# Some weird integration by parts to derive momentum operator

1. Mar 20, 2013

### 71GA

In a Griffith's book (page 15-16) an author derives a momentum operator. In the derivation he states that he used a integration by parts two times. He starts with this equation which i do understand how to get to.
$$\begin{split} \frac{d \langle x \rangle}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^{\infty} x \cdot \frac{d}{dx} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx \end{split}$$

After 1st integration by parts he gets:
$$\begin{split} \frac{d \langle x \rangle}{dt} &= -\frac{i\hbar}{2m} \int\limits_{-\infty}^{\infty} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx \end{split}$$

After 2nd integration by parts he gets:
$$\begin{split} \frac{d \langle x \rangle}{dt} &= -\frac{i\hbar}{m} \int\limits_{-\infty}^{\infty} \frac{d \Psi}{dx}\Psi^* \, dx \end{split}$$

I tried to repeat the procedure but I get lost as soon as i try to do first intergation by parts. Only thing i managed to do is the procedure below which i have no clue what it means nor do i know how to continue. Could anyone shom e mow to do it?

$$\frac{d \langle x \rangle}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^{\infty} x \cdot \frac{d}{dx} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx = x \cdot x \cdot \frac{d}{dx} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) - \int\limits_{-\infty}^{\infty} x \,d \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right)$$

2. Mar 20, 2013

### DimReg

For expressions like this, you can do integration by parts in your head as follows: integration by parts moves a derivative from one factor to the other, picking up a minus sign, and adds a "surface" term. In the first integration by parts, the d/dx moves to the x, and just becomes 1. The "surface" term disappears by the behavior of the wave functions as they approach infinity. For most physics applications the surface term vanishes, but it never hurts to be careful.

To work it out explicitly, try this. Integration by parts, in the form you are most likely familiar with from calculus, is as follows:

$\int u dv = uv - \int v du$

And your expression is of the form:

$\int x \frac{d}{dx} f(x) dx$

So if u = x, and dv = df/dx dx, then du = dx and v = f. So:

$\int x \frac{d}{dx} f(x) dx = xf(x) - \int f(x) dx$

So xf(x) is the "surface" term.

Last edited: Mar 20, 2013
3. Mar 20, 2013

### andrien

In first one,you will have to do by part by taking x as differentiating function and utilize the fact that ψ vanishes at infinity.In the second one,also utilize the fact that ψ*ψ vanishes at infinity.

4. Mar 20, 2013

### 71GA

I ll still need some guidance but here we go! If i obey standard calculus formula $\int u dv = uv - \int v du$ i can mark what is $u$ and what is $dv$ in my equation like this:
$$\frac{d \langle x \rangle}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^{\infty} \underbrace{x}_{=u} \cdot \underbrace{\frac{d}{dx} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx}_{=dv}$$

If i understood right now i have to calculate $du$ and $dv$. Here i go:
$$x = u \longrightarrow \frac{dx}{dx} = \frac{du}{dx} \longrightarrow 1 = \frac{du}{dx} \longrightarrow \boxed{du = dx}$$
$$dv = \frac{d}{dx} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx \longrightarrow \int dv = \int \frac{d}{dx} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx \longrightarrow \boxed{v = \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right)}$$

By using standard calculus formula $\int u dv = uv - \int v du$ i can now write that:
$$\frac{d \langle x \rangle}{dt} = -\frac{i\hbar}{2m} \left[ \underbrace{x \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right)}_{\text{surf. pt. - how can this be 0?}} - \underbrace{\int\limits_{-\infty}^{\infty} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx}_{\text{int. by party 1 more time ???}} \right]$$

Please confirm that i am doing OK until now. What i don't understand at the moment is how can surface part equal 0? I can't just put $x$ inside a bracket to get $\frac{dx}{dx}=1$. I don't understand this. Do i have to integrate right part 1 more time?

5. Mar 20, 2013

### DimReg

Ok good, you're doing it right now. Remember though, for a definite integral you have to evaluate the surface term at the bounds of integration.

Yes, you will need to integrate by parts one more time, but you already knew that because that's what Griffiths does. What you did is the first integration by parts.

For the surface term, you have to use some knowledge of the wavefunction to find out if it vanishes. You know that psi is square integrable, which implies that it must fall at least at a certain rate (I think it's 1/sqrt(x), but I don't remember exactly). Whatever the exact details are (they are almost certainly explained in Griffiths), the point is that the surface term goes to zero at +/- infinity, so all that remains in the second term.

6. Mar 20, 2013

### DimReg

Oh, and when you integrate by parts the second time, you will do it to only one of the two terms under the integral. Again you'll find that the surface term goes to zero at infinity.

7. Mar 20, 2013

### 71GA

I added an eval symbol to the surface term but i still don't know how to evaluate it in order for surface term to equal 0:
$$\begin{split} \frac{d \langle x \rangle}{dt} &= -\frac{i\hbar}{2m} \left[ x \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right)\Bigg|_{-\infty}^{\infty} - \int\limits_{-\infty}^{\infty} \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \, dx\right]\\ \end{split}$$

I will do this now and leave evaluation of surface part for later if you could explain a bit more why it is 0:

$$\begin{split} \frac{d \langle x \rangle}{dt} &= -\frac{i\hbar}{2m} \left[ x \left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right)\Bigg|_{-\infty}^{\infty} - \int\limits_{-\infty}^{\infty} \underbrace{\left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right)}_{=u} \, \underbrace{dx}_{=dv}\right]\\\\ u&=\left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \longrightarrow \frac{du}{dx} = \frac{d}{dx}\left( \frac{d \Psi^*}{dx} \Psi - \frac{d \Psi}{dx}\Psi^* \right) \longrightarrow \frac{du}{dx} = \underbrace{\frac{d}{dx}\left( \frac{d \Psi^*}{dx} \Psi\right)}_{\text{product rule}} - \underbrace{\frac{d}{dx}\left(\frac{d \Psi}{dx}\Psi^* \right)}_{\text{product rule}} = \frac{d^2 \Psi^*}{dx^2} \Psi + \frac{d\Psi^*}{dx}\frac{d \Psi}{dx} - \frac{d^2 \Psi}{dx^2} \Psi^* - \frac{d\Psi}{dx}\frac{d \Psi^*}{dx} = \frac{d^2 \Psi^*}{dx^2} \Psi - \frac{d^2 \Psi}{dx^2} \Psi^* \longrightarrow \underbrace{\boxed{\frac{du}{dx} = \frac{d^2 \Psi^*}{dx^2} \Psi - \frac{d^2 \Psi}{dx^2} \Psi^*}}_{\text{i got lost here... how can i get du???}} \\\\ dx &= dv \longrightarrow \int dv = \int dx \longrightarrow \boxed{v = x} \end{split}$$

I got lost when deriving $du$ so i can't finish this. Could you point me a bit please? I hope my calculation for $v$ is ok. When i ll get these two i ll be able to finish 2nd integration by parts.

8. Mar 20, 2013

### Staff: Mentor

In order for $\Psi$ to be normalizable, it has to approach 0 as $x \rightarrow \pm \infty$. Its derivative will also approach 0.

9. Mar 20, 2013

### dextercioby

This is wrong. There are functions in the L^2 which do not go to 0 when x goes to infinity. But the authors tacitly assume they are working in S(R) for it's an invariant domain of all x and p operators.

10. Mar 20, 2013

### 71GA

I still don't understand 2 things:
- Why surface part equals 0? Is there any mathematical rule u can use to justify this? I know the normalisation rule $\int\limits_{-\infty}^{\infty} |\Psi|^2 dx = 1$...
- How can i finish second integration by parts.

Arguing about some spaces won't help me as i am a beginner. So please stay on the beginners level.

Last edited: Mar 20, 2013
11. Mar 21, 2013

### DimReg

For the surface term, are you sure you've looked at the book thoroughly? I used that book as an undergrad and I seem to recall reading why that term has to be zero.

Throwing mathematical rigor away, the basic argument is that as it approaches infinity, the wavefunction needs to go to zero at a sufficiently fast rate in order to be normalisable (and it has to be zero at infinity, because otherwise it could be "at infinity" which doesn't make sense physically). This essentially implies that $|\Psi|^2$ will be zero at +/- infinity. Intuitively, you can think of the derivative as lowering the power of the function (e.g. x^n goes to x^n-1), so $x \frac{d}{dx}\Psi$ would go at the same rate as $|\Psi|^2$, so the surface term is 0 at +/- infinity. I wouldn't ask for more help on this part on the forum, it's not the easiest thing to explain, but it's explained in a lot of other places (I'm pretty sure there are other posts on this forum explaining about that part. Look for them, also look in your book).

For the integration by parts, you once again made a mistake deciding what is u and dv. You should probably brush up on your integration by parts, since you'll use it a LOT in QM. But a good tip is that if it doesn't work at first, try assigning different things to u and dv, see if you can get a better result. A good tip is that whatever you set as dv, make sure it can be integrated. Also, dv will "go up in power", while u will "go down". So if you have an x^n term, you probably want that to be u, so that it becomes x^(n-1), and as you keep repeating the process eventually x^0 = 1, which will hopefully be a simpler integral.

As for the actual integral, you have:

$\int_{-\infty}^\infty (\psi \frac{d\psi^*}{dx} - \psi^* \frac{d\psi}{dx} )dx = \int_{-\infty}^\infty \psi \frac{d\psi^*}{dx} dx - \int_{-\infty}^\infty \psi^* \frac{d\psi}{dx} dx$

The trick here is to pick only one of the integrals to do integration by parts on:

$\int_{-\infty}^\infty \psi^* \frac{d\psi}{dx} dx$

Choose u = ψ*, dv = dψ/dx dx. The integral should be simple after that, with a surface term that is zero (trust me on that until you understand why, it sounds like you need more integration practice to understand it anyways). After the integration by parts, you should have two identical integrals, so you can add them together to get what Griffiths gets.