# What does coherent superposition mean?

1. There is only one coherent state: $$|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle$$

2. Also, a pure state does not mean a coherent state.

But what does one mean when they talk about a coherent superposition of ground and excited state:

$$c\left|g\right> + d \left|e\right>.$$ Drawing on the bloch sphere, it is on the surface, but so does just being a superposition. But what does one mean and what does it imply?

I also see a Phys.SE post: 262052

The word "coherent" is used in Physics in a rather sloppy way. Your first state is a linear combination of harmonic oscillator eigenvectors that turns into a gaussian in momentum/position representations. In a more general background, a coherent state is just a state where coherences (off-diagonal terms in the density matrix) are non-zero, which means the state can skipp from one stationary state to another.

Now, a coherent superposition is quite like a coherent state: a superposition is said to be coherent if there's an observable that, if applied to one state, can turn it into another also present in the superposition. As an example, consider the $z$-axis spin up and spin down states of the electron in a Stern-Gerlach experiment. Then there is one spin operator, namely $S_x$, that can turn one into the other. This means they form a coherent superposition. As a counter-example consider the ground and the first excited states of the harmonic oscillator: the creation operator can turn the former into the latter, but this operator is not an observable. The superposition is a non-coherent one, meaning that off-diagonal elements in the density matrix are irrelevant to the problem at hand.

• Thank you. But could you explain the last sentence to me? If I am in the ground state of harmonic oscillator, it is diagonal in the fock state basis, so does being in first excited states. Also, I remember that $\left<\hat{a}\right>$ mean the amplitude of radiation even though it's not hermitian.
– diff
Jun 24, 2016 at 3:38
• Remember: I'm in a superposition of the ground and first excited states. If this example is confusing consider the superposition of two position eigenstates belonging to two particles in different extremes of the Universe. You can't turn one into the other, so the superposition is non-coherent. Jun 24, 2016 at 3:43
• Oh understand now. Can I say the superposition of the ground and first excited states of SHO is non-coherent because $a$ can turn one to another but then $a$ cannot turn it back?
– diff
Jun 24, 2016 at 3:51
• Not because of that. It is non-coherent because there's no observable that can turn one into the other. Anihilation and creation operators are not observables: you can't measure them. What you can do is to build products with them that are observables, but these products cannot turn ground states into excited states, since the same number of anihilation and creation operators must be present, otherwise the resulting operator is still non-hermitian. Jun 24, 2016 at 3:58
• Should that observable operator be able to turn it back or be unitary?
– diff
Jun 24, 2016 at 4:06

Coherence has many faces. See Quantum coherence - what is it's definition?

First state refers to a state of the field. Originally Glauber developed this formalism to give quantum description of laser fields. Later it was adopted in other fields.

Second state refers to the state of a 2-level system (in your case). You can also get superposition states with incoherent light, but those are not very useful. The word coherent is used to describe the superposition states that you create with coherent fields. Usually you don't have to quantize the field, but can still work in what is know as "semi-classical" approximation. This means that you have aclassical field and a quantized system. This is the more common experimental situation that one encounters. The field dealing with this is called "coherent control". Check out P.L Knight or N.V. Vitanov, they have planty of papers there.