I have studied the annihilation and creation operators and number operator $N$ in relation with the simple harmonic oscillator that is governed by: $\ H = \hbar\omega(N+ \frac{1}{2})$.
I don't understand the relation between the harmonic oscillator and, for example, this Hamiltonian $\ H = \hbar\omega_0a^{\dagger}a+\hbar\omega_1a^{\dagger}a^{\dagger}aa $ that I have found in an example in a lecture notes. They calculate the energies of this system.

They use the annihilation operator that is defined from the simple harmonic oscillator to solve that system. what is physically this system? why I can use the SHO to calculate the energies? I feel that I am confused with the a operator. I thought that it was defined from the Hamiltonian of the simple harmonic oscillator ,,,, isn't it ?

thanks in advance please see the attached corresponding images

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    $\begingroup$ First, do you understand the meaning of the Hamiltonian of a system? $H=\hbar\omega(N+\frac{1}{2})$ is the reduced form of the Hamiltonian of a harmonic oscillator or electromagnetic field. $\endgroup$ – TBBT Sep 15 '15 at 6:07
  • $\begingroup$ Thanks for your response. Yes i understand , N here is the number operator equal to N=aa+ , but here they use the simple harmonic oscillator to solve this problem? why the a, a+ and N are the same ? i thought that those operators are ONLY for the simple harmonic oscillator. $\endgroup$ – Mati Sep 15 '15 at 6:13
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    $\begingroup$ That is not what I meant. $\endgroup$ – TBBT Sep 15 '15 at 6:14
  • $\begingroup$ TBBT please can you explain? thanks in advance $\endgroup$ – Mati Sep 15 '15 at 6:17
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    $\begingroup$ I am writing an answer as we speak. $\endgroup$ – TBBT Sep 15 '15 at 6:18

Let's quickly review the quantum harmonic oscillator. We have a single particle moving in one dimension, so the Hilbert space is $L^2(\mathbb{R})$: the set of square-integrable complex functions on $\mathbb{R}$. The harmonic oscillator Hamiltonian is given by

$$H= \frac{P^2}{2m} + \frac{m\omega^2}{2}X^2$$

where $X$ and $P$ are the usual position and momentum operators: acting on a wavefunction $\psi(x)$ they are $X \psi(x) = x\psi(x)$ and $P \psi(x) = -i\hbar\ \partial \psi / \partial x$. Of course, we can also think of them as acting on an abstract vector $|\psi\rangle$.

By letting $P \to -i\hbar\ \partial/\partial x$ we could solve the time independent Schrödinger equation $H \psi = E \psi$, but this is a bit of a drag. So instead we define operators $a$ and $a^\dagger$ as in your post. Notice that the definition of $a$ and $a^\dagger$ has nothing whatsoever to do with our Hamiltonian. It just so happen that these definitions are convenient because the Hamiltonian turns out to be $\hbar \omega (a^\dagger a + 1/2)$.

For convenience we define the number operator $N = a^\dagger a$; at this stage number is just a name with no physical interpretation. Using the commutation relation $[a,a^\dagger] = 1$ and some algebra we notice that $N$ has a nondegenerate spectrum given by the natural numbers. In other words, the eigenvalues of $N$ are $\{0,1,2,\dots\}$, and to each eigenvalue $n$ there corresponds a single state $|n\rangle$ with $N|n\rangle = n |n\rangle$. Notice that, again, $N$ is independent of our Hamiltonian. However, because the Hamiltonian turns out to be $\hbar \omega (N+1/2)$ we immediately know that the states $|n\rangle$ are its eigenvectors, with energies $\hbar \omega (n + 1/2)$.

Now you are given a different Hamiltonian. The Hilbert space is still exactly the same, and so are $a$, $a^\dagger$ and $N$, because their definition had nothing to do with the original Hamiltonian. You can still use their properties to find energies, eigenvectors, and so on. The states $|n\rangle$ are still the eigenstates of $N$, though a priori they might not be eigenstates of the new $H$ (exercise 31 asks you to prove that they in fact are eigenstates of the new $H$). The important point here is that operators are (usually) defined independently of the Hamiltonian. They characterize the physical system. After all, you know that there are operators $X$ and $P$, and you have no qualms about using them with different Hamiltonians. The Hamiltonian gives the energy and the time evolution, but the observables and related operators are independent of your choice of Hamiltonian.

About the physical interpretation... exercise 31 asks you to prove that $H=\hbar\omega_0 N + \hbar \omega_1 (N^2-N)$; notice that we have gotten rid of $\hbar\omega_0 /2$ since it is just a constant. I would usually expect $\omega_1$ to be smaller than $\omega_0$ so this is a small perturbation (for small $n$ at least), but we don't really care about that right now. You can see that $|n\rangle$ are still the eigenstates of the Hamiltonian; all we did is shift the energies by an amount $\hbar \omega_1 (N^2-N)$.

  • $\begingroup$ Just a comment, i was born in BSAS :) $\endgroup$ – Mati Sep 19 '15 at 6:13

The Annihilation and Creation Operators are NOT specific to any particular Hamiltonian. They are defined through X and P which are position and momentum operators for ANY system that you are studying. It just so turned out that these operators facilitate the calculations in SHO specifically. (In fact, I believe these operators originated from Dirac's study of SHO)

Your new Hamiltonian is different from SHO formally, but really the OPERATORs have NOT Changed! N is still the same N! Since arguments of H ONLY involve one operator N and arbitrary scalars, H and N commute and thus eigenstates of N are also eigenstates of H. So if H acts on an eigenstate of N, the output is just the energy. Here notice that although n no longer COUNTS the excitation level of a single oscillator, it is still a sufficiently good label for the energy levels of your new system.

To summarize, N is just a tool. You could have expressed your Hamiltonian in terms of other "ingenious" (or stupid) operators and solve the problem in terms of eigenstates of those operators! N is by NO MEANS unique. It can exist on its own without reference to SHO. I guess that is the biggest point to make.

  • $\begingroup$ Yes, this $a^\dagger$ is related with the harmonic oscillator! The relation is very close as it creates eigenstates of this Hamiltonian. (Of course you can write other Hamiltonians in terms of these operators, but those Hamiltonians are in general not diagonalized by them). As you are generally interested in diagonalizing the Hamiltonian (as you can then read off the spectrum, ...) this counts for a lot. For another example, the operators creating Cooper pairs are related to the BCS Hamiltonian, they create the fundamental excitations of the system (as they diagonalize the Hamiltonian). $\endgroup$ – Sebastian Riese Sep 16 '15 at 18:36
  • $\begingroup$ Hi Sebastian Riese, I guess my answer was somewhat misleading. I am not saying that N is not associated with SHO. All I am saying is that it CAN exist on its own! And Mati's original confusion was "why can we apply results of SHO to another Hamiltonian that seems unrelated." My answer is, we are not applying results of SHO, but rather just results derived from the operator N, which is universally applicable. $\endgroup$ – Zhengyan Shi Sep 17 '15 at 3:35
  • $\begingroup$ Your edit fixes the issues with the answer, so +1. $\endgroup$ – Sebastian Riese Sep 17 '15 at 10:58
  • $\begingroup$ @SebastianRiese , Zhengyan Shi thank you very much! $\endgroup$ – Mati Sep 19 '15 at 6:14

Hamiltonian of different systems is different. Hamiltonian of different systems do not have to be related. You are taught to study the dynamics of the harmonic oscillator in Quantum Mechanics is because it related to the electromagnetic field quantisation, which you will learn later on. Basically said, the Hamiltonian of the harmonic oscillator is the same with light field.

Anyway, the exercise, which you are presented us, is asking you to study a quantum system with a total Hamiltonian of $H=\hbar\omega_{0}a^{\dagger}a+\hbar\omega_{1}a^{\dagger}a^{\dagger}aa$. Therefore, you should be using this one to solve the exercise.

The annihilation operator is also given in the exercise. In case you wonder, that is the same as the relation you find in the harmonic oscillator system. I'll show you, $$ a=\sqrt{\frac{m\omega}{2\hbar}}x+i\sqrt{\frac{1}{2m\omega\hbar}}p $$ and its Hermitian conjugate, $$ a^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}x-i\sqrt{\frac{1}{2m\omega\hbar}}p $$ From these to relations, you can be able to find $x$ and $p$.

I can tell you are confusing with the fundamental concept of Quantum Mechanics. I suggest you to go back and read more materials. If you are not clear on the operators, as you said, you should learn more about Mathematics for Quantum Mechanics.

I think this exercise is not very hard. Since you only asked us to clarify the confusion you have with the Hamiltonian, I won't go any further that this. If you find the exercise challenging, please ask us for specific assistant. Please note that this site is not for homework-solving but homework-helping. If you find my answer not helpful, please comment below.

  • $\begingroup$ Thanks. I understand that i need to use this Hamiltonian to solve the problem. I went back and read a lot before i wrote this question. Specifically i learned that a and a+ operators derive from the hamiltonian of the simple harmonic oscillator: H = p^2/2m + mw^2x^2/2 so why i can use them without adapting them to this problem ? $\endgroup$ – Mati Sep 15 '15 at 6:49
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    $\begingroup$ You can. Please read my edit answer. $\endgroup$ – TBBT Sep 15 '15 at 7:04
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    $\begingroup$ Oh! One more thing, please learn how to write appropriate math equation. You can use latex code. It works in here. $\endgroup$ – TBBT Sep 15 '15 at 7:06
  • $\begingroup$ Thanks, i am sorry but i know that i am confused. i know the annihilation operator is given there. But why this operator can be used with this hamiltonian that is very different from the simple oscilator harmonic ? thanks in advance $\endgroup$ – Mati Sep 15 '15 at 7:26
  • $\begingroup$ I find this answer to be slightly condescending. Specifically, the last paragraph: "I think this exercise is not very hard". Every exercise is extremely hard to someone who doesn't understand it. $\endgroup$ – Javier Sep 16 '15 at 21:11

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