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For a homework problem we are given the wave function

$$ \Psi(x) = \frac{N}{x^2 + a^2},\ a > 0 $$

and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function

$$ \Psi(x) = \frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)} $$

Then, following my notes computed

$$ \int_{-\infty}^\infty \Psi(x)^* \langle x \rangle \Psi(x)\,\mathrm dx $$

Given that the wave equation is currently in the $x$ basis I then computed

$$ \int_{-\infty}^\infty \left(\frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)}\right)^* x \left(\frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)}\right)\,\mathrm dx $$

by plugging it into Wolfram Alpha. Though this results in a solution of $0$.

This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$?

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    $\begingroup$ Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.) $\endgroup$
    – David Z
    Oct 14, 2018 at 2:33
  • $\begingroup$ @DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future. $\endgroup$
    – KDecker
    Oct 14, 2018 at 3:28
  • $\begingroup$ Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough. $\endgroup$
    – David Z
    Oct 14, 2018 at 3:40
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    $\begingroup$ Your gut instinct is half-right; dimensional analysis implies the answer should be $ka$ for some dimensionless $k$. It's just that it turns out $k=0$, so the "dependence" on $a$ is removed. $\endgroup$
    – J.G.
    Oct 14, 2018 at 9:12
  • $\begingroup$ This is exactly the reason why relying exclusively on calculators makes you worse at physics. At any of the four steps above you could've immediately seen the answer was $0$ if you used some intuition! $\endgroup$
    – knzhou
    Oct 14, 2018 at 17:11

2 Answers 2

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You've got the right answer! The wave-function is symmetric about $0$, while the operator $\langle x \rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$! Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.

Edit: I meant to write $\hat{x} = x$, not $\langle x \rangle = x$. The correct expression for the expectation value is $\langle x \rangle = \int\psi^*\hat{x}\psi dx$.

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  • $\begingroup$ Ahh, this makes sense. Working through the rest of the problem does confirm this too. Thanks! $\endgroup$
    – KDecker
    Oct 14, 2018 at 15:51
  • $\begingroup$ Actually, this might not be right. If the area below zero of xf(x) is infinite and the area above zero is infinite, even though it 'should' be zero, it is really infinity minus infinity which is undetermined (it's a divergent interval). There's a discussion of these here. $\endgroup$
    – user142547
    Oct 14, 2018 at 22:00
  • $\begingroup$ @user11599 Ah sure maybe I could have been more clear that I'm assuming the function is well behaved! Thanks for pointing that out though $\endgroup$ Oct 14, 2018 at 22:56
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Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|\psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.


I also wanted to address some of your notation and understanding of it.

It seems like you are familiar with bases. The expectation value of the position operator can be written without specifying any basis: $$\langle X\rangle=\langle\psi|X|\psi\rangle$$

Since we are given the wavefunction in the position basis $\psi(x)=\langle x|\psi\rangle$, it makes sense to work in the position basis.

Therefore, we can use the identity $\int|x\rangle\langle x|dx=1$ to write our expectation value as $$\langle X\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\langle\psi|x\rangle\langle x|X|x'\rangle\langle x'|\psi\rangle dxdx'$$

Then, knowing that $X$ in its own eigenbasis is $\langle x|X|x'\rangle=x'\delta(x'-x)$, the integral becomes $$\langle X\rangle=\int_{-\infty}^{\infty}\psi^*(x)x\psi(x)dx$$

This is the integral you want. You don't want your expectation value inside the integral (and you also need to specify which variable you are integrating over). Otherwise you have an integral that is not very useful:

$$\langle X\rangle=\int_{-\infty}^{\infty}\psi^*(x)\langle X\rangle\psi(x)dx=\langle X\rangle\int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx=\langle X\rangle$$

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    $\begingroup$ That last equation is really an equality. It may not be the correct way to find the expectation value, but it's a valid equality because it simply becomes $\langle X\rangle = \langle X\rangle\int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx$ as $\langle X\rangle$ has no dependency on $x$ and is not an operator that can affect $\psi$. $\endgroup$ Oct 14, 2018 at 6:08
  • $\begingroup$ @StephenG You are exactly right. Will edit accordingly. $\endgroup$ Oct 14, 2018 at 11:41

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