# Integral of (psi_0 * d/dx psi_1)

• Aziza
In summary, a student needs help determining the validity of a mathematical statement involving stationary states of the SHO. They share their approach in which they cancel out dx's and change the limits of integration, but ask for assistance in finding the flaw as the result is not correct. The student also mentions that they were able to correctly solve a similar problem using a different stationary state. A discussion follows in which the student shares an equation and others provide examples and suggestions for further evaluation. The discussion ends with a clarification that the wavefunction does indeed vanish at infinity.
Aziza
I need help determining if the following statement is mathematically valid, where psi_0 and psi_1 are the first two stationary states of the SHO:

Basically what I did was cancel the dx's as if it was algebraic equation, and then since my integration became wrt psi_1, I changed my limits from 0->0, since at +/- infinity, psi_1 must be zero. And of course if your limits of integration are same, result must be zero. Specifically, I would like to know where exactly (at which of my above steps) is the flaw? (there must be a flaw since I do not get correct solution when I use this to calculate other stuff...this integral should in fact not be zero).

If I replace psi_1 by psi_0, for example, then I think this is correct, since I did it on my recent QM test and got full credit.

Any help with this is appreciated!

Any help is appreciated!

Hello,

$\psi_{1}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}2\xi e^{-\frac{\xi^{2}}{2}}$. And $\xi = \sqrt{\frac{m\omega}{\hbar}}x$

So it does not vanish at minus infinity.

PhysicsGente said:
Hello,

$\psi_{1}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}2\xi e^{-\frac{\xi^{2}}{2}}$. And $\xi = \sqrt{\frac{m\omega}{\hbar}}x$

So it does not vanish at minus infinity.

How does it not vanish at minus infinity? first, it is a wavefunction, so by the QM postulates it must vanish at +/- inifinity for it to be a physical solution.

second, just looking at the formula above you can see it does vanish, since at -∞ you basically have the quantity (-∞)*exp(-∞)...this is equal to zero as you can prove with l'hopital rule

Aziza, we can't see the image attached to an email hosted at mail.ru. Please post it in the thread.

sorry i did not realize that my picture could not be seen. Here is the equation:

∫ψ0$\frac{d}{dx}$ψ1dx = ∫ψ01 = ψ0ψ1|$^{0}_{0}$ = 0

Where the first integral runs from -inf to +inf, and the second from 0 to 0, since my integration is now wrt dψ1 and ψ1 is zero at +/- inf

Here's a similar specific example of what you did:
$$\int \cot\theta\,d\theta = \int \frac{1}{\sin\theta}\cos \theta\,d\theta = \int \frac{1}{\sin\theta}\,\frac{d}{d\theta}\sin \theta\,d\theta = \int \frac{1}{\sin\theta}\,d(\sin\theta).$$ So far so good, but from here, you claimed that
$$\int \frac{1}{\sin\theta}\,d(\sin\theta) = \frac{1}{\sin\theta}{\sin\theta}=1.$$

vela said:
Here's a similar specific example of what you did:
$$\int \cot\theta\,d\theta = \int \frac{1}{\sin\theta}\cos \theta\,d\theta = \int \frac{1}{\sin\theta}\,\frac{d}{d\theta}\sin \theta\,d\theta = \int \frac{1}{\sin\theta}\,d(\sin\theta).$$ So far so good, but from here, you claimed that
$$\int \frac{1}{\sin\theta}\,d(\sin\theta) = \frac{1}{\sin\theta}{\sin\theta}=1.$$

no...then using my method above i would say sinθ=u and so integrate wrt u...so the answer is ln(sinθ) by my method.

the reason i just pulled out the ψ0 was because the integration is wrt ψ1...so ψ0 is treated as constant...is there an error here?

I did exactly what you did where ##\psi_0 = \frac{1}{\sin \theta}## and ##\psi_1 = \sin\theta## and where ##\theta## plays the role of ##x##.

vela said:
I did exactly what you did where ##\psi_0 = \frac{1}{\sin \theta}## and ##\psi_1 = \sin\theta## and where ##\theta## plays the role of ##x##.

ohh I am sorry yes i see what you mean...so basically if i don't know (or actually just don't plug in, in this case) the actual psi's and their dependence on x, then there is no way to do this integral

The Hermite polynomials have certain properties that might be helpful to evaluate the integral. You could try looking into that.

Aziza said:
How does it not vanish at minus infinity? first, it is a wavefunction, so by the QM postulates it must vanish at +/- inifinity for it to be a physical solution.

second, just looking at the formula above you can see it does vanish, since at -∞ you basically have the quantity (-∞)*exp(-∞)...this is equal to zero as you can prove with l'hopital rule

Yes, I'm sorry I overlooked the $^2$. My bad.

## 1. What is the meaning of "Integral of (psi_0 * d/dx psi_1)"?

The integral of (psi_0 * d/dx psi_1) is the mathematical operation of finding the area under the curve formed by multiplying the functions psi_0 and d/dx psi_1. It is a commonly used concept in calculus and is denoted by the symbol ∫.

## 2. How is the integral of (psi_0 * d/dx psi_1) calculated?

The integral of (psi_0 * d/dx psi_1) is calculated using integration techniques such as substitution, integration by parts, and partial fractions. The fundamental theorem of calculus is also used to evaluate the integral.

## 3. What is the significance of calculating the integral of (psi_0 * d/dx psi_1)?

Calculating the integral of (psi_0 * d/dx psi_1) helps in determining the total change in the values of the two functions over a given interval. It is also used in various physics and engineering applications, such as calculating work and displacement.

## 4. What are the conditions for the integral of (psi_0 * d/dx psi_1) to exist?

The integral of (psi_0 * d/dx psi_1) exists if both the functions psi_0 and d/dx psi_1 are continuous over the given interval and d/dx psi_1 is differentiable. Additionally, the integral may exist if the functions have certain types of discontinuities or infinite limits at a finite number of points.

## 5. How does the order of integration affect the integral of (psi_0 * d/dx psi_1)?

The order of integration does not affect the value of the integral of (psi_0 * d/dx psi_1). However, it may affect the difficulty of evaluating the integral and the techniques required to do so.

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