# Can we solve the particle in an infinite well in QM using creation and annihilation operators?

The particle in an infinite potential well in QM is usually solved by easily solving Schrodinger differential equation. On the other hand particle in the harmonic oscillator oscillator potential can be solved elegantly algebraically using the creation and annihilation operators to find its spectrum.

Is it possible to do the particle in a box problem using creation and annihilation operator and how?

• Well, you can solve it in matrix mechanics, and susy QM which comes closer to creators and annihilators. Jun 30, 2018 at 17:58

Probably not, but you could always define $$a=\sum_n \sqrt{n}\left|n-1\right>\left which acts on the eigenstates, and write $$H=E(a^\dagger a)^2.$$ This doesn't help you solve the problem to begin with, but it does let you write $$H$$ with $$a,a^\dagger$$ once you know the spectrum.

Not really...

Firstly, "creation" and "annihilation" have more sense in the second quantization language for many body systems, while both SHO and infinite well discussed here are single body problems.

Secondly, indeed you can use certain operator method for SHO, which looks like "creation"/"annihilation" operators. They actually have the meaning of "creation"/"annihilation" if you forget the oscillator but deem the ground state as certain vacuum and excited states as states with more "particles" carrying the same amount of energy.

Thirdly, the existence of operator method for SHO is associated with the fact that the energy spacing of arbitrary two adjacent states is a constant, which allows you to write the Hamiltonian as a linear function of $$a^{\dagger}a$$, i.e. $$(H-H_0)\propto a^{\dagger}a$$. And in practice, this is because the Hamiltonian of SHO in the 1-dim space is purely a second order one (in the operator sense) which allows you to do a square-decomposition, while an infinite well problem is not: the potential energy vanishes outside the well, which is a purely non-linear constraint. For sure, as Sal suggested, you may try a decomposition of a forth order (in the operator sense) Hamiltonian (we know that from the result of the infinite well problem), which is not easy at all -- to find the explicit form of the operator....

One has to distinguish creation/annihilation operators and raising/lowering operators.

Second quantization is the method for introducing creation and annihilation operators for any system. This is however not what is meant in the OP, which really talks about raising and lowering operators. These again can be introduced for any system, e.g., using the prescirpition in the answer by @SalElder. The special thing about harmonic oscillator is that the two pairs of operators have the same matrix representation (and consequently the result of their action on the eigenstates, commutation relation, etc.) Moreover, for a HO they can be represented in simple differential form, which is not possible for an arbitrary potential.

Another useful example is a two-level system, also called sometimes *fermionic harmonic oscillator. Given a Hamiltonian $$H=\begin{bmatrix}E_1&0\\0&E_2\end{bmatrix}$$ we have raising and lowering operators $$a=\begin{bmatrix}0&0\\1&0\end{bmatrix}, a^\dagger=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ with commutation relation $$aa^\dagger + a^\dagger a=1.$$