How do you think of quantum states? I like to think of their order of symmetry. It is a very hard job to characterize the order of symmetry of quantum states in general; but there’s a special subclass of quantum states that exhibit very clear symmetry. They are called bosonic states.

A brief introduction

An order-$N$ state is a $+1$ eigenstate of the discrete rotation operator $\hat{R}_N$, \[ \hat{R}_N = \exp(i\dfrac{2\pi}{N}\hat{n}), \] where $\hat{n}=\hat{a}^\dagger\hat{a}$. As usual, $\hat{a}$ ($\hat{a}^\dagger$) is the annihilation (creation) operator.

Now, it’s natural to write down these symmetrical states in terms of Fock number states, $\hat{n}$ (these are the eigenstates of $\hat{n}$). Because of their inherent symmetry, their Fock space structure will have something special. Let us see what that thing is.

Let us assume this arbitrary state has $N$-fold symmetry, then it follows that the state has support only on $N$-th number states in Fock space.

This link can logically derived as follow. Let us have an arbitrary state $|\psi\rangle = \sum_n a_n|n\rangle$. If such a state is a $+1$ eigenstate of $\hat{R}_N$, then \[\begin{align} e^{i(2\pi/N)\hat{n}} \sum_n a_n |n\rangle = \sum_n e^{i(2\pi/N)n}a_n|n\rangle = \sum_n a_n |n\rangle \end{align}\] For this to be true, $2\pi n/N = 2k\pi$, or $n=kN$.

No symmetry at all: coherent state

Now, the most trivial case is when $N=1$. The state has an order-1 symmetry, which means it has no symmetry at all! Think about a coherent state $|\alpha$. They’re the eigenstates of the annihilation operator $\hat{a}$. If you were to write them down in the Fock basis, \[ |\alpha\rangle = e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\dfrac{\alpha^n}{\sqrt{n!}}|n\rangle, \] you’ll see that it is a state with support on an infinite number of number states $|n\rangle$. But the cardinality of this infinite set is definitely going to be smaller than that of a $N>1$-order state. Now, let us plot the state in the phase space. Phase space is the dual space of Fock space, and by plotting the so-called Wigner function of the states, we can appreciate their symmetry a bit more.

import qutip as qt
import matplotlib.pyplot as plt

dim_fock = 100
a = qt.destroy(dim_fock)
alpha = 1.5 * np.exp(1j*np.pi/4) # amplitude

vacuum_state = qt.fock(dim_fock, 0)
D_op = (alpha*a.dag()-np.conj(alpha)*a).expm()
coherent_state = D_op * vacuum_state

fig, ax = plt.subplots(ncols=1, figsize=(5, 5))
qt.plot_wigner(coherent_state, fig=fig, ax=ax)

For readers with background in elementary quantum optics, you can tell that the Python code above created a coherent state by displacing a vacuum state $|0\rangle$. The state is displaced to the north-east of the phase space (of an angle $\pi/4$).

Now you can do all sort of crazy things with the space, but its inherent symmetry won’t change. For example, you can squeeze the state, $|\alpha’\rangle = \hat{S}(\beta)\hat{D}(\alpha)|0\rangle$, where \[ S(\beta)=\exp\left(\dfrac{\beta}{2}\hat{a}^2-\dfrac{\beta^*}{2}(\hat{a}^\dagger)^2\right)\quad\text{and}\quad D(\alpha)=\exp\left(\alpha\hat{a}^\dagger -\alpha^{*} \hat{a}\right) \] Note that both $\alpha=|\alpha|e^{i\phi_{\alpha}}$ and $\beta=|\beta|e^{i2\phi_{\beta}}$ (note the factor 2) are complex-defined. Their amplitudes define the amount at which the state is being “displaced” or “squeezed”, while the argument, i.e. $\phi_\alpha$ and $\phi_\beta$ tells about the “direction of displacement” and “axis of squeezing”. For the latter case, the squeezing effect is happening perpendicular to the axis. For example, we can displace the coherent state from the vacuum state to the same $\pi/4$ angle, then squeeze it along the $\pi/4+\pi/2$ angle. It should look something like this.

dim_fock = 100
a = qt.destroy(dim_fock)
alpha = 1.5  * np.exp(1j*np.pi/4) # cat size

vacuum_state = qt.fock(dim_fock, 0)

# Displacement
D_op = (alpha*a.dag()-np.conj(alpha)*a).expm()
coherent_state = D_op * vacuum_state

epsilon = 0.75 * np.exp(1j*(2*(np.pi/4+np.pi/2)))
# Squeezing
S_op = (np.conj(epsilon)/2*a**2 - epsilon/2*(a.dag())**2).expm()
squeezed_coherent_state = S_op * coherent_state

fig, ax = plt.subplots(ncols=1, figsize=(5, 5))
qt.plot_wigner(squeezed_coherent_state, fig=fig, ax=ax)

Squeezing is however very costly. If you intuit it, you see that it requries a large number of Fock number states to accommodate itself, meaning a large average number of photon $\bar{n}$. State-of-the-art experiments have realized $\bar{n}=100$ photon states already, but I suppose there’s a catch about it. An experimental colleague of mine suggested a maximum of $\bar{n}=6$.

Coming back to the original point. Regardless of how many crazy actions that you’ve done to the state, it still requires support on all possible Fock number states. In other words (and should be clear from the Wigner plot), the state still has no symmetry at all!

$N=2$: Schrodinger cat states

A Schrodinger cat state is a superposition of two coherent states $\mathcal{N}(|\alpha\rangle + |-\alpha\rangle)$,

dim_fock = 100
a = qt.destroy(dim_fock)

def coherent_state(size, angle):
    '''
    Create a coherent state
    '''
    vacuum_state = qt.fock(dim_fock, 0)
    phi_alpha = angle
    alpha_val = size * np.exp(1j*phi_alpha)

    return (alpha_val*a.dag()-np.conj(alpha_val)*a).expm() * vacuum_state

cat = (coherent_state(2, 0)+coherent_state(2, np.pi)).unit()

fig, ax = plt.subplots(ncols=1, figsize=(5, 5))
qt.plot_wigner(cat, fig=fig, ax=ax)

It is quite intuitive to see that the cat state is invariant under a $\pi$ rotation. This also means that the state is an $+1$ eigenstate of $\exp(i\pi\hat{n})$. Is this true? Well, recall that $|\alpha\rangle$ has support on all Fock number states, it also follows that $|-\alpha\rangle$ also has support on all Fock number states. However, consider a superposition of them, \[ \begin{align} |\alpha\rangle +|-\alpha\rangle \propto \sum_{n} a_n|n\rangle + (-1)^n a_n|n\rangle, \end{align} \] it is clear that for $n$ even, the two coherent states add constructively, while for $n$ odd, they sum up to zero. This simply means that $|\alpha\rangle +|-\alpha\rangle$ has support on only even number of Fock number states, i.e. $n=2k$ only. This means $|\alpha\rangle +|-\alpha\rangle=\sum_{k=0}^\infty a_{2k}|2k\rangle$, and therefore being a $+1$ eigenstate of the discrete rotation $\pi$. Because $|\alpha\rangle +|-\alpha\rangle$ has only support on even Fock number states, they are typically referred to as even cats, as opposed to odd cats, $|\alpha\rangle - |-\alpha\rangle$.

even_cat = (coherent_state(2, 0)+coherent_state(2, np.pi)).unit()
odd_cat = (coherent_state(2, 0)-coherent_state(2, np.pi)).unit()

fig, (ax1, ax2) = plt.subplots(ncols=2, figsize=(10, 5))
qt.plot_wigner(even_cat, fig=fig, ax=ax1)
qt.plot_wigner(odd_cat, fig=fig, ax=ax2)

You can tell the difference of them from the interfering pattern in the middle. While an even cat has a blue nose, an odd cat has a red nose. What a nice feature!

$N=3$: Binomial states

Ok. You can generalize the cat state to an arbitrary order of symmetry. We’ve done with the $N=2$ case. I don’t want to do the $N=3$ case for cat states; jesus, you cannot have a three-eyed cat: that would be absolutely terrifying.

For those who still wanna do it (you pervert). Here’s how. Very simple: since they are states with $3$-fold symmetry, they are invariant under $2\pi/3$ rotation. What does it mean? It means that $e^{i(2\pi/3)\hat{n}}|\psi\rangle = +1|\psi\rangle$, i.e. $|n\rangle=|3k\rangle$. They are supported by only the $3k$ Fock number states. That’s the Fock space picture. If you turn into the phase space picture, things get much more intuitive. You start with a coherent state, then you displace it $2\pi/3$, then you displace it $2\pi/3$ again. Then you superimpose them together. What you would have is a state with support on only $|3k\rangle$ Fock number states.

I am going to introduce the binomial states. They are unambiguously defined as

\[ \begin{align} |\text{bin}\rangle =
\end{align} \]