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$$\hat{p}^2=\hat{p}_r\hat{p}_r=(-i\hbar)^2\frac{1}{r}\frac{\partial}{\partial r}r\frac{1}{r}\frac{\partial}{\partial r}r=-\hbar^2\frac{1}{r}\frac{\partial^2}{\partial r^2}r$$

I'm trying to understand what is happening here. I understand the part $(-i \hbar)^2=-\hbar^2$, but not the rest. I would expect the result to be something like $-\hbar^2 * \frac1 r^2 * \frac{d^2} {dr^2} * r^2 $.

The way it seems to me now, is that multiplying operators isn't the same as multiplying numbers? But instead it means applying the first operator to the result of the later operator? That would mean (here I leave out the constants) applying $\frac1 r \frac{d} {dr} r$ to $\frac1 r \frac{d} {dr} r$ and $r * \frac1 r $ is $1$ and then we get the result.

But then, why is $\nabla * \nabla$ a dot product, considering that multiplication doesn't work the simply anymore? For example, it seems like when multiplying operators, you can't switch around order anymore. So why does the dot product rule still apply? After all, $ \Delta$ yields a number, but $\nabla$ is a vector and still yields a vecor right? Simply applying $\nabla$ twice instead of $\nabla * \nabla = \Delta$ would still yield a vector, right?

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  • $\begingroup$ For your last paragraph: you are applying $\Delta$ to a scalar $f$ and getting a scalar, right? Well, let's see what's happening here. $\nabla f$ is a vector (the gradient of a scalar field $f$). Then we apply another $\nabla$ from the left, we are getting the divergence of the vector field $\nabla f$, which is again, a scalar. $\endgroup$ – Yuriy S Sep 29 '18 at 21:50
  • $\begingroup$ Will the identity $\nabla f(r)=\frac{\vec r}{r} \partial_r f(r)$ help you? $\endgroup$ – Cosmas Zachos Sep 30 '18 at 0:02
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Nothing special has happened. To get from the second to last, to the last equation all you have to do is realize $$ r\frac{1}{r}=1 $$ and it follows immediately. In general, operator algebra is associative which means you may apply operators products of operators in any order you like as long as you preserve their sequence i.e. $$A(BC)=(AB)C$$ I see no reason to expect the result to be $$-\hbar^2 \frac{1}{r^2} \frac{\partial^2} {\partial r^2} r^2$$ as you suggest. Also remember that your result is still an operator, so you must apply $$\left(\frac{\partial^2} {\partial r^2} r^2\right)$$ to whatever is acted upon by $\hat{p}^2$.

As for your confusion with the Laplacian or $\nabla^2$, you have to recognize that this is all just shorthand. The Laplacian is defined as $\nabla\cdot\nabla$ so that's just what it is.

If you want $\nabla \left(\nabla f\right)$ to be defined then you must take some care. $\nabla f$ represents the gradient of a scalar function which is itself a vector. So if you try to apply $\nabla$ to it again, how should you apply it? The gradient is defined acting on a scalar field, but now you are trying to apply it to a vector. Normally when you apply $\nabla$ to a vector, you use the divergence or $\nabla\cdot\vec{f}$ so then we'd have $\nabla\cdot \left(\nabla f\right)$ or $\left(\nabla\cdot \nabla\right) f$ which is exactly what we've defined the Laplacian as.

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  • $\begingroup$ I just understood why you were thinking the result would be $$-\hbar^2 \frac{1}{r^2} \frac{\partial^2} {\partial r^2} r^2$$ You just moved the $r$'s through the derivatives. But you know this can't work from basic calculus. $$r \frac{\partial}{\partial r} r = \frac{\partial}{\partial r} r^2 = 2r?$$ Or even more extreme $$r^2\frac{\partial}{\partial r} 1 = \frac{\partial}{\partial r} r^2 = 2r?$$ You can only pull something through a derivative when it's a constant w.r.t. the derivative. $\endgroup$ – bRost03 Sep 30 '18 at 0:59
  • $\begingroup$ That makes sense! So basically multiplying operators means applying one operator to the other, and not squaring the result of one Operator? Because the second one was what I thought had to happen. $\endgroup$ – MathyPulse Sep 30 '18 at 7:56

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