Suppose we have a quantum state, well described by its time-independent wave function Psi. And we have a well-defined Hermitian (self-adjoint) operator $A$. We successfully evaluate the expectation value of the operator $A$. Next we derive the general formula for the higher moments of $A$ (i.e. the expectation value of $A^n$ for $n=2,3,4\ldots $).

In this situation, is it permitted to regard each of the $A^n$ for $n=1,2,3,\ldots$ as a proper operator by itself?

For example, should every $A^n$ have a positive variance and other statistical properties (as long as we restrict ourselves to the state $\Psi$)?

Can one make linear combinations of different powers to construct a new operator, e.g. $B = A + A^2$?

Is it allowed to construct new operators acting on $\Psi$, that are defined in terms of their series expansion in $A^n$? For example, $D = \exp(CA)$ where $C$ is a constant?


Most of your question is unclear to me, but I the answer to what I think is the core of your question is yes:

  • For any hermitian operator $A$ and any well-behaved-enough function $f:\mathbb R\to\mathbb R$, it is possible to construct a new operator $f(A)$ which acts in essentially all important respects as the action of $f$ on $A$.

There are different ways to construct $f(A)$, and they depend on exactly what $f$ is, what "well-behaved-enough" means, and how nice $A$ is. In general, the keyword to search for is function of an operator.

For example, if $f$ is analytic in a large enough region - one that includes all the spectrum of $A$, for example, then you can define it as $$ f(A)=\sum_{n=0}^\infty \frac{f^{(n)}(a_0)}{n!}(A-a_0)^n. $$

Note, in particular, that this includes functions of the form $A+A^2$, which are perfectly allowed. If $A$ is a linear operator then so is $A^2$, and adding two linear operators is bread and butter in linear algebra. There are a few caveats - for example, if the domain of $A$ is smaller than the Hilbert space then the domain of $A^2$ will typically be smaller, so the sum only makes sense in that restricted domain - but this is only the standard measure of care one needs to take when infinite-dimensional spaces are involved.

Alternatively, if $A$ has an eigenvector expansion as $A=\sum_k a_k|a_k\rangle\langle a_k|$, then you can define $$f(A)=\sum_k f(a_k)|a_k\rangle\langle a_k|.$$ If everything behaves well, then both definitions will match.

Finally, note that one should keep an eye on the dimensional analysis of the whole thing. If $A=x$ has dimensions of position, then it does not make sense to add $x+x^2$, any more than it does to do this in classical mechanics. This is particularly the case with, say, exponentials of operators, like the displacement operator $$ e^{ix_0\hat p}=\sum_{n=0}^\infty i^n\frac{x_0^n}{n!}\hat p^n $$ which only makes sense in units where $\hbar=1$. (Otherwise, you need to add in the $\hbar$ explicitly, as $\exp(ix_0\hat p/\hbar$.)

| cite | improve this answer | |
  • $\begingroup$ There is another possibility to define $f(A)$, which afaik needs less requirements on $f$ and $A$. Specifically, it works for Borel-measurable functions $f: \mathbb R \rightarrow \mathbb C$ and self-adjoint operators $A$. $f(A)$ is then given by the Helffer-Sjöstrand formula: $f(A)=\frac{1}{\pi}\int_{\mathbb C} \frac{\partial \tilde f}{\partial \bar z}(z)(A-z)^{-1}\mathrm{d}x \mathrm{d}y$. Here, $\tilde f$ is the almost analytic continuation of $f$, $z=x + \mathrm{i}y$ and $(A-z)^{-1}$ the resolvent of $A$. $\endgroup$ – Simeon Carstens Jun 13 '14 at 8:52
  • $\begingroup$ It's been years that I learned about this and I have to admit I never really understood completely. So to the more mathematically inclined people here: please correct me if I'm mistaken in the above comment. $\endgroup$ – Simeon Carstens Jun 13 '14 at 8:56
  • 1
    $\begingroup$ @Emilio Pisanty Hi Emilio. Unfortunately your recipe based on Taylor series generally fails unless $A$ is bounded (which is quite uncommon in general QM). If restricting to very particular cases of $f$ like the exponential, it works applying bot sides to very particular vectors called "analytic vectors". The only always working procedure is the spectral one: $f(A) := \int_{\sigma(A)} f(\lambda) dP^{(A)}(\lambda)$ for every Borel measurable function $f: \sigma(A) \to \mathbb C$, where $P^{(A)}$ is the spectral measure of $A$ (supposed to be self-adjoint or at least normal). $\endgroup$ – Valter Moretti Jun 13 '14 at 10:00
  • $\begingroup$ @V.Moretti thanks for the precision. I was deliberately hazy (hence "well-behaved-enough") partly because I don't know the precise results off the top of my head but also to avoid dizzying the OP with spectral measures. But point taken. $\endgroup$ – Emilio Pisanty Jun 14 '14 at 0:49

Sure. Anything that maps one state to another is an operator. If $A$ satisfies this definition, namely that when applied to a state it gives you a state, then so does repeated application of $A$.

For example, suppose you have a set of quantum states $\lvert i\rangle$ for various values of $i$, parametrized so that $A\lvert i\rangle = \lvert i+1\rangle$. Then

$$A^2\lvert i\rangle = A(A\lvert i\rangle) = A\lvert i+1\rangle = \lvert i+2\rangle$$

Hopefully you can see how this generalizes, so that $A^2$ is the operator that takes $\lvert i\rangle\to\lvert i+2\rangle$.

And yes, you can generalize this to construct an operator as a power series of other operators. This is how the exponential of an operator is defined, for example.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much for your clear reply! I am worried that this great freedom to create new operators in terms of linear combinations of other operators can lead to odd results, that are in conflict with statistical rules. $\endgroup$ – M. Wind Jun 13 '14 at 13:07
  • $\begingroup$ In fact I have just posted a new thread "Significance of an operator with a negative variance". It is an example of an operator A with fairly normal moments evaluated in some state. From this a new operator B is constructed which has very odd properties. $\endgroup$ – M. Wind Jun 13 '14 at 14:28

For a continuous (linear) operator $A:H\to H$, aka. bounded operator, it is always possible to construct well-defined powers $A^2$, $A^3$, $\ldots.$ Here $H$ denotes a (complex) Hilbert space.

However, the situation changes drastically for a general unbounded operator $A$. Unbounded operators often appear in quantum mechanics, see e.g. this, this and this Phys.SE posts.

The domain $D(A)\subsetneq H$ of an unbounded operator is never the full Hilbert space! Therefore if the image ${\rm Im}(A)\subseteq H$ is not a subset of the domain $D(A)$, then the square operator $A^2$ does not necessarily make sense on the full domain $D(A)$ of $A$. In other words, the domain $D(A^2)$ of the square operator $A^2$ is in general different from the domain of $A$! Similar for higher powers of $A$.

The topic of unbounded operators is a huge subject in functional analysis, which is impossible to cover in a single Phys.SE post. For a student of operator theory, the natural next couple of questions to ponder is:

  1. Is it possible to extend a domain $D$ of an unbounded operator $A$?

  2. If yes, is there a natural way to partially order the set of all possible domains of an unbounded operator $A$?

  3. Is there a canonical choice of a domain for an unbounded operator $A$?

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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