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Archive for May 23rd, 2010

Effect of sinusoid operators

Posted by peeterjoot on May 23, 2010

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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.


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

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Time evolution of some wavefunctions (incomplete)

Posted by peeterjoot on May 23, 2010

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In [1] is problem 3.14, Describe the time evolution of the following wavefunctions

\begin{aligned}\psi_1 &= A \sin \omega t \cos k (x + c t) \\ \psi_2 &= A \sin 10^{-5} k x \cos k (x - c t) \\ \psi_3 &= A \cos k ( x - c t ) \sin 10^{-5} k (x - c t)\end{aligned}

This isn’t really a QM problem, but seems worthwhile anyways, because it isn’t obvious looking at the functions what this is.


Let’s start in reverse order with \psi_3, but in a slightly more general form that is less error prone to manipulate. These wavefunctions can be viewed as superpositions, and expanding out as exponentials temporarily gets us to a form that makes this more obvious.

\begin{aligned}\psi &= A \sin k_1 ( x + v_1 t) \cos k_2 ( x + v_2 t) \\ &= \frac{A}{4i} \left( e^{ i k_1 ( x + v_1 t)} - e^{ -i k_1 ( x + v_1 t)} \right) \left( e^{ i k_2 ( x + v_2 t)} + e^{ -i k_2 ( x + v_2 t)} \right) \\ &= \frac{A}{2} \left( \sin ((k_1 + k_2) x + (k_1 v_1 + k_2 v_2 ) t) + \sin ((k_1 - k_2) x + (k_1 v_1 - k_2 v_2 ) t) \right) \\ &= \frac{A}{2} \left( \sin \left( (k_1 + k_2) \left(x + \frac{k_1 v_1 + k_2 v_2 }{k_1 + k_2} t\right) \right)+\sin \left( (k_1 - k_2) \left(x + \frac{k_1 v_1 - k_2 v_2 }{k_1 - k_2} t\right) \right)\right)\end{aligned}

Now the problem is simplified to observing how a wave of the form \phi = \sin \kappa (x + v t) propagates, or really the interaction of two such waves moving together or against each other, depending on the signs of the constants. Let’s now put in the constants for \psi_3 to get a better feel for it

\begin{aligned}\psi_3&= \frac{A}{2} \left( \sin \left( 1.00001 k \left(x - c t\right) \right)-\sin \left( 0.99999 k \left(x - c t\right) \right)\right)\end{aligned}

It wasn’t obvious from the original product of sinusoids form that the question asked about that the resulting wave form stays in phase for its time propagation, but we see that to be the case above. This really just leaves some thought about the standing wave itself to understand what is happening. For that, at time 0, we have a destructive interference superposition of two almost identical period standing waves. That near perfect cancellation will likely leave an envelope, and a Mathematica plot gives a better feel for this waveform. This in turn will propagate at light speed down the x-axis. Because k_1 is so small we have a nearly linear, and nearly flat, envelope for the \cos k x, as can be expected near the origin since we have there

\begin{aligned}\psi_3(x, 0) \approx A 10^{-5} k x \cos k x\end{aligned}

Comparing to a smaller range plot, one sees that it is necessary to increase the plot range significantly before seeing the oscillatory nature of the envelope.


For \psi_2 we have almost the same wave function, but out sine term has no time variation. What does this do to the waveform? Let’s see if a sum and difference of angles form sheds some light on that. We have

\begin{aligned}\psi &= A \sin k_1 x \cos k_2 ( x + v_2 t) \\ &= \frac{A}{4i} \left( e^{ i k_1 x } - e^{ -i k_1 x } \right) \left( e^{ i k_2 ( x + v_2 t)} + e^{ -i k_2 ( x + v_2 t)} \right) \\ &= \frac{A}{2} \left( \sin ((k_1 + k_2) x + k_2 v_2 t) + \sin ((k_1 - k_2) x - k_2 v_2 t) \right) \\ &= \frac{A}{2} \left( \sin \left( (k_1 + k_2) \left(x + \frac{k_2 v_2 }{k_1 + k_2} t\right) \right)+\sin \left( (k_1 - k_2) \left(x - \frac{k_2 v_2 }{k_1 - k_2} t\right) \right)\right)\end{aligned}

Specifically for k_1 = 10^{-5} k, and k_2 = k, we have

\begin{aligned}\psi_2&= \frac{A}{2} \left( \sin \left( 1.00001 k \left(x - \frac{1}{1.00001} c t\right) \right)-\sin \left( 0.99999 k \left(x + \frac{1}{0.99999 } c t\right) \right)\right)\end{aligned}

Again at t=0 we have a very widely spread envelope with rapid oscillations within it. There is a very small difference in the rate that the two components waveforms will go out of phase, and each of the component waveforms is moving in the opposite directions. What does that phase change do to the evolution? Looking back to the original product of sinusoids form for \psi_2 I believe this will just mean we have the phase shifting with time within the envelope.


Finally for the first wave function we have both of the sinusoid factors with time variation. What does that expand out to in terms of superposition of fundamental frequencies?

\begin{aligned}\psi_1&= A \sin \omega t \cos k ( x + c t) \\ &= \frac{A}{4i} \left( e^{ i \omega t} - e^{ -i \omega t } \right) \left( e^{ i k ( x + c t)} + e^{ -i k ( x + c t)} \right) \\ &= \frac{A}{2} \left( \sin ( k (x + c t) + \omega t )-\sin ( k (x + c t) - \omega t )\right),\end{aligned}


\begin{aligned}\psi_1&= \frac{A}{2} \left( \sin ( k x + ( k c + \omega ) t )-\sin ( k x + ( k c - \omega) t )\right)\end{aligned}

We have the superposition of two \sin k x wave forms, destructively interfering with each other, one with phase changing at the angular rate k c + \omega, and the other at the rate k c - \omega. What is the overall waveform? It’s still not obvious what this is. I actually have the inclination to try to not treat these analytically, but pull out some graphing software. Something like the real mathematica software would be nice since it would allow for the use of sliders to vary parameters and then animate the graphs as time varied.


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

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