Algebraic solution of the Quantum Harmonic Oscillator

The Quantum Harmonic Oscillator (QHO) was a key topic in a course on Quantum Physics I took (definitely one of the most interesting lectures I ever took). We solved it using the algebraic approach, which involves the factorization of the Hamiltonian using ladder operators. While this method is elegant, the derivation of the ladder operators in the lecture felt somewhat arbitrary to me, and I struggled to grasp the deeper motivation behind choosing this particular approach to solve the problem. However, the part on deriving the eigenvalues of the number operator was presented quite intuitively and self-contained. This is why I wrote this little collection of some of the steps and intuitions in the derivation, which made this whole approach sound more natural to me. I suppose most parts of the following derivation are part of standard textbooks on quantum mechanics. Prerequisites are the basics of the mathematical model of quantum mechanics.

Motivation

The Hamiltonian of a single-particle system in 1D is given by:

$$ \hat{H} = \frac{1}{2 \mu} \hat{P}^2 + V(\hat{X}) $$

where $V$ is some arbitrary potential and $\mu$ is the mass. The operators $\hat{P}$ and $\hat{X}$ represent the momentum and position operator respectively.

Oftentimes we want to consider systems which are close to a stable equilibrium state (say at $x_0 \in \mathbb{R}$), which is just a fancy term for a local minimum in the potential function. We can consider the second order Taylor expansion of the potential function at $x_0$:

$$ V(x) = V(x_0) + \underbrace{V’(x_0)}_{= 0}(x-x_0) + \frac{1}{2}V^{(2)}(x)(x-x_0)^2 + o((x-x_0)^2) $$

The first derivative equals $0$ because we consider a local minimum. We can translate our coordinate system accordingly to obtain approximately

$$ V(x) \propto x^2 $$

This motivation also feels somewhat heuristic, but it should just be a coarse motivation of why we are interested in solving a Hamiltonian of the following form (one could also just say that we have the following Hamiltonian, for some reason and want to solve the dynamics):

$$ \hat{H} = \frac{\hat{P}^2}{2 \mu} + \frac{1}{2} \mu \omega^2 \hat{X}^2 $$

Note that the constant in front of the $\hat{X}^2$ is just chosen for convenience like this (we could also just write an arbitrary constant $a$, however, later one can see that it makes sense to split it up like this. Physicists apparently rely on dimensional analysis :)).

Now, we want to solve the dynamics of this system. More precisely, we want to solve the Schrödinger equation for this Hamiltonian:

$$ \hat{H} \ket{\psi(t)} = i \hbar \frac{\partial}{\partial t} \ket{\psi(t)} $$

Because the Hamiltonian is clearly time-independent, the reduces to finding an eigenbasis of the Hamiltonian (time-independent Schrödinger equation):

$$ \hat{H} \ket{\psi} = E \ket{\psi} $$

With the Hamiltonian given above, this becomes:

$$ \frac{\hat{P}^2}{2 \mu}\ket{\psi} + \frac{1}{2} \mu \omega^2 \hat{X}^2 \ket{\psi} = E \ket{\psi} $$

and in position basis (using $\bra{x}\hat{P}\ket{\psi} = - i \hbar \frac{\partial }{\partial x} \psi(x)$):

$$ \frac{-\hbar^2}{2 \mu} \frac{\partial^2}{\partial x^2} \psi(x) + \frac{1}{2} \mu \omega^2 x^2 \psi(x) = E \psi(x) $$

We could try to solve this differential equation, but it could get very annoying. Another approach is to find the eigenbasis of the Hamiltonain algebraically (ladder operator method) - I think one can not really generalize exactly this approach, but it is still super satisfying to see how we can solve the Hamiltonian here.

Factorization of the Hamiltonian

The idea is that we somehow rewrite the Hamiltonian in another way and will then see that this leads to the problem of finding an eigenbasis of another operator for which we have some nice properties which we can use properly. First of all, we can make several observations.

Observing things

As we have seen, diagonalizing the Hamiltonian by solving the Schrödinger equation by hand requires a lot of work, which is why we do it algebraically. We want to do clever algebraic tricks, to obtain some easier form of the Hamiltonian, which somehow gives us the eigenbasis of $\hat{H}$ for free (or at least some single eigenvector). I think there is no real argument why this should work in the first place. However, if one starts to look at the Hamiltonian, and searches for some interesting structure, we clearly see that somehow

$$ \hat{H} \simeq a^2 + b^2 $$

for some $a$ and $b$ (ignoring that these are operators right now). Now we might want to play around and define the complex number $z = a + ib$ which implies that

$$ a^2 + b^2 = |z|^2 = z^* z = (a - bi)(a + bi) $$

Something like $z^* z$ would be more interesting than $a^2 + b^2$, because we would somehow have a factorization of the operator (we didn’t yet justify how all of this works for operators - this is just the intuition).

Almost ladder operators

Now going to the level of operators and becoming more concrete, we still have:

$$ \hat{H} = \frac{\hat{P}^2}{2 \mu} + \frac{1}{2} \mu \omega^2 \hat{X}^2 $$

With the intuition above, we can define¹:

$$ \hat{Z} = \sqrt{\frac{\mu \omega^2}{2}} \hat{X} + i \sqrt{\frac{1}{2\mu}} \hat{P} $$

We also have:

$$ \hat{Z}^{\dagger} = \sqrt{\frac{\mu \omega^2}{2}} \hat{X} - i \sqrt{\frac{1}{2\mu}} \hat{P} $$

What we essentially do here is a change of basis on the operator space. We change our basis for the space of linear operator from the $\hat{X}$, $\hat{P}$ basis into the $Z$, $Z^\dagger$ basis.

Now we would like to know if

$$ \hat{H} \overset{?}{=} Z^\dagger Z $$

as this would hold for normal complex numbers. We obtain:

where in the last step $[\hat{X}, \hat{P}] = \hat{X}\hat{P} - \hat{P}\hat{X} = i \hbar \mathbb{I}$ is the well-known commutator of position and momentum operator. Thus, we can see that our approach didn’t really work out, because the the real and imaginary part of our complex number (a.k.a. operator in our case) fail to commute. We therefore obtain after rearranging:

$$ \hat{H} = Z^\dagger Z + \frac{\hbar \omega}{2} \mathbb{I} $$

However, this is still very useful and an interesting step in the right direction. Our goal is to find the eigenstates of the Hamiltonian. With the form above, we can directly see that the eigenstates of $\hat{H}$ must also be eigenstates of $Z^\dagger Z$, as we add a multiple of the identity operator. It follows that $Z^\dagger Z$ and $\hat{H}$ have the same eigenbases. This is the key observation. For the eigenvalues, we can observe that there is an offset of $+ \hbar \omega / 2$ obtained through the addition of $\hbar \omega \mathbb{I} /2$.

But how do we find the eigenbasis of $Z^\dagger Z$?

The probably most (at least for me) unintuitive thing happens here. We take a look at the commutator $[Z^\dagger Z, Z]$ and $[Z^\dagger Z, Z^\dagger]$. Why would one do this to find the eigenvalues? Well I think it is still a bit of a headache, but the commutator is always useful when trying to uncover insights about the algebraic structure of an opeartor - so why not consider it? Especially because there are connections to eigenvalues and eigenvectors like:

If $[A,B] = cB$ for some constant $c$, then if $\ket{\psi}$ is an eigenstate of $A$ with eigenvalue $a$, the state $B\ket{\psi}$ is an eigenstate with eigenvalue $a + c$. One can easily check this, by direct computation: $$ A B \ket{\psi} = (BA + cB) \ket{\psi} = (a+c) B \ket{\psi} $$

Commutation relations

First of all, we could for example investigate the commutator:

It would be very convenient for possible future computation if this just would be $\mathbb{I}$ - just for convenience - and we eliminate the factor of $\hbar \omega$ somehow.

We can do so, by instead of $Z$ define the ladder operators. We simply define

Note that we have

$$ \hat{a} = \frac{1}{\sqrt{\hbar \omega}}Z $$

We could do all the computation from with $\hat{a}$ instead of $Z$ and obtain:

$$ \hat{H} = \hbar \omega \left(\hat{a}^\dagger \hat{a} + \frac{1}{2}\mathbb{I}\right) $$

and the thing we want, namely

$$ [a, a^\dagger] = \mathbb{I} $$

This can be verified by direct computation. Intuitively, this means that the cost of swapping $\hat{a}$ and $\hat{a}^\dagger$ brings one identity. In the following expression for example:

$$ \hat{a} \hat{a}^\dagger \hat{a} = ( \hat{a}^\dagger \hat{a} + \mathbb{I})\hat{a} = \hat{a}^\dagger \hat{a} \hat{a} + \hat{a} $$

Note that neither $\hat{a}$ is not Hermitian or anything (neither was $Z$). It’s just some operator. We still somehow want to find the eigenbasis of $\hat{H}$, where the eigenbasis of $\hat{a}^\dagger \hat{a}$ is sufficient, as seen above.

Number Operator

For more convenience, we define the number operator:

$$ \hat{n} = \hat{a}^\dagger \hat{a} $$

this is essentially just renaming (we will see where this naming as number operator comes froms). So we have:

$$ \hat{H} = \hbar \omega \left(\hat{n} + \frac{1}{2} \mathbb{I}\right) $$

Commutators

We obtain the following properties:

Well, what does this tell us? As we have stated above, this commutator implies that for an eigenstate $\ket{v}$ of $\hat{n}$ with eigenvalue $v$, the state $\hat{a} \ket{v}$ would be an eigenstate with eigenvalue $v-1$. Similarly, $\hat{a}^\dagger \ket{v}$ would be an eigenstate with eigenvalue $v+1$.

Solving eigenvalue problem for the number operator

Let $\{\ket{n}\}_{n}$ be the eigenbasis of the number operator (spectral decomposition, as $\hat{n}$ is Hermitian). Let $n$ be the corresponding eigenvalue, such that we have:

$$ \hat{n}\ket{n} = n \ket{n} $$

Note that $n$ can be arbitrary at this point. We only consider $\ket{n}$ to be the eigenstate corresponding to eigenvalue $n$.

First of all, we can clearly see that $\hat{n}$ must be positive semi-definite. This follows from:

$$ \bra{\psi} \hat{n} \ket{\psi} = \bra{\psi} \hat{a}^\dagger \hat{a} \ket{\psi} = |\hat{a} \ket{\psi} |^2\geq 0 $$

As discussed above, $\hat{a}$ and $\hat{a}^\dagger$ have a special effect on an eigenstate.

$$ \hat{n} \hat{a} \ket{n} = (n-1) \hat{a} \ket{n} $$

This means that $\hat{a} \ket{n} \propto \ket{n-1}$. We cannot just assume equality, because the $\hat{a}$ operator might do anything to the norm of $\ket{n}$. Remember that $\hat{a}$ is just an operator and is neither Unitary nor Hermitian. Similarly, we obtain $\hat{a}^\dagger \ket{n} \propto \ket{n+1}$. To compute this norm factor, we can do the following:

Therefore

We can use this to reason about possible eigenvalues of $\hat{n}$. Let $n\in \mathbb{R}^+_0$ be an arbitrary eigenvalue of $\hat{n}$ with the corresponding eigenvector $\ket{n}$. Now imagine, we apply the $\hat{a}$ operator over and over again, until we either reach $\ket{m}$ for $m \leq 0$. Assume that we don’t reach $0$ (i.e. $m < 0$), this would exactly be the case if $n \notin \mathbb{N}_0$. Because we know that $\ket{m}$ is the eigenstate with the eigenvalue $m$, this would mean that $\hat{n}$ has a negative eigenvalue, which is a contradiction to the fact that $\hat{n}$ is positive semi-definite. Therefore, it must be the case that $n \in \mathbb{N}_0$.

The same argument can essentially be used to directly determine an eigenvalue of $\hat{n}$. Because we know that $\hat{n} = \hat{a}^\dagger \hat{a}$. More precisely, $\ket{0}$ must necessarily be an eigenstate of $\hat{n}$. We can obtain this state by just applying $\hat{a}$ to an arbitrary eigenstate a sufficiently large amount of times. If this state would not exist. Then we had this contradiction from above, because the eigenvalues get negative and this state can indeed exists, because as we reach $\ket{0}$, we can check that :

$$ \hat{n}\ket{0} = \hat{a}^\dagger \hat{a} \ket{0} = \hat{a}^\dagger \sqrt{0} \ket{\dots} = 0 $$

Now the property of $\hat{a}^\dagger$ (yielding an eigenstate with $+1$ eigenvalue), can be used to obtain an arbitrary eigenstate:

These algebraic arguments are quite interesting (there is the rabbit hole of Fock states).

We have shown that only natural numbers can be eigenvalues and that every natural number is indeed an eigenvalue. This concludes: The number operator $\hat{n}$ has natural numbers (including $0$) as it’s eigenvalues.

Conclusion

As the eigenbases of $\hat{H}$ and $\hat{n}$ are the same and we have²:

$$ \hat{H} = \hbar \omega \left(\hat{n} + \frac{1}{2} \mathbb{I}\right) $$

the spectrum of $\hat{H}$ turns out as:

$$ \frac{1}{2} \hbar \omega, \frac{3}{2} \hbar \omega, \dots, \frac{2n + 1}{2} \hbar \omega, \dots $$

which correspond to the eigenstates

$$ \ket{0}, \ket{1}, \dots, \ket{n}, \dots $$

How would we now solve for the position wave function so the energy eigenstates?

Well, one can insert $\ket{0}$ into the time-independent Schrödinger equation, which yields:

$$ \hbar \omega (\hat{n} + \frac{1}{2} \mathbb{I}) \psi_{0}(x) = E_0 \psi_{0}(x) = 0 $$

we would need to write $\hat{n}$ in terms of $\hat{X}$ and $\hat{P}$, solve the differential equation and then use the $\hat{a}^\dagger$ operator to generate the position wave functions for the other states. One can show that this will indeed turn out to give the recursion for the Hermite Polynomials.