Ladder operators and creation & annihilation operators - different between $a$, $b$ and $c$ [closed]

Question

Usually, the ladder operator denoted by $a$ and $a^\dagger$. In some case, people talk about the creation operator and denote it by $c$ and $c^\dagger$. Recently I see another notation, $b$ and $b^\dagger$, and I don't understand why in each case we use with different letters. I believe that there a reason for this.

I try to understand the different between the ladder and creation operators, I thought maybe this what make the difference, but I really confused. Maybe the difference is due to the difference between the ladder operators and the creation/annihilation operators? If there is any difference at all..

Someone can tell me the different between the notation $a$ or $b$ or $c$? and if this connect to the difference between the ladder and creation operators? If it's really caused by the difference, then I'd love to clarify the difference between the two, because I'm not sure I understand the difference.

Answer 1

When we talk about the quantum harmonic oscillator, $a$ and $a^{\dagger}$ are usually used.($a$ may stands for "annihilation".) Note that we have not discuss second quantization here, so $a$ and $a^{\dagger}$ are not field operators.

In the area of many-body physics, we usually use field operators to write Hamiltonian. For bosonic system, we tend to write $b$ and $b^{\dagger}$ where I believe $b$ stands for "boson" here. For fermionic system, both $c,c^{\dagger}$ and $f, f^{\dagger}$ are widely used. ($f$ may stand for fermion while I don't know what $c$ stands for.)

Note that letter itself is not important, you can use whatever letter to represent the annihilation operator $a, b, c, f, ...$. All you have to focus is their (anti-)commutation relation.

Answer 2

Someone can tell me the different between the notation $a$ or $b$ or $c$?

The name of these operators doesn't matter. The important thing are the commutation relations between these two operators (let us call them $a$ and $a^\dagger$) and a third operator (let us call it $n$):

$$\begin{align} [a,n]&=-ca \\ [a^\dagger,n]&=+ca^\dagger \end{align}$$

When these relations are satisfied, then

• $a$ is a step-down operator
  (meaning $a|n\rangle$ is either the null-vector or an eigenvector of $n$ with eigenvalue $n-c$),
• $a^\dagger$ is a step-up operator
  (meaning $a^\dagger|n\rangle$ is either the null-vector or an eigenvector of $n$ with eigenvalue $n+c$)

Maybe the difference is due to the difference between the ladder operators and the creation/annihilation operators? If there is any difference at all.

Step-up/down operators, ladder operators and creation/annihilation operators all mean the same thing. But the names "creation/annihilation operators" seem to be the preferred ones, if $n$ is actually the particle-number operator.