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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:

  1. first prove that all operators would produce ladders with the same bottom rung;
  2. 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.

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  • $\begingroup$ Just slightly related to your question, but I thought you might be interested. physics.stackexchange.com/q/256013 $\endgroup$ – user181180 Jan 17 '18 at 23:22
  • $\begingroup$ And, simply because I read over Griffith's book today on this topic, there is a second, more formal discussion, again I apologise that this is not as direct a reply as your straightforward question deserves: sciencedirect.com/science/article/pii/S0377042702006131 $\endgroup$ – user181180 Jan 17 '18 at 23:30
  • $\begingroup$ @countto10 Yes, but to a reader who is following the text in linear fashion from the start, it is point worth discussing in relation to the ladder operators method. $\endgroup$ – user135626 Jan 17 '18 at 23:33
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    $\begingroup$ Related: physics.stackexchange.com/q/23028/2451 and links therein. $\endgroup$ – Qmechanic Jan 18 '18 at 7:48
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Jahan Claes Jan 29 '18 at 23:24
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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 $A$).

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Your argument is right, of course: there's no guarantee there won't be many more states that our ladder operators don't reach. The simplest possibility is another equivalent ladder in parallel, but states in between the ladder states can occur too.

In the case of the quantum harmonic oscillator, there are more specific arguments you can make. Valter Moretti points out here that if you assume the Hilbert space is $L^2(\mathbb{R})$, you know the standard ladder of states is enough because it gives a complete basis. I think that rules out your proposal with fractional derivatives, but you can still get a hidden ladder if the Hilbert space is actually larger, like $L^2(\mathbb{R}) \otimes L^2(\mathbb{R})$. An explicit model here is two parallel ladders both with spacing $\hbar \omega/2$, where the raising operator we've given raises on both simultaneously.

But from a phenomenological view this is beside the point. Of course there are many models you can use; the key is finding what fits experiment. We know that a transition between states with energy difference $\hbar \Delta \omega$ releases a photon of frequency $\Delta \omega$. We also know that a classical particle oscillating with frequency $\omega$ emits radiation of frequency $\omega$ (and harmonics thereof). That tells us that our quantum model needs energy spacings of $\hbar \omega$ to fit with experiment, and no smaller.

Of course you can postulate other states and make some rule that they can never radiate, but Occam's razor means such a model is worse. Experiment pushes us towards the Hilbert space $L^2(\mathbb{R})$ and the Hamiltonian $H = p^2/2m + kx^2/2$, where the ladder is complete.

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  • $\begingroup$ So, if I understand you correctly, you are saying that the standard ladder used is sufficient representation because it is a complete set in the assumed problem space ($L^{2}$). This doesn't, on it own, give it uniqueness, because I think we may find other complete sets for the same space, but you are saying that we have settled for this particular set because it agrees with experiment. Is that what you meant? $\endgroup$ – user135626 Jan 18 '18 at 13:30
  • $\begingroup$ @user135626 actually, for this particular situation, the basis is unique. We said we wanted a basis of eigenstate of H, which uniquely specifies this particular basis. Any other basis wouldn't be a basis of eigenstates. $\endgroup$ – Jahan Claes Jan 29 '18 at 21:10
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The operator (observable) factorization into a product does not lead necessarily to ladder operators. Suppose the Hamiltonian operator, for example, is written as $H=XY$, then $$ [H,X]=[XY,X]=X[Y,X]. $$ Unless $[X,Y]$ is a constant, as for the traditional ladder operators for the simple harmonic oscillator, $X$ and $Y$ are not ladder operators. In fact, an operator $M$ is a ladder operator for an operator $H$ when it is an eigenoperator for the adjoint action $\text{ad}(H)$: $$ \text{ad}(H)M=[H,M]=\lambda M, $$ where $\lambda$ is a complex number. Suppose $|\psi\rangle$ is an eigenstate for $H$ with eigenvalue $c$, then $$ HM|\psi\rangle=(\lambda M+MH)|\psi\rangle=(\lambda+c)M|\psi>. $$ It does not matter if $M$ "divides" $H$. Obviously $\lambda$ must be real if the eigenvalues are to be interpreted as the possible observable measurements. Notice that this definition of ladder operator as an eigenoperator for $\text{ad}(H)$ is limited to spectra with equaly spaced eigenvalues. For general spectra $\lambda_0<\lambda_1<\ldots<\lambda_n<\ldots$ and associated eigenstates $\psi_0,\psi_1,\ldots,\psi_n,\ldots$, a ladder operator $M$ (no relation with the former $M$) is defined by $M_{\pm}|\psi_n\rangle\propto |\psi_{n\pm 1}\rangle$ for all $n$. The two definitions coincide for the simple harmonic oscillator, but not for the Hydrogen atom, where only the last one is applicable.

As you noticed, not only $a_+$ and $a_-$ are eigenoperators for $\text{ad}(H)$ and, simultaneously, for the number operator $\text{ad}(N)$, where $N=a_+a_-$ is the number operator, but any power of them. More generally, defining the grade of a monomial $a_+^na_-^m$ as $$ \text{grade}(a_+^na_-^m)=n-m, $$ any monomial is an eigenoperator for $N$ with eigenvalue equal to its grade: $$ \text{ad}(N)(a_+^na_-^m)=(n-m)a_+^na_-^m. $$ Any sum of same grade operators is also an eigenoperator. Even more: if inverses can be defined, the ratio between same grade operators are also eigenoperators. As an example, let $M=2a_+^7a_-^5+7a_+^3a_-$. Thus $$ [N,M]=2M, $$ and $$ [N,M^{-1}]=-\frac{1}{2}M^{-1}, $$ where the fact that if $[N,M]=\lambda M$, then $[N,M^{-1}]=-\frac{1}{\lambda}M^{-1}$ was used. The last identity can be obtained from $[N,MM^{-1}]$.

Operators can be expressed as a function of the position and momentum operators or, equivalently, in terms of $a_+$ and $a_-$: the transformation between them is simply a change of coordinates. As any operator can be written as a function of $a_+$ and $a_-$ and any eigenoperator for $\text{ad}(H)$ is a sum or ratio of same grade monomials, your point 2) can be rephrased as: must the grade necessarily be an integer? The answer is no. The operator $\sqrt{a_+}$ (if it does exist), for example, is an eigenoperator for $\text{ad}(H)$: $$ [H,\sqrt{a_+}]=[a_+a_-,\sqrt{a_+}]=a_+[a_-,\sqrt{a_+}]=\frac{1}{2}\sqrt{a_+}. $$ It was used the formal rule $[a_-,f(a_-,a_+)]=\frac{\partial f}{\partial a_+}$. If $N|\psi\rangle=c|\psi\rangle$ then $$ N\sqrt{a_+}|\psi\rangle=(c+\frac{1}{2})|\psi\rangle. $$

By now should we conclude that new solutions to the quantum simple harmonic oscillator can be obtained by repeated application of an operator such as $a_+^{1/n}$, with $n$ an integer, on the ground state $|0\rangle$? No, we shouldn't because:

  1. Operators such as $\sqrt{a_-}$ could not exist. For example, the matrix $\left({\begin{smallmatrix}\\0&1\\0&0\end{smallmatrix}}\right)$ has no square root. In fact, it is well established that the simple harmonic oscillator spectrum has a unique spetrum, thus operators with non-integer grade should not exist.

  2. Even if they can be defined in some way, they could (and given the spectrum uniqueness, should) produce states that are not square-integrable.

I've looked for a definition of $(x-\frac{d}{dx})^{1/2}$, but I couldn't find one. The plan was apply this operator to a Gaussian function and see if the result satisfies Schrödinger equation.

In conclusion, the spectrum uniqueness for the simple harmonic oscillator forbids the existence of non-integer grade operators. Nonetheless, it would be very instructive to have a nonexistence proof that does not resort to the spectrum.

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  • $\begingroup$ Thanks for this illuminating answer. I think assumed in your discussion on monomials is that $a_{\pm}$ have $\lambda=\pm1$, right? The two conclusion points aren't very clear to me, though: why are we talking about this matrix? & why cannot they give sequare-integrable states? Also, if we forget about my example of using fractional powers (grade), isn't there any other forms? We are assuming here that all operators can be expressed in terms of the traditional $a_{\pm}$ operators, but we haven't proved that they're smallest "unit" of such operators: $\lambda$ was $\hbar\omega$ in SHO and not 1. $\endgroup$ – user135626 Jan 29 '18 at 0:45
  • $\begingroup$ Yes, $\lambda=\pm$ for $a_{\pm}$, but the difference with $\hbar\omega$ is unimportant: of you multiply the number operator by any factor, the eigenvalue will be multiplied by this factor. What is important is getting a new eigenvalue for the same operator, what would be the case for $\sqrt{a_{\pm}}$. The matrix was just an example that operators might not have fractional powers and in point 2 I've made a mistake: I should have written "they could produce states that are not square-integrable". $\endgroup$ – jobe Jan 29 '18 at 11:16
  • $\begingroup$ Answering your question I've became so intrigued about the existence of fractional powers of ladder operators that I asked this question: physics.stackexchange.com/questions/382002/… $\endgroup$ – jobe Jan 29 '18 at 11:46
  • $\begingroup$ I see your point, but to clarify: yes, multiplying operator $M$ by factor $k$, as in $kM$, will give $k\lambda$ as the associated eignevalue, but I am talking about powers of $M$, such as $M^{k}$ giving $k\lambda$, and hence working as ladder with a step equal to $\lambda$ (Indeed, since $M$ and $N$ make up $H$, any multiplicative factors will eventually be restricted) So, for $M$ equal to the traditional operators $a_{\pm}$, $\lambda$ was $\pm\hbar\omega$, & I am asking if there isn't other operators definitions that would give smaller $\lambda$, hence my suggestion of fractional powers $\endgroup$ – user135626 Jan 29 '18 at 14:34
  • $\begingroup$ @user135626, I made some edits to my answer that - hopefully - will eliminate some ambiguities. For example, at the beginning of my answer I wrote about factorization and ladder operators only to emphasize that they are two different things that happens to be associated for the simple harmonic oscillator, but not in general. Unfortunately, I used the same letter $M$ for a factorization term and an eigenoperator for $H$, what led to some confusion in your last comment. $\endgroup$ – jobe Jan 29 '18 at 16:18

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