# The harmonic oscillator - ladder operators

Reading from Griffiths. I have got two questions. First, the halmiltonian operator that used to find the energy eigenvalue in only harmonic oscillator is: $$H={\hbar}w(a_-a_+-\frac{1}{2})$$ and $$H={\hbar}w(a_+a_-+\frac{1}{2})$$ Correct?

Second, to prevent negative energy,we use $a_-{\psi_0}=0$. So we get and take ${\psi_0}$ as ground state. Then the next state is ${a_+}{\psi}$, this is logic because ${a_-}$ and ${a_+}$ are the solution of Hamiltonian. But why suddenly the other state is ${a_+^2}$? Is that just because when we apply Hamiltonian operator on ${a_+^2}{\psi}$ then we can get another eigenvalue so we can say ${a_+^2}{\psi}$ this is another state? How do we get this formula,$${\psi_n}(x)= A{({a_+})^n}{\psi_0}(x)$$

• The basics are explained in this chapter of Wikipedia – Trimok Nov 8 '13 at 10:15

Reading from Griffiths. I have got two questions. First, the halmiltonian operator that used to find the energy eigenvalue in only harmonic oscillator is: $$H=ℏw(a_−a_+−1/2)$$ and $$H=ℏw(a_+a_−+1/2)$$ Correct?

## Yes

Second, to prevent negative energy, we use $a_−ψ_0=0$. So we get and take $ψ_0$ as ground state. Then the next state is $a_+ψ$, this is logic because $a_−$ and $a_+$ are the solution of Hamiltonian. But why suddenly the other state is $a^2_+$? Is that just because when we apply Hamiltonian operator on $a^2_+ψ$ then we can get another eigenvalue so we can say $a^2_+ψ$ this is another state? How do we get this formula, $$ψ_n(x)=A(a_+)^nψ_0(x)$$

Mostly from the book by Cohen-Tanoudji: The only ingredient you need here is the commutation relation $$[a,a^\dagger]=1$$, which is easily computed from $[x,y]=i\hbar$ and the definition of $a$.

• Let's define $N=a^\dagger a$ and write $n$ its eigenvalues and $|n\rangle$ the corresponding eigenvectors. The eigenvalues $n$ of $N$ are positive since, for a given eigenstate $|n\rangle$, $$n=\langle n|a^\dagger a |n\rangle = \lVert a |n\rangle \rVert^2$$ and a norm is positive.

• For $n=0$ one has $$n=0 \quad\Longrightarrow\quad \lVert a |0\rangle \rVert^2 = 0 \quad\Longrightarrow\quad a |0\rangle = 0 \quad .$$

• From the commutation relation, for any eigenvector $|n\rangle$ or $N$, $$N a |n\rangle = a^\dagger a a|n\rangle=\left(a^\dagger a-1\right)a| n\rangle=aN|n\rangle - a |n\rangle=(n-1)a|n\rangle \quad .$$ Hence $a|n\rangle$ is an eigenvector of $N$ with eigenvalue $n-1$.

From the two previous properties we know $n\in\mathbb{N}$, because if $|n\rangle$ is an eigenvector of $N$, so $a|n\rangle$ is, with eigenvalue $n-1$; also $a^2|n\rangle$ is an eigenstate with eigenvalue $n-2$, and so on. All possible $n-m$ with $m$ integer thus are part of the spectrum, except if the condition $n=0$ is fulfilled, in which case $a|n\rangle=0$ and $n-1$ is not an eigenstate of $N$. Since we know from the first property that eigenvalues $n$ are positive, $n$ can only be chosen as an integer.

Thus the spectrum of $N$ is $\mathbb{N}$. Since $$H=\hbar\omega\left(N+\frac12\right) \quad,$$ $H$ and $N$ have the same eigenvectors.

To generate $|n\rangle$ from $|0\rangle$, one can see, using the commutation relation, that $$N \left(a^\dagger\right)^n|0\rangle = a^\dagger a \left(a^\dagger\right)^{n} |0\rangle =\left[a^\dagger \left(a^\dagger\right)^n a-a^\dagger\left(a^\dagger\right)^{n-1}\right]|0\rangle =\left(a^\dagger\right)^n|0$$

using the relation $a \left(a^\dagger\right)^{n} =\left(a^\dagger\right)^{n}a + [a,\left(a^\dagger\right)^n]=\left(a^\dagger\right)^{n}a+n\left(a^\dagger\right)^{n-1}$. Thus $\left(a^\dagger\right)^n|0\rangle$ is an eigenvector of $N$ is eigenvalue i.e. it is $|n\rangle$.

Hope it is clear enough, otherwise go directly to Cohen-Tanoudji's book, but you will find a longer presentation there.

The square of an operator means that the operator is applied twice: $$a_+^2\psi_0=a_+(a_+\psi_0)=a_+\psi_1=\psi_2$$ Hence, we obtain further eigenfunctions by repeatedly applying the raising operator $a_+$ to $\psi_0$.

• I think I should ask why do we know the next state of ${a_+}$ is ${{\a_+}^2}$ ? From the Hamiltonian equation ,we have ${\a_-} and {\a_+}$ – Outrageous Nov 8 '13 at 15:11
• @Outrageous following the equations I give here: $a_+^2$ means by definition using the operator twice, nothing more. – gigacyan Nov 8 '13 at 19:17