# Kinetic Energy operator in Quantum Mechanics

Firstly, I'd like to point out clearly that I'm not a physicist but I'm a nano engineer studying quantum mechanics so that I understand my work on surface sciences better, so please don't presume my knowledge because I lack the stringent background of a typical physicist.

Having said that, if I start form Born's interpretation $$\int_{-\infty}^{+\infty}\psi\psi^{*}\,\mathrm{d}x=1 \tag{1}$$ I understand the most expectant value $<x>$ could be defined as $$\left<x\right>=\int_{-\infty}^{+\infty}\psi^{*}x\psi\,\mathrm{d}x.\tag{2}$$

From here, were I to derive the momentum operator which as I understand, only from the mathematics (remember I'm an absolute illiterate in physics), is a **mass* scaled value of* the rate of change of the most expected value of the wavefunction or which I would call the time evolution of the wave function which is $$\left<p\right>=m\frac{\partial\left<x\right>}{\partial t}\tag{3}.$$ Hence,

$$\left<p\right>=\int_{-\infty}^{+\infty}x\left(\psi^{*}\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^{*}}{\partial t}\right)\,\mathrm{d}x.\tag{4}$$ Now, looking at .the TDSE, multiplying with the conjugate wavefunction and taking complex conjugates and double integrating twice, I land with $$\left<p\right>=\frac{i\hbar}{2}\int_{-\infty}^{+\infty}\left(\psi\frac{\partial\psi^{*}}{\partial x}-\psi^{*}\frac{\partial \psi}{\partial x}\right)\,\mathrm{d}x.\tag{5}$$

I also figure that $$\int_{-\infty}^{+\infty}\psi\frac{\partial \psi^{*}}{\partial x}\,\mathrm{d}x=-\int_{-\infty}^{+\infty}\psi^{*}\frac{\partial \psi}{\partial x}\,\mathrm{d}x.\tag{6}$$ Hence, $$\left<p\right>=\int_{-\infty}^{+\infty}\psi^{*}\frac{\hbar}{i}\frac{\partial \psi}{\partial x}\,\mathrm{d}x.\tag{7}$$ Thus the momentum operator could be written as $$\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}.\tag{8}$$ Now, the question is when I try to derive the kinetic energy operator, is the below equation the right way to go $$\left<KE\right>=\frac{1}{2m}\left<p\right>\left<p\right>\tag{9}$$ or should I rather aim for $$\hat{KE}=\frac{1}{2m}\hat{p}\hat{p}~?\tag{10}$$ The bottom approach doesn't make any sense at all firstly, neither mathematically nor physically. The top approach does make mathematical sense, but I don't see the physical sense in it AT ALL. I mean, we call the mass scaled rate of change of the most expectant value of the distribution as momentum and then take the square of that and call its scaled value as kinetic energy. It doesn't seem to make any sense, but what's more annoying is that the bottom approach directly gives results and makes neither physical nor mathematical sense to me. I'm unable to join any dots logically. Kindly help.

There is nothing to "derive" for the kinetic energy operator. By definition, classical kinetic energy is $\frac{p^2}{2m}$, and so $\hat{E}_\text{kin} = \frac{\hat{p}^2}{2m}$ quantumly. It's not exactly clear why you think this doesn't make sense mathematically, but it does: In words, it says "apply the momentum operator twice, then divide the result by $2m$".
Note that $\langle p^2\rangle \neq \langle p\rangle^2$, the difference is precisely what the standard deviation is defined as and what is usually called the "uncertainty" $\Delta p$ in most physics texts.
• @ACuriousMind: The reason I presume it dosen't make mathematical sense is because I first derived, as shown in the question, the value of $\left< p\right>$ from $\left< x\right>$ and then went on to say that the operator $\hat{p}=\frac{\hbar}{i}\partial/\partial x$. Hence, I'd be satisfied if I could derive $\left<KE\right>$ from $\left<p\right>$ and then go on to write/simplify/extract the value of $\hat{KE}$ from it. Is there a way to derive $\left<KE\right>$ from $\left<p\right>$? – ubuntu_noob May 9 '17 at 1:02