Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Given the following pair of operators $a$ and $a^{\dagger}$ that satisfy the usual bosonic CCR:
$$[a,a]=[a^{\dagger},a^{\dagger}] = 0;\ [a,a^{\dagger}] = 1$$
For what values of $\alpha \in\mathbb C$ are the following expressions well defined?
$$a^{\alpha} \ \text{and}\ (a^{\dagger})^{\alpha}$$
For integer $\alpha$, I know that these expressions are well defined, but I am interested in knowing what kind of constraints are imposed on $\alpha$, if such an operator must be appropriately definable. If the above expressions can be defined in precise manner, then please also provide the appropriate construction of these operators.

Some meandering background:
The actual problem I'm trying to solve has nothing to do with quantum mechanics! It is instead a reformulation of the dynamical equations of a polymer (hydrodynamically interacting with a solvent) using the rouse modes as "bosons" [JCP(45), 1966, M. Fixman, Polymer Dynamics:Boson Representation and Excluded Volume forces]. So there are no nice hamiltonians and unitary operators to work with. The dynamical equation is of the following form:
$$\dfrac{\partial{\left|\rho\right\rangle}}{\partial t}+\mathcal{L}\left|\rho\right\rangle=0$$
Now the operator $\mathcal{L}$ has a representation in the boson operators as
$$\mathcal{L}=g(a^{\dagger}a+\zeta a^{\dagger}a^{\dagger})$$
where $g$ and $\zeta$ are real constants. The kets live in a standard bosonic Fock space, and a complete basis of number states is present constructed in a fashion equivalent to the case of the harmonic oscillator. On trying to solve the eigenvalue equation for $\mathcal{L}$, one can write down the eigenstates, indexed by the eigenvalue as
$$\left|\psi;\lambda\right\rangle=exp\left(-\frac{\zeta}{2}a^{\dagger}a^{\dagger}\right)(a^{\dagger})^{\lambda}\left|0\right\rangle$$
such that $\mathcal{L}\left|\psi;\lambda\right\rangle=\lambda\left|\psi;\lambda\right\rangle$. This is where my original question comes into the forefront - Deciding on how these operators behave on being raised to arbitrary powers, I can fix the possible eigenvalues of $\mathcal{L}$. If it is possible to arrive at a conclusion on the eigenvalues of the given $\mathcal{L}$ operator in any other fashion, without resorting to this approach, then please go ahead and tell me.

share|improve this question
    
Re your edit: That's one ugly operator, and it looks like a job getting from $\mathcal L$ to $|\psi;\lambda\rangle$. If all you want is to make sense of the eigenstate, why not use the representation over coherent states to provide an expression directly for $|\psi;\lambda\rangle$, and then check where it's well defined and where not, and see directly if it's an eigenstate of $\mathcal L$. –  Emilio Pisanty Nov 19 '13 at 22:05
    
I would actually be quite a bit more wary of the manipulations that led you to say that that that state is an eigenstate; whatever commutation properties of $(a^\dagger)^\nu$ with $a$ that you assumed are probably the "properties would you expect of it" that I was talking about. I'd be interested in seeing more of those inner workings; hopefully they'll make it clearer what the situation is. –  Emilio Pisanty Nov 19 '13 at 22:15
    
Ok, so the commutation properties of $a^{\nu}$ and $(a^{\dagger})^{\nu}$ used and hence expected are $$[a,(a^{\dagger})^{\nu}]=\frac{\partial (a^{\dagger})^{\nu}}{\partial a^{\dagger}}=\nu (a^{\dagger})^{\nu -1}$$ and similarly for the case of $[a^{\nu},a^{\dagger}]$ –  surajshankar Nov 20 '13 at 4:57
    
The other thing used is the properties of the inverse operators (they are only pseudo-inverses): $a^{\dagger}(a^{\dagger})^{-1}=1$ and $(a^{\dagger})^{-1}a^{\dagger}=1-\left|0\right\rangle\left\langle 0\right|$. –  surajshankar Nov 20 '13 at 5:07
    
I've expanded my answer. Unfortunately my construction does not obey that commutation relation. I do offer some thoughts on how you could phrase that particular problem, though. –  Emilio Pisanty Nov 20 '13 at 21:58

1 Answer 1

up vote 7 down vote accepted

This is eybrow-raisingly tricky to answer. The short answer is: you can define them, in a complicated way that's not really useful, but why would you want such a thing?

There's two main reasons why this is complicated, which hold for integer and non-integer powers respectively.

  • For one, the two operators will behave quite differently. Because $a$ annihilates the vacuum state, it is not invertible, and its inverse $a^{-1}$ will not behave as expected. (Note that $n^{-1}a^\dagger$ is a left inverse, but not on the right; $a^{-1}$ ought to commute with $a$.) The most you can hope for is a Moore-Penrose pseudoinverse, which will have a rank 1 kernel. (Similarly, further negative powers will increase the kernel dimension.) The creation operator $a^\dagger$ has the opposite problem, as there's no $|\psi\rangle$ such that $a^\dagger|\psi\rangle=|0\rangle$, so again you can only hope for a rank-deficient pseudoinverse.

  • Further, these operators do have eigenvalues, but they're complex: there's one coherent state $|\alpha\rangle$ for each $\alpha\in\mathbb C$ which obeys $a|\alpha\rangle=\alpha|\alpha\rangle$. Thus to make sense of $a^\nu$ for noninteger $\nu$ you need to make sense of $\alpha^\nu$ for all $\alpha$, and this means braving the branch cuts. While this can be dealt with in some arbitrary fashion, it will never hold up under the time evolution $a\mapsto a e^{-i\omega t}$, $\alpha\mapsto \alpha e^{-i\omega t}$ that's generated in phase space by the harmonic hamiltonian $H=\hbar\omega a^\dagger a$: coherent states rotate around the origin and they will meet within a period any branch cut you set up. While this is in principle OK, it will never have enough "nice" properties to be useful in practice.

That said, it is possible to construct such operators. Since $a$ has eigenvalues, we would expect $a^\nu$ to behave well with respect to that structure, which means that if it didn't satisfy $$a^\nu|\alpha\rangle=\alpha^\nu|\alpha\rangle$$ then we would not find the construction useful at all. This expression is not particularly well-defined, though, and you need to fix a branch cut to make sense of $\alpha^\nu:=e^{\nu\log(\alpha)}$. Once you do that, this fixes the action of $a^\nu$ on any state $|\psi\rangle$, because the coherent states, while non-orthogonal and not a basis, do obey the resolution of identity $$ \int\frac{\text d^2\alpha}{\pi}|\alpha\rangle\langle\alpha|=1. $$ Thus, for any vector $|\psi\rangle$ you must have $$ a^\nu|\psi\rangle =\int\frac{\text d^2\alpha}{\pi}a^\nu|\alpha\rangle\langle\alpha|\psi\rangle =\int\frac{\text d^2\alpha}{\pi}\alpha^\nu|\alpha\rangle\langle\alpha|\psi\rangle .\tag1 $$

So there, you've defined it! You might even go as far as saying that the domain of this new operator is all those $|\psi\rangle$ such that the integral above converges. Now what do you want to do with it? It's even hard to define exactly why the construction above is unsatisfactory, and that's because there's really no call to use such operators in any physical situation, so it's hard to think of properties that it ought to satisfy but doesn't.

I pointed out one example above: we'd like $a^\nu$ to evolve to $a^\nu (e^{-i\omega t})^\nu=a^\nu e^{-i\nu\omega t}$ under the harmonic motion that takes $a$ to $a e^{-i\omega t}$, but in general this is not the case. If $U(t)=e^{-i\hat{n}\omega t}$, so $U(t)|\alpha\rangle=|\alpha e^{-i\omega t}\rangle$, then $a^\nu$ will evolve to $U(t)^\dagger a^\nu U(t)$, which, as defined above, would give $$ \begin{align} U(t)^\dagger a^\nu U(t)|\psi\rangle & =\int\frac{\text d^2\alpha}{\pi}\alpha^\nu U(t)^\dagger|\alpha\rangle\langle\alpha|U(t)|\psi\rangle \\ & =\int\frac{\text d^2\alpha}{\pi}\alpha^\nu |\alpha e^{+i\omega t}\rangle\langle\alpha e^{+i\omega t}|\psi\rangle \\ & =\int\frac{\text d^2\alpha'}{\pi}\left(\alpha'e^{-i\omega t}\right)^\nu |\alpha \rangle\langle\alpha |\psi\rangle. \end{align} $$ This would simplify to $e^{-i\nu\omega t}a^\nu|\psi\rangle$, if $\left(\alpha'e^{-i\omega t}\right)^\nu $ simplified to $\alpha'^\nu e^{-i\nu\omega t}$, but it doesn't. Instead, you must define it as $\left(\alpha'e^{-i\omega t}\right)^\nu :=e^{\nu\log\left(\alpha'e^{-i\omega t}\right)}$. On the complex plane, the logarithm no longer splits products into sums; all you can say is equality up to a multiple of $2\pi i$: $$\log(\alpha\beta)\equiv\log(\alpha)+\log(\beta) \mod 2\pi i.$$ This means that $\left(\alpha'e^{-i\omega t}\right)^\nu=\alpha'^\nu e^{-i\nu\omega t}e^{2\pi i \nu n(\alpha,t)}$, for some integer $n$ which, in general, depends on $\alpha$ and $t$. If $\nu$ is integer, this is not a problem. Otherwise, though, you've introduced a new, complicated function of $\alpha$ into your integrand, and it will not generally integrate to the same thing.

Other than this, though, it depends on what you want to do with it. One is always free to define whatever one wants, but this right is earned a posteriori by doing something useful with the definition. Thus, where would you expect such a monster to occur, and what properties would you expect of it for it to be useful there?


Addendum: some thoughts on your edit.

OK, so what you really care about isn't so much a power of $a$, but instead an operator that will obey $$[a^\nu,a^\dagger]=\nu a^{\nu-1}.$$ I'm not completely sure, but I think the construction (1) will not, in general, satisfy that constraint. I would look at it from the opposite perspective: what you're really trying to solve is the operator-valued functional equation $$ [A(\nu),a^\dagger]=\nu A(\nu-1) $$ for some function $A:\mathbb C\to \rm{End}(\mathcal H)$. Equivalently, you can take the matrix elements of this equation to get a countable infinity of coupled functional equations, $$ \nu A_{mn}(\nu-1)=\sqrt{n+1}A_{m,n+1}(\nu)-\sqrt{m}A_{m-1,n}(\nu), \tag2 $$ where $A_{m,n}(\nu)=\langle m|A(\nu)|n\rangle$. You know already one solution to this equation, for integer $\nu$: $$ A_{mn}(\nu)=\sqrt{\frac{n!}{(n-\nu)!}}\delta_{m-n,\nu}\ \text{for }\nu=0,1,2,\ldots. $$ The trick is extending these solutions.

Now this doesn't get you very far in that it's only a restatement of your problem (which is often, of course, very useful). I can only offer some thoughts as to how my construction (1) fares in this regard. If you take that, then it means postulating $$ A_{mn}(\nu) =\int\frac{\text d^2\alpha}{\pi} e^{-|\alpha|^2} \frac{\alpha^m{\alpha^\ast}^n}{\sqrt{m!n!}} e^{\nu \log(\alpha)} $$ as a solution. Plugging this into the equation (2), and using the fact that $\alpha^{\nu-1}=e^{(\nu-1)\log(\alpha)}=\alpha^\nu \alpha^{-1}$ (i.e. sums of powers do map into products) you can reduce the mess to the equation $$ \int e^{-|\alpha|^2} {\alpha^m{\alpha^\ast}^n} e^{(\nu-1) \log(\alpha)} \left[ |\alpha|^2-m-\nu \right] {\text d^2\alpha} =0. $$ The reason this vanishes in the integer-$\nu$ case is because the angular integration, over the argument of $\alpha$, averages out to zero, unless all the phases cancel out exactly (in which case the radial integral fixes $\nu$). This will in general, I think, not happen.

To see whether it does, write $\alpha=r e^{i\theta}$ and $\log(\alpha)= \ln(r) +i\theta +2\pi i n(\theta)$, where $n(\theta)$ is an integer-valued function such that $n(\theta+2\pi)=n(\theta)-1\ \forall\theta$, and which embodies your branch cut. The integral above then reduces to $$ \iint e^{-r^2} r^{m+n+\nu-1} e^{i(m-n+\nu-1)\theta} e^{2\pi i n(\theta)(\nu-1) } \left[ r^2-m-\nu \right] r\text d\theta\text dr =0. $$ Thus, you want the product of integrals $$ \int e^{i(m-n+\nu-1)\theta} e^{2\pi i n(\theta)(\nu-1) } \text d\theta \times \int_0^\infty e^{-r^2} r^{m+n+\nu-1} \left[ r^2-m-\nu \right] r\text dr $$ to vanish. For concreteness, take $n(\theta)$ to be zero for $\theta\in(-\pi, \pi]$, and integrate over that interval. This yields, after reducing the radial integral to $$ \Gamma\left(\frac{m+n+\nu+1}{2}+1\right) -(m+\nu)\Gamma\left(\frac{m+n+\nu+1}{2}\right) \propto \left[ \frac{m+n+\nu+1}{2}-(m+\nu) \right], $$ the integral $$ 0\stackrel{?}{=} i\left(m-n+\nu-1\right) \int_{-\pi}^{\pi} e^{i(m-n+\nu-1)\theta} \text d\theta = \left. e^{i(m-n+\nu-1)\theta} \right|_{-\pi}^{\pi} = e^{i(m-n+\nu-1)\pi} -e^{i(m-n+\nu-1)\pi}. $$ Thus, then: for (1) to be a solution of (2), you need $\sin\left((m-n+\nu-1)\pi\right)=0$ or, equivalently, $$\boxed{\sin(\nu\pi)=0}$$ to hold. This means that the construction in (1) only obeys the commutation relations for integer $\nu$. Bummer!

share|improve this answer
    
Great answer! maybe i'm mistaken, but i think in the coherent state resolution of unity you missed a factor $\exp(-\alpha\alpha^*)$, exactly because they are not mutual orthogonal and in fact overcomplete. –  nephente Nov 19 '13 at 18:02
    
@Emilio: Just as I had thought, and I had no clue how to handle the arbitrary branch cuts when I tried defining these previously. I will have to go through your entire argument carefully before making any further remarks. I have edited my question, so I shall wait for further answers before accepting a final solution. –  surajshankar Nov 19 '13 at 18:26
    
@nephente No, the form is correct; see e.g. Wikipedia. I was off, though, by a factor of 2. Thanks for pointing it out! –  Emilio Pisanty Nov 19 '13 at 21:52
    
@EmilioPisanty Sorry. I went by the convention introduced in Altland/Simons "Condensed Matter Field Theory". They define the bosonic coherent states as $\exp(\alpha a^\dagger)$, not $\exp(\alpha a^\dagger-\alpha^* a)$. I think the latter one is the more commonly used, so that's a bit unconventional. –  nephente Nov 20 '13 at 15:06
1  
@EmilioPisanty I agree. Normalizing to unity makes more sense, but I was taught differently. Doesn't really matter. –  nephente Nov 20 '13 at 16:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.