An Easy QM normalisation question

[FaraDazed]

Homework Statement

Given a wave function psi, $\psi (x) = A \sqrt{|x|} e^{- \beta x^2}$ where $\beta$ is a constant (take the positive square root) . Normalise the wave function and hence find A.

Homework Equations

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The Attempt at a Solution

This is my first attempt at a problem like this, and I have missed around a 1/3 of the lectures on this this semester due to being ill.

Where it says that beta is a constant, I assumed that mean real, but even if its imaginary I still can't do it.

First if it is real then
$$\int_{-\infty}^{\infty} |\psi|^2 \> dx= 1 \\ \int_{-\infty}^{\infty} |A|^2 |x| e^{- 2 \beta x^2} \> dx = 1 \\ |A|^2 \int_{-\infty}^{\infty} |x| e^{- 2 \beta x^2} \> dx = 1$$
But in a formula book we were given it says that that integral equals 0 , so I tried if it was complex then
$$\int_{-\infty}^{\infty} |\psi|^2 \> dx= 1 \\ \int_{-\infty}^{\infty} |A|^2 |x| e^{- \beta x^2} e^{ \beta x^2} \> dx = 1 \\ |A|^2 \int_{-\infty}^{\infty} |x| \> dx = 1$$
Then how do I proceed from there, as surely $[\frac{1}{2}|x|^2]_{-\infty}^{\infty}$ is just infinite?

I don't even know if if my method for the two cases is even correct. Please any help/advice is much appreciated, I have been looking at this problem for the past 3 hours, looking at the lecture notes, and online and I am really lost!

 

[blue_leaf77]

Please check again the region within which the state $\psi (x) = A \sqrt{|x|} e^{- \beta x}$ is defined because this expression goes to infinity for $x$ approacing $-\infty$.

 

[FaraDazed]

Please check again the region within which the state $\psi (x) = A \sqrt{|x|} e^{- \beta x}$ is defined because this expression goes to infinity for $x$ approacing $-\infty$.

It does not say.

The question has four parts, A,B,C,D. The above is part A, and I have written it out exactly as it comes. There are no preceding questions.

Part B asks us to sketch the normlised wavefunction and the probability distribution
Part C asks us to indicate on our sketch where the particle is most likely to be found and to find the expectation value [itex]<x>/[itex] of position.
Part D asks us what the probability of finding the particle in the region 0<x<infinity is

 

[FaraDazed]

Please check again the region within which the state $\psi (x) = A \sqrt{|x|} e^{- \beta x}$ is defined because this expression goes to infinity for $x$ approacing $-\infty$.

Sorry there was a typo, it is $\psi (x) = A \sqrt{|x|} e^{- \beta x^2}$ with the x in the exponential squared. But there are still no regions given. Apart from that typo (that was carried through my latex) the question is as I have stated. I doubt that typo fixes the the issue you were concerned with though.

 

[Samy_A]

First if it is real then
$$\int_{-\infty}^{\infty} |\psi|^2 \> dx= 1 \\ \int_{-\infty}^{\infty} |A|^2 |x| e^{- 2 \beta x^2} \> dx = 1 \\ |A|^2 \int_{-\infty}^{\infty} |x| e^{- 2 \beta x^2} \> dx = 1$$
But in a formula book we were given it says that that integral equals 0

As the function is positive, the integral can't be 0. I think the formula in the book didn't have $|x|$ in it, but $x$.
You should be able to evaluate that integral, and see for which real $\beta$ you get a finite value.

 

[FaraDazed]

You should be able to evaluate that integral, and see for which real $\beta$ you get a finite value.

How do I do that? Do I have to use integration by parts, and then plug +/- infinity in as the limits?

 

[blue_leaf77]

I see, so you missed the square on the exponent. In that case, the wavefunction and its modulus square are even functions and you can evaluate the integral only in the half of the coordinate, e.g. from 0 to $\infty$ and multiply the result with 2 to get the actual value.

 

[vela]

Use the fact that the integrand is an even function to say
$$\int_{-\infty}^\infty \lvert x \rvert e^{-2 \beta x^2}\,dx = 2\int_0^\infty \lvert x \rvert e^{-2 \beta x^2}\,dx = 2\int_0^\infty x e^{-2 \beta x^2}\,dx.$$ You don't want to use integration by parts. You can try it, of course, but you'll quickly run into problems. This integral can be easily evaluated using a simple substitution.

 

[FaraDazed]

I see, so you missed the square on the exponent. In that case, the wavefunction and its modulus square are even functions and you can evaluate the integral only in the half of the coordinate, e.g. from 0 to $\infty$ and multiply the result to get the actual value.

Use the fact that the integrand is an even function to say
$$\int_{-\infty}^\infty \lvert x \rvert e^{-2 \beta x^2}\,dx = 2\int_0^\infty \lvert x \rvert e^{-2 \beta x^2}\,dx = 2\int_0^\infty x e^{-2 \beta x^2}\,dx.$$ You don't want to use integration by parts. You can try it, of course, but you'll quickly run into problems. This integral can be easily evaluated using a simple substitution.

Thanks for the help, so this is what I have done now...

$|A|^2 \int_{-\infty}^{\infty} |x| e^{- 2 \beta x^2} \> dx \\ 2|A|^2 \int_{0}^{\infty} |x| e^{- 2 \beta x^2} \> dx \\$

Then just looking at the integrand $\int_{0}^{\infty} |x| e^{- 2 \beta x^2} \> dx$
Let u = $-2 \beta x^2$ so that $dx = \frac{du}{-4 \beta x}$
$\int_{0}^{\infty} \frac{|x|}{e^{u}} \> \frac{du}{-4 \beta x} \\ -\frac{1}{4 \beta} \int_{0}^{\infty} e^{u} du = -\frac{1}{4 \beta} [e^{- 2 \beta x^2}]_0^{\infty} \\ = -\frac{1}{4 \beta} \cdot (-1) = \frac{1}{4 \beta}$

Because $[e^{- 2 \beta x^2}]_0^{\infty}$ surely equals (0)-(1) right ? If so, then...

$2|A|^2 \int_{0}^{\infty} |x| e^{- 2 \beta x^2} \> dx = 1 \\ \frac{2|A|^2}{4 \beta} = 1 \\ |A| = \sqrt{2 |\beta|}$

So therefore the normalised wavefunction is $\psi (x) =\sqrt{2 |\beta| |x|} e^{-2 \beta x^2}$

 

[blue_leaf77]

You are lucky to get the final answer correct even though you messed up when calculating

Let u = $-2 \beta x^2$ so that $dx = \frac{du}{-4 \beta x}$
$\int_{0}^{\infty} \frac{|x|}{e^{u}} \> \frac{du}{-4 \beta x} \\ -\frac{1}{4 \beta} \int_{0}^{\infty} e^{-u} du = -\frac{1}{4 \beta} [e^{- 2 \beta x^2}]_0^{\infty} \\ = -\frac{1}{4 \beta} \cdot (-1) = \frac{1}{4 \beta}$

Because you have defined that $u = -2 \beta x^2$ yet you still wrote $e^{-u}$ in the integrand.

 

[FaraDazed]

You are lucky to get the final answer correct even though you messed up when calculating

Because you have defined that $u = -2 \beta x^2$ yet you still wrote $e^{-u}$ in the integrand.

Oh yeah, that was just a typo, it was obviously meant to just be $\int_{0}^{\infty} e^u du$. So its not luck as on paper when working it out I did it with just the integral of e^u. But thanks for clarifying that it is correct, thanks!

 

[vela]

Oh yeah, that was just a typo, it was obviously meant to just be $\int_{0}^{\infty} e^u du$. So its not luck as on paper when working it out I did it with just the integral of e^u. But thanks for clarifying that it is correct, thanks!

That integral diverges. You need to get the signs right.

 

[FaraDazed]

That integral diverges. You need to get the signs right.

Im confused. Which bit has the sign error? Does $[e^{- 2 \beta x^2}]_0^{\infty}$ = -1? As $e^{-\infty}$ surely is 0? and e^0=1? Or is it somewhere else? This is one of (if not the) first problem I have had with integrals with infinity's as limits, so I really am not trying to be obtuse. I am just a bit clueless with this stuff at the mo after missing so many lectures.

 

[vela]

The integral from $u=0$ to $u=\infty$ of $e^u$ diverges. The exponential grows without bound. Either fix the limits or fix the integrand.

 

[FaraDazed]

The integral from $u=0$ to $u=\infty$ of $e^u$ diverges. The exponential grows without bound. Either fix the limits or fix the integrand.

Ah, I see (I think)

So should I have set $u=2 \beta x^2$ to get $\int_{0}^{\infty} e^{-u} du$ ? But surely then it would be $-\frac{1}{4 \beta}$ making $|A|$ complex? or is that correct? Or is that what the question meant by taking the positive square root?