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Effect of sinusoid operators

Posted by Peeter Joot on May 23, 2010

Problem 3.19.

[1] problem 3.19 is

What is the effect of operating on an arbitrary function $f(x)$ with the following two operators

\begin{aligned}\hat{O}_1 &\equiv \partial^2/\partial x^2 - 1 + sin^2 (\partial^3/\partial x^3)+ cos^2 (\partial^3/\partial x^3) \\ \hat{O}_2 &\equiv + cos (2 \partial/\partial x) + sin^2 (\partial/\partial x)+ \int_a^b dx\end{aligned} \hspace{\stretch{1}}(1.1)

On the surface with $\sin^2 y + \cos^2 y = 1$ and $\cos 2y + 2 \sin^2 y = 1$ it appears that we have just

\begin{aligned}\hat{O}_1 &\equiv \partial^2/\partial x^2 \\ \hat{O}_2 &\equiv 1 + \int_a^b dx\end{aligned} \hspace{\stretch{1}}(1.3)

but it this justified when the sinusoids are functions of operators? Let’s look at the first case. For some operator $\hat{f}$ we have

\begin{aligned}\sin^2 \hat{f} + \cos^2 \hat{f}&=-\frac{1}{{4}} \left( e^{i\hat{f}} -e^{-i\hat{f}}\right)\left( e^{i\hat{f}} -e^{-i\hat{f}}\right)+\frac{1}{{4}} \left( e^{i\hat{f}} +e^{-i\hat{f}}\right)\left( e^{i\hat{f}} +e^{-i\hat{f}}\right) \\ &=\frac{1}{{2}} \left(e^{i\hat{f}} e^{-i\hat{f}} +e^{-i\hat{f}} e^{i\hat{f}}\right)\end{aligned}

Can we assume that these cancel for general operators? How about for our specific differential operator $\hat{f} = \partial^3/\partial x^3$? For that one we have

\begin{aligned}e^{i \partial^3/\partial x^3} e^{-i \partial^3/\partial x^3} g(x)&=\sum_{k=0}^\infty \frac{1}{{k!}} \left(\frac{\partial^3}{\partial x^3}\right)^k\sum_{m=0}^\infty \frac{1}{{m!}} \left(\frac{\partial^3}{\partial x^3}\right)^m g(x)\end{aligned}

Since the differentials commute, so do the exponentials and we can write the slightly simpler

\begin{aligned}\sin^2 \hat{f} + \cos^2 \hat{f} = e^{i\hat{f}} e^{-i\hat{f}} \end{aligned}

I’m pretty sure the commutative property of this differential operator would also allow us to say (in this case at least)

\begin{aligned}\sin^2 \hat{f} + \cos^2 \hat{f} = 1\end{aligned}

Will have to look up the combinatoric argument that allows one to write, for numbers,

\begin{aligned}e^x e^y = \sum_{k=0}^\infty \frac{1}{{k!}} x^k \sum_{m=0}^\infty \frac{1}{{m!}} y^m =\sum_{j=0}^\infty \frac{1}{{j!}} (x+y)^j = e^{x+y}\end{aligned}

If this only assumes that $x$ and $y$ commute, and not any other numeric properties then we have the supposed result 1.3. We also know of algebraic objects where this does not hold. One example is exponentials of non-commuting square matrices, and other is non-commuting bivector exponentials.

References

[1] R. Liboff. Introductory quantum mechanics. 2003.