Context:
Griffith's book on Quantum Mechanics (QM), in Section 2.3.1, tries to solve for the stationary states $\psi(x)$ of a harmonic oscillator by solving the Time-Independent Schrodinger Equation (TISE),
$$\frac{1}{2m}[p^{2}+(m\omega x)^{2}]\psi=E\psi,$$
using the method of ladder operators. The ladder operators’ method starts from a postulated definition of $a_{+}$ and $a_{-}$ (eq. 2.47):
$$ a_{\pm}\equiv \frac{1}{\sqrt{2\hbar\omega m}} (\mp i p+m\omega x),$$
which then was shown to work in terms of factorizing the Hamiltonian $H=[p^{2}+(m\omega x)^{2}]/(2m)$, and thus used to raise/lower energy in discrete steps. The discussion comes to a head in p.46 and footnote 21, where it is concluded that “we can construct all the stationary states” by repeated application of this operator $a_{+}$ starting from the lowest energy level (rung) on the ladder, $E_{0}$, and that such ladder is unique because two ladders with the same step size ($\hbar \omega$) and common first rungs will completely overlap and therefore be identical.
However, from a logical point of you, there might be a little issue for the reader up to this point as it was not proven (or discussed) that such operators ($a_{+}$,$a_{-}$) and their ensuing ladder (with $\pm \hbar \omega$ step size) were the only possible ones that could represent the Hamiltonian (i.e. no uniqueness was discussed), and therefore one might start to imagine an example of another set of possible operators that produce steps of half the size of the ones discussed (so, $\hbar \omega/2$, instead of $\hbar \omega$), and produce a different ladder, that will still overlap with the original ladder, even for common bottom rung ($E_{0}$), but with twice the number of rungs. By extension, an infinite number of such ladders of step size equal to $\hbar \omega/n$, where $n$ is integer, might be imagined in this sense, and they would still not conflict each other. Such uniqueness is not rigorously discussed.
The issue:
I think that the key to concluding that we only have one unique ladder is to:
- first prove that all operators would produce ladders with the same bottom rung;
- and then to show that the original operators ($a_{\pm}$) actually give the smallest energy step size (resolution) among all admissible operators/ladders (and thus be our only choice for operators, because other operators would be simply just integer multiples of them). But there is an issue here, as described below.
Point (1) is easy to prove: any new operators like $a^{2}_{-}$, $a^{2}_{+}$, or indeed any generalization thereof ($a^{m}_{-}$, $a^{m}_{+}$, for $m\in \mathbb{Z}$), share the same bottom rung with the original ladder operators ($a_{-}$ and $a_{+}$). I can prove this rigorously as follows: say that we have $m=2$, and we wish to find its lowest rung state (call it $\bar{\psi_{0}}$ for this case, to distinguish it from the original case $\psi_{0}$), then we find it by putting:
\begin{eqnarray}
a_{-}a_{-}\bar{\psi_{0}}&=&0 \nonumber\\
\Rightarrow a_{+}a_{-}(a_{-}\bar{\psi_{0}})&=&0 \nonumber\\
\Rightarrow (a_{-}a_{+}-1)(a_{-}\bar{\psi_{0}})&=&0 \nonumber\\
\Rightarrow a_{-}\bar{\psi_{0}}&=&a_{-}\underbrace{(a_{+}a_{-}\bar{\psi_{0}})}_{=(0)\bar{\psi_{0}}=0} \nonumber\\
\Rightarrow a_{-}\bar{\psi_{0}}&=& 0 \nonumber
\end{eqnarray}
\begin{equation}
\Rightarrow \boxed{\bar{\psi_{0}} \equiv \psi_{0}}, \nonumber
\end{equation}
which proves that all such ladders definitely share the same rung (I have used the fact that $a_{+}a_{-}\psi_{n}=n\psi_{n}$ in the above reduction to zero). Similar proofs can be done for higher $m$.
However, point (2) is proving to be more subtle and addresses the possibility of constructing new operators that are not of the form $a^{m}_{-}$, $a^{m}_{+}$, for $m\in \mathbb{Z}$. We note that the entire choice of operators $a_{-}$, $a_{+}$ as equal to: $(\text{constant})[\pm\hbar D + m\omega x]$, where $D\equiv\frac{d}{dx}$, was originally based simply on them producing a second-order derivative ($D^{2}\equiv\frac{d^{2}}{dx^{2}}$) when multiplied together, to match the Time-Independent Schrodinger Equation (TISE). Since TISE is second-order, the choice of $a_{\pm}$ was made to have each operator with a first-order derivative operator ($D$), which is a natural choice. Then it was proved that their product is of the form:
$$b_{-}b_{+}\ \ \ \varpropto \ \ \ \left[\underbrace{-\hbar D^{2}+(m\omega x)^{2}}_{=2 m H} + \text{some constant} \right],$$
(here I chose letter $b_{\pm}$ to generalize the discussion and to later distinguish it from the original operator $a_{\pm}$) which is later written in $H$ in the following form
$$ H= \hbar\omega\left( b_{-}b_{+} – \phi \right)\ \ \ ;\ \ \ H= \hbar\omega\left( b_{+}b_{-} + \phi \right),$$
where $\phi$ is some constant. And this will give us later
$$ \text{Commutator}[b_{-},b_{+}]=2\phi, $$
which later gives us the key conclusions that energy jumps in steps as
$$ \boxed{H(b_{-}\psi)=[E-(2\phi)\hbar\omega]\ \psi \ \ \ ;\ \ \ H(b_{+}\psi)=[E+(2\phi)\hbar\omega]\ \psi}. $$
Now, yes, clearly, if we choose operators $b_{\pm}$ like before to be some higher integer order of the original operators $a_{\pm}$ (such as $a^{m}_{\pm}$, with $m\in\mathbb{Z}$), then clearly they will have LARGER steps on the ladder, and therefore $a_{\pm}$ are the operators with the finest allowed resolution of steps (smallest allowed step). And since we have proved (above) that the first rung is shared by all such operators ($a^{m}_{\pm}$, with $m\in\mathbb{Z}$), then the original $a_{\pm}$ operators are unquestionably brilliant and unique. But, what if we chose operators $b_{\pm}$ that are not of the form $a^{m}_{\pm}$, with $m\in\mathbb{Z}$ ? What if I chose fractional derivatives, for example, such as $D^{1/2}$ or $D^{3/2}$ (which are also formal operators in applied mathematical analysis— e.g. see wiki page), which when multiplied will still give second-order ($D^{2}$), and therefore may factorize $H$ and represent the TISE equation? In fact, their application to this problem might be especially convenient because we have the $x$ dependence of the form $x^{k}$, which lends itself relatively easily to fractional derivative operators.
Griffith's text doesn't discuss whether this is doable or not, and therefore leaves the door open to the reader's imagination (or unease) about uniqueness here. For example, what if we chose say:
$$ b_{-} \ \varpropto \ \ (\hbar D)^{1/2} + (m\omega x)^{3/2}\ \ \ ;\ \ \ b_{+} \ \varpropto\ \ – (\hbar D)^{3/2} + (m\omega x)^{1/2}$$
or some other similar definitions that, when multiplied, could lead us (with the help of Gamma function identities that usually result from fractional derivatives) again to the sought form of:
$$b_{-}b_{+}\ \ \ \varpropto \ \ \ \left[\underbrace{-\hbar D^{2}+(m\omega x)^{2}}_{= 2 m H} + \text{some constant} \right],$$
(assuming we could make wise algebraic choices to produce the constant in this expression), and we could then find
$$ H= \hbar\omega\left( b_{-}b_{+} – \phi \right)\ \ \ ; \ \ \ H= \hbar\omega\left( b_{+}b_{-} + \phi \right),$$
with some new $\phi$ that is less than $1/2$ (that is $\phi<0.5$), and therefore produce a new ladder with "legal" energy steps that are SMALLER than $\hbar\omega$ (namely, energy step size $\boxed{2\phi\hbar\omega}$) ?
Any help in settling this idea of uniqueness would be appreciated.
Best Answer
Okay, time to gather all this up in an answer.
What does Griffiths show in his book? Boiled down, he shows that there exists an operator $\hat a$ such that:
The SHO Hamiltonian can be written as $\hat H = A\hat a^\dagger\hat a+B$
$[\hat a,\hat a^\dagger]=1$
The combination of these two properties imply that the Hamiltonian has a set of ladder eigenstates. These eigenstates are defined as follows:
We define $|0\rangle$ to be a state satisfying $\hat{a}|0\rangle=0$.
We define $|n\rangle\equiv\frac{1}{\sqrt{n!}}(\hat{a}^\dagger)^n|0\rangle$
Using the two properties of $\hat a$ above, we can prove that each $|n\rangle$ is an eigenstate $\hat H$ with energy $nA+B$.
We know there exists at least ONE such $\hat{a}$, because Griffiths explicitly writes it down in his book and shows it obeys both of the required properties. The question is, can there exist TWO operators with this property?
Let's say $\bar a$ is an operator, and that $\bar{a}$ obeys both properties. I.e., we have:
$\hat{H}=\bar{A}\bar{a}^\dagger\bar{a}+\bar{B}$
$[\bar{a},\bar{a}^\dagger]=1$
We follow the identical procedure as above to develop a new set of ladder states $|\bar{n}\rangle$ with energies $\bar{A}\bar{n}+\bar{B}$. There are three possible cases here: either $A\neq\bar A$, or $B\neq \bar B$, or $A=\bar A$ and $B=\bar B$ but somehow still $a\neq \bar a$. We need to show each of these possibilities are impossible.
The proofs will be proofs by contradiction. In each case, we'll select a state, and show that by acting on the state with some operator, we can construct another state with lower energy. We'll show that this process doesn't terminate (doesn't result in the zero vector) no matter how many times we do it. Thus, if we continue long enough, we get negative-energy states. Since we know the simple harmonic oscillator does not have negative energy states, we'll reach a contradiction.
Let's start with $A\neq \bar A$. Without loss of generality, assume $A>\bar{A}$. Then for some $\bar{n}$, we'll have that $|\bar{n}\rangle$ has an energy that cannot be written as $An+B$. Then, by acting with $\hat{a}$, we can generate a whole ladder of states with lower energies. The state $|\phi_m\rangle\equiv(\hat a)^m|\bar n\rangle$ has energy $E_m=\bar A\bar n+\bar B-Am$. We know that $|\phi_m\rangle$ will never equal $|0\rangle$, because $|\phi_m\rangle$ never has the same energy as $|0\rangle$. But since $|0\rangle$ is the unique vector that satisfies $\hat a |0\rangle=0$, that means this process can never terminate; every $m$ gives a nonzero vector in Hilbert space. By making $m$ large, we can make $|\phi_m\rangle$ have negative energy. But the SHO is strictly positive, so this is impossible. We conclude that we cannot have $A\neq \bar A$.
Now, let's assume $A=\bar{A}$, but $B\neq \bar{B}$. WLOG, assume $B<\bar{B}$. Then $|0\rangle$ has energy $B$, while $|\bar{0}\rangle$ has energy $\bar{B}$. In particular, since $|\bar{0}\rangle$ is the unique state which satisfies $\bar{a}|\bar 0\rangle=0$, acting on $|0\rangle$ with $\bar{a}$ produces states of arbitrarily negative energy. The state $|\phi_m\rangle\equiv(\bar a)^m|0\rangle$ has energy $B-Am$, which can be arbitrarily negative if we pick large $m$. We thus conclude that we can't have $B\neq \bar B$.
Finally, say $A=\bar{A}$ and $B=\bar{B}$. We want to show that $a$ and $\bar{a}$ are essentially the same.
We know the matrix elements of $\hat{a}$ are given by $$ \langle m|\hat{a}|n\rangle = \sqrt{n}\delta_{n-1,m} $$
Because $\bar{a}$ generates the same ladder of energies, and the spectrum of $H$ is non-degenerate, $\bar{a}$ connects the same states as $a$, up to phases:
$$ \langle m|\bar{a}|n\rangle = e^{i\theta_n}\sqrt{n}\delta_{n-1,m} $$ where $\theta_n$ possibly depends on the state $n$.
That's as good as you can do: you ARE allowed to pick some random ladder operator that adds phases to your states as you go up and down the ladder. But that's the only freedom you have. Phases aren't really important to the story, so you should consider the ladder operators essentially unique. In particular, you CAN'T have a ladder operator with a different lowest rung (different $B$), and you CAN'T have a ladder operator with a different spacing (different $A$).