Derivation of kinetic energy operator QM

Question

In griffiths introduction to quantum mechanics, he starts off by writing

$$\langle x\rangle = \int x|\psi(x,t)|^2 dx$$ $$\langle x \rangle = \int \psi^* [x] \psi(x,t) dx$$

He then says that the operator $\hat x = x$ , 'represents position' sandwiching the operator between the complex conjugate and the wave function yields the expectation value of position.

He then postulates that:

$$\langle p \rangle = m\frac{d\langle x\rangle}{dt}$$

$$m\frac{d\langle x\rangle}{dt} = m\frac{d}{dt}\int x|\psi(x,t)|^2 dx$$

He then shows that given schrodingers equation is true that we can express this quantity as

$$\langle p \rangle= \int \psi^* [-i\hbar\frac{\partial}{\partial x}]\psi$$

Where the operator $\hat p =-i\hbar\frac{\partial}{\partial x}$ 'represents momentum' sandwiching the operator between the complex conjugate and the wave function yields the expectation value of momentum.

I am all fine upto this point. However how would this logic continue on to deriving the expectation value for the kinetic energy?

I would intuitively say that $$\langle Ke \rangle = \frac{\langle p\rangle^2}{2m}$$ However I know that from probability theory the expectation value of the square and the square of the expectation value are not the same, And hence would expect $$\langle Ke\rangle = \frac{\langle p^2\rangle}{2m}$$

I know the standard derivation is to apply the momentum operator twice, however is there a proof that applying the operator twice, and finding the expectation value, is the actual expectation value of the quantity p^2?

I recall something I have read about constructing a momentum^2 probability density and then doing a fourier transform, however I am not that familiar with fourier transforms.

Can someone point me in the right direction on this last part as I feel this is the correct method, rather than blindly applying the operator twice, when the operator only represents such quantity if sandwiched between the wave functions

Answer 1

I recall something I have read about constructing a momentum^2 probability density and then doing a fourier transform, however I am not that familiar with fourier transforms.

The single-particle probability density in real space is $P(\vec r) = |\Psi(\vec r)|^2$.

The single-particle probability density in momentum space is: $$ \Pi(\vec p)\;, $$ but what is this in terms of the wave function?

Well, consider the following Fourier transforms. (You must understand Fourier transforms well if you want to proceed in your studies of quantum mechanics): $$ \Phi(\vec p)\equiv \int \frac{d^3 r}{(2\pi\hbar)^{3/2}}\Psi(\vec r)e^{-i\vec p\cdot \vec r/\hbar}\; $$

Thus, by well known properties of Fourier transforms, $$ \Psi(\vec r)= \int \frac{d^3 p}{(2\pi\hbar)^{3/2}}\Phi(\vec p)e^{+i\vec p\cdot \vec r/\hbar}\;, $$ so that we now see that $\Psi(\vec r)$ can be viewed as a linear combination of waves with well-defined momenta $\vec p$ weighted by the $\Phi(\vec p)/(2\pi\hbar)^{3/2}$ coefficients. (You can check for yourself that $e^{+i\vec p_0\cdot \vec r/\hbar}$ is an eigenfunction of the momentum operator with eigenvalue $\vec p_0$.)

If there was only one $e^{+i\vec p_0 \cdot \vec r/\hbar}$ in the linear combination for $\Psi(\vec r)$ (e.g., if $\Phi(\vec p)$ was a delta function at $\vec p_0$) then the momentum would be $\vec p_0$ with certainty. If $\Phi(\vec p)$ was peaked about $\vec p_0$ with some small width, the measured value of $\vec p$ would likely be $\vec p_0$ with the uncertainty being related to the width of $\Phi(\vec p)$.

Thus, we are led to define: $$ \Pi(\vec p)\equiv |\Phi(\vec p)|^2\;, $$ where one can show that $\int d^3p \Pi(\vec p) = 1$ and clearly $\Pi(\vec p)$ is non-negative, and these properties are all consistent with $\Pi(\vec p)$ being the probability density in momentum space (so our definition of $\Pi(\vec p)$ is reasonably and workable).

Given the probability density $\Pi(\vec p)$, the expectation value of any function of the momentum $f(\vec p)$ is: $$ \int d^3p f(\vec p)\Pi(\vec p) = \int d^3p \Phi^*(\vec p)f(\vec p)\Phi(\vec p)\;. $$

Thus, the expectation value of the kinetic energy is: $$ \langle K \rangle = \int d^3p \frac{p^2}{2m}\Pi(\vec p) $$ $$ =\langle \frac{\hat p^2}{2m}\rangle\;, $$ regardless of which basis you work in. (Regardless of if you work in real space with $\Psi(\vec r)$ or in momentum space with $\Phi(\vec p)$).

For example, if you want to transition back to the $\vec r$ basis: $$ \langle \frac{\hat p^2}{2m}\rangle = \int d^3p \frac{p^2}{2m}\Pi(\vec p) $$ $$ =\int d^3p \frac{p^2}{2m}\Phi^*(\vec p)\Phi(\vec p) $$ $$ \int d^3p \frac{p^2}{2m}\int \frac{d^3r d^3 r'}{(2\pi \hbar)^3}\Psi^*(\vec r')e^{+i\vec p\cdot \vec r'/\hbar}\Psi(\vec r)e^{-i\vec p\cdot \vec r/\hbar} $$ $$ =\int d^3p \frac{1}{2m}\int \frac{d^3r d^3 r'}{(2\pi \hbar)^3}\Psi^*(\vec r')e^{+i\vec p\cdot \vec r'/\hbar}\Psi(\vec r)(-\nabla^2\hbar^2e^{-i\vec p\cdot \vec r/\hbar}) $$ $$ =\int d^3p \frac{1}{2m}\int \frac{d^3r d^3 r'}{(2\pi \hbar)^3}\Psi^*(\vec r')e^{+i\vec p\cdot \vec r'/\hbar}e^{-i\vec p\cdot \vec r/\hbar}(-\nabla^2\hbar^2\Psi(\vec r))\;, $$ where the last step above uses integration by parts on $\vec r$. $$ =\frac{1}{2m}\int \frac{d^3r d^3 r'}{(2\pi \hbar)^3}\Psi^*(\vec r')(2\pi\hbar)^3\delta^3(\vec r - \vec r')(-\nabla^2\hbar^2\Psi(\vec r)) $$ $$ =\frac{1}{2m}\int d^3r\Psi^*(\vec r)(-\nabla^2\hbar^2\Psi(\vec r)) $$ $$ =\int d^3r\Psi^*(\vec r)(\frac{-\nabla^2\hbar^2}{2m}\Psi(\vec r)) $$ $$ =\langle \Psi|\frac{\hat p^2}{2m}|\Psi\rangle $$

Answer 2

Your assumption about the expectation of kinetic energy is incorrect. The correct formulation would be to just take the expectation value by sandwiching the entire kinetic energy operator inside of the wave function bracket, \begin{gather*} \langle K \rangle = \langle \psi | \frac{\hat{p}^2}{2m} | \psi \rangle = \frac{\langle p^2 \rangle}{2m} \end{gather*} based on the definition of a quantum expectation value. You are correct that in general there is no relation between $\langle p \rangle$ and $\langle p^2 \rangle$, but that is no matter because the form of the kinetic energy operator is simply $\frac{\hat{p}^2}{2m}$. Now as to why the momentum operator has the peculiar form $\hat{p} = - i\hbar \frac{\partial}{\partial x}$, this is a question of finding Hermitian operators that serve as the generators of spacetime translations of the wave function. The time translation generator is the Hamiltonian, and we make the association with the momentum and total energy explicit by showing that operators $\hat{x}$, $\hat{p}$, and $\hat{H}$ obey the quantum versions of Hamilton's equations, as in the question I asked.

Answer 3

It seems like you are trying to derive a relation from an approximation of that relation.

The laws of classical mechanics were developed to agree with experiments done at the time. But those experiments only provided results according to the precision of the measuring instruments of the time. The difference between $\langle p^2\rangle$ and $\langle p\rangle^2$ is the squared standard deviation $\sigma_p^2$. I don't know how accurate they could measure in the 1800's, but the uncertainty principle states gives a lower bound for uncertainty of about $\sigma_x\sigma_p \approx 10^{-34}Js$ (very tiny). In other words there was no way they could possibly know there was something wrong with the formula $T = \frac{p^2}{2m}$ which, according to the retrospective knowledge that all classical measurements of $p$ are actually measurements of $\langle p \rangle$ and similarly for $T$ and $\langle T\rangle$, wrongly equates $\langle T\rangle = \frac{\langle p\rangle^2}{2m}$.

The point being, we can't use classical mechanical formulas to choose between different quantum mechanical formulas if the difference in experimental predictions between them is classically undetectable.