Quantum harmonic oscillator

Ladder operator method

The spectral method solution, though straightforward, is rather tedious. The "ladder operator" method, due to Paul Dirac, allows us to extract the energy eigenvalues without directly solving the differential equation. Furthermore, it is readily generalizable to more complicated problems, notably in quantum field theory. Following this approach, we define the operators a and its adjoint a^†

${\begin{matrix}a&=&{\sqrt {m\omega \over 2\hbar }}\left(x+{i \over m\omega }p\right)\\a^{\dagger }&=&{\sqrt {m\omega \over 2\hbar }}\left(x-{i \over m\omega }p\right)\end{matrix}}$

The operator a is not Hermitian since it and its adjoint a^† are not equal.

The operator a and a^† have properties as below:

${\begin{matrix}a\left|\phi _{n}\right\rangle &=&{\sqrt {n}}\left|\phi _{n-1}\right\rangle \\a^{\dagger }\left|\phi _{n}\right\rangle &=&{\sqrt {n+1}}\left|\phi _{n+1}\right\rangle \end{matrix}}$

We can also define a number operator N which has the following property:

$N=a^{\dagger }a$

$N\left|\phi _{n}\right\rangle =n\left|\phi _{n}\right\rangle$

In deriving the form of a^†, we have used the fact that the operators x and p, which represent observables, are Hermitian. These observable operators can be expressed as a linear combination of the ladder operators as

${\begin{matrix}x&=&{\sqrt {\hbar \over 2m\omega }}\left(a^{\dagger }+a\right)\\p&=&i{\sqrt {{\hbar }m\omega \over 2}}\left(a^{\dagger }-a\right)\end{matrix}}$

The x and p operators obey the following identity, known as the canonical commutation relation:

$\left[x,p\right]=i\hbar$.

The square brackets in this equation are a commonly-used notational device, known as the commutator, defined as

$\left[A,B\right]\ {\stackrel {\mathrm {def} }{=}}\ AB-BA$.

Using the above, we can prove the identities

$H=\hbar \omega \left(a^{\dagger }a+1/2\right)$

$\left[a,a^{\dagger }\right]=1$.

Now, let $\left|\psi _{E}\right\rangle$ denote an energy eigenstate with energy E. The inner product of any ket with itself must be non-negative, so

$\left(a\left|\psi _{E}\right\rangle ,a\left|\psi _{E}\right\rangle \right)=\left\langle \psi _{E}\right|a^{\dagger }a\left|\psi _{E}\right\rangle \geq 0$.

Expressing a^†a in terms of the Hamiltonian:

$\left\langle \psi _{E}\left|{H \over \hbar \omega }-{1 \over 2}\right|\psi _{E}\right\rangle =\left({E \over \hbar \omega }-{1 \over 2}\right)\geq 0$,

so that $E\geq \hbar \omega /2$. Note that when ($a\left|\psi _{E}\right\rangle$) is the zero ket (i.e. a ket with length zero), the inequality is saturated, so that $E=\hbar \omega /2$. It is straightforward to check that there exists a state satisfying this condition; it is the ground (n = 0) state given in the preceding section.

Using the above identities, we can now show that the commutation relations of a and a^† with H are:

${\begin{matrix}\left[H,a\right]&=&-\hbar \omega a\\\left[H,a^{\dagger }\right]&=&\hbar \omega a^{\dagger }\end{matrix}}$.

Thus, provided ($a\left|\psi _{E}\right\rangle$) is not the zero ket,

${\begin{matrix}H(a\left|\psi _{E}\right\rangle )&=&(\left[H,a\right]+aH)\left|\psi _{E}\right\rangle \\&=&(-\hbar \omega a+aE)\left|\psi _{E}\right\rangle \\&=&(E-\hbar \omega )(a\left|\psi _{E}\right\rangle )\end{matrix}}$.

Similarly, we can show that

$H(a^{\dagger }\left|\psi _{E}\right\rangle )=(E+\hbar \omega )(a^{\dagger }\left|\psi _{E}\right\rangle )$.

In other words, a acts on an eigenstate of energy E to produce, up to a multiplicative constant, another eigenstate of energy $E-\hbar \omega$, and a^† acts on an eigenstate of energy E to produce an eigenstate of energy $E+\hbar \omega$. For this reason, a is called a "lowering operator", and a^† a "raising operator". The two operators together are called ladder operators. In quantum field theory, a and a^† are alternatively called "annihilation" and "creation" operators because they destroy and create particles, which correspond to our quanta of energy.

Given any energy eigenstate, we can act on it with the lowering operator, a, to produce another eigenstate with $\hbar \omega$-less energy. By repeated application of the lowering operator, it seems that we can produce energy eigenstates down to E = −∞. However, this would contradict our earlier requirement that $E\geq \hbar \omega /2$. Therefore, there must be a ground-state energy eigenstate, which we label $\left|0\right\rangle$ (not to be confused with the zero ket), such that

$a\left|0\right\rangle =0{\hbox{(zero ket)}}$.

In this case, subsequent applications of the lowering operator will just produce zero kets, instead of additional energy eigenstates. Furthermore, we have shown above that

$H\left|0\right\rangle =(\hbar \omega /2)\left|0\right\rangle$

Finally, by acting on $\left|0\right\rangle$ with the raising operator and multiplying by suitable normalization factors, we can produce an infinite set of energy eigenstates $\left\{\left|0\right\rangle ,\left|1\right\rangle ,\left|2\right\rangle ,...,\left|n\right\rangle ,...\right\}$, such that

$H\left|n\right\rangle =\hbar \omega (n+1/2)\left|n\right\rangle$

which matches the energy spectrum which we gave in the preceding section.

This method can also be used to quickly find the ground state wave function of the quantum harmonic oscillator.

${\begin{matrix}\left\langle x\left|a\right|\psi _{0}\right\rangle &=&\left\langle x\left|{1 \over {\sqrt {2m\hbar \omega }}}\left(m\omega x+ip\right)\right|\psi _{0}\right\rangle =0\\&=&{1 \over {\sqrt {2m\hbar \omega }}}\left(m\omega x+\hbar {d \over dx}\right)\psi _{0}(x)\\&=&\int {-m\omega x \over \hbar }dx-\int {d\psi _{0}(x) \over \psi _{0}(x)}\\&=&{-m\omega x^{2} \over 2\hbar }-ln\left(\psi _{0}(x)\right)+Const.\\\psi _{0}(x)&=&Const.\times e^{-m\omega x^{2} \over 2\hbar }\end{matrix}}$

Which by normalization of the wave functions leads us to the following ground state wave function.

<math>\psi_0(x)= \left({m\omega \over \pi\hbar}\right)^{1 \over 4}e^{-{m\omega \over 2\hbar}x^2}