Quantum Mechanics time derivative of operators [duplicate]

Ok so Griffiths, Introduction to Quantum Mechanics page 32. If you look at equation 1.29, what he is trying to do is essentially take the time derivative of the expectation value of the position operator. That is all good until the part right above equation 1.30 where he says "This can be simplified using Integration by Parts". Can someone explain to me how he simplifies equation 1.29 to 1.30, because I've tried it and it is not giving me the same answer.

• Can you show what you’ve tried? Jun 16, 2022 at 2:41
• Integration by parts with $u = x$ and $dv = (\psi^* \frac{\partial \psi}{\partial x} - \frac{\partial \psi^*}{\partial x} \psi)dx$. From this, $du = dx$ and $v = \psi^* \frac{\partial \psi}{\partial x} - \frac{\partial \psi^*}{\partial x} \psi$ by the fundamental theorem of calculus. Jun 16, 2022 at 2:42
• @CameronWilliams your $dv$ is missing a $\frac{\partial}{\partial x}$. Jun 16, 2022 at 3:06
• Does this answer your question? Integration by parts to derive $d\langle x \rangle / dt$ Jun 16, 2022 at 3:21
• @J. Murray Look at the 2 answers, I got it from then. I just didn't figure out what to do with that second term whilst perfoming IBP so yeah, was a bit unsure on that but it just goes to 0 because if you imagine normalizing the wave function, a similar term pops up there, and it goes to 0 if you apply limits of integration, and if you multiply that by the expectation value of position, you would get 0. Jun 16, 2022 at 4:19

I am not 100% sure why we go from derivatives involving time to derivatives involving position. I will nevertheless thake the rightmost term in (1.29) and start from there. Let us ignore the constants.

$$$$\int x\frac{\partial}{\partial x}\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)dx$$$$ let us use integration by parts where $$x = U$$ and $$\frac{\partial}{\partial x^\prime}\Big( \psi^{*}\frac{\partial\psi}{\partial x^\prime} - \frac{\partial\psi}{\partial x^\prime}\psi^{*} \Big)dx = dV$$. Integrtion by parts says that

$$\int U dV = UV\Big|_{-\infty}^{\infty} -\int V dU$$ where we are integrating over the entire real line.

Notice that $$dU = dx$$ and that $$V = \int\frac{\partial}{\partial x^\prime}\Big( \psi^{*}\frac{\partial\psi}{\partial x^\prime} - \frac{\partial\psi}{\partial x^\prime}\psi^{*} \Big)dx =\Big( \psi^{*}\frac{\partial\psi}{\partial x^\prime} - \frac{\partial\psi}{\partial x^\prime}\psi^{*} \Big)$$ Ignore the constant of integration. $$\int x\frac{\partial}{\partial x}\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)dx = x\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)\Big|_{-\infty}^{\infty}- \int VdU =$$

$$x\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)\Big|_{-\infty}^{\infty}- \int VdU =$$ $$x\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)\Big|_{-\infty}^{\infty}- \int \Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big) dx$$

the sneaky assumption being made is that the wave fucntion $$\psi$$, its derivative and ints comple conjugate all go to zero at infinity faster than the linear term $$x$$ blows up. If you use a gaussian for $$\psi$$ you will see that this is so. Therefore the boundary term is just zero, i.e. $$x\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)\Big|_{-\infty}^{\infty} = 0$$

hence $$\int x\frac{\partial}{\partial x}\Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big)dx = - \int \Big( \psi^{*}\frac{\partial\psi}{\partial x} - \frac{\partial\psi}{\partial x}\psi^{*} \Big) dx.$$

This should do it. I was not too careful with the integration dummy variables and a few other details but the framework of what you are trying to show is presented here.

• yup thanks, got it Jun 16, 2022 at 4:15

Usually we implement integration by parts as $$\int_{a}^{b} u dv = [uv]_{a}^{b} - \int_{a}^{b} v du$$ So he takes this whole part $$\frac{d}{dx} (\Psi^* \frac{d \Psi}{d x} - \frac{d \Psi^*}{d x} \Psi)dx$$ to be $$dv$$. Hence, $$\frac{dv}{dx} = \frac{d}{dx} (\Psi^* \frac{d \Psi}{d x} - \frac{d \Psi^*}{d x} \Psi ) \implies v = (\Psi^* \frac{d \Psi}{d x} - \frac{d \Psi^*}{d x} \Psi)$$ On the other hand, $$u = x$$ and hence $$du = dx$$.

Finally, since $$\Psi$$ (and hence $$\Psi^{*}$$) must go to zero at $$x = \pm \infty$$ for the wavefunction to be normalizable, evaluating the $$[uv]_{-\infty}^{\infty}$$ object shows it vanishes.

In QM then, quite often integration by parts simpifies to $$\int_{-\infty}^{\infty} u \frac{d}{dx} (v)dx = - \int_{\infty}^{\infty} v \frac{d}{dx} (u)dx$$ Swap which term the negative acts on for the cost of a negative sign!

• Thanks, I just didn't now what to do with the 'uv' term because one one hand you have the position operator, spatial derivative of the wave function and the wave function and the conjugate itself so multiplying all those together, I just couldn't figure it out. Thanks alot though! Jun 16, 2022 at 2:55