}^{\dagger}, b_{k}^{\dagger}\right]=0, \quad\left[\hat{b}_{i}, \hat{b}_{k}\right]=0, \\ \left[b_{i}, b_{j}^{\dagger}\right]=\delta\left(x-x^{\prime}\right)$There are three anti-commutation rules for Fermions:

$\left\{c_{i}^{\dagger}, c_{k}^{\dagger}\right\}=0, \quad \Big \{ \hat{c}_{i}, \hat{c}_{k}\Big \}=0, \\ \left\{c_{i}, c_{j}^{\dagger}\right\}=\delta\left(x-x^{\prime}\right)$

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