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I know that

$\begin{gather*} \langle x\rangle = \int_{-\infty}^{\infty} P(x)x\,\mathrm{d}x = \int_{-\infty}^{\infty}\psi^*(x)\psi(x)x\,\mathrm{d}x \end{gather*}$,

however

$\begin{gather*} \int_{-\infty}^{\infty}P(x)\mathrm{d}x = 1 \end{gather*}$

by definition. Working out the integral for $\langle x\rangle$ I get

$\begin{gather*} \langle x\rangle = x|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \mathrm{d}x = x|_{-\infty}^{\infty} - x|_{-\infty}^{\infty} = 0. \end{gather*}$

I assume that $\langle x\rangle = 0$ is not true of all quantum states, but I don't see what I'm doing wrong. It seems like it should be pretty simple, but I'm lost. Can anyone show me my mistake?

EDIT:

Integration by parts:

$\begin{gather*} \int u\,\mathrm{d}v = uv-\int v\,\mathrm{d}u. \end{gather*}$

Let

$\begin{gather*} u=x, \mathrm{d}u=\mathrm{d}x, \end{gather*}$

$\begin{gather*} \mathrm{d}v=\psi^*(x)\psi(x)\mathrm{d}x = P(x)\mathrm{d}x, v=1, \end{gather*}$

then

$\begin{gather*} \langle x\rangle = \int_{-\infty}^{\infty}\psi^*(x)\psi(x)x\,\mathrm{d}x = \int_{-\infty}^{\infty}P(x)\mathrm{d}x = (x-\int \mathrm{d}x)|_{-\infty}^{\infty} = (x-x)|_{-\infty}^{\infty} = 0. \end{gather*}$

Also, upon searching the internet I find plenty of derivations of my initial expression for $\langle x\rangle$, but I don't find any examples of the integral being evaluated. Why was this such a bad question to ask?

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closed as off-topic by Chris, Cosmas Zachos, ZeroTheHero, Kyle Kanos, Jon Custer May 27 '18 at 0:12

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  • $\begingroup$ Please explain step by step why each equal sign in your third equation is true, and then you will see the mistake. $\endgroup$ – user110373 May 19 '18 at 0:01
  • $\begingroup$ @user110373 I edited the post to clarify my thought process using integration by parts. I still don't see my mistake. $\endgroup$ – The Ledge May 19 '18 at 0:32
  • $\begingroup$ Are you assuming the probability density is uniform? If so, this needs to be stated explicitly. $\endgroup$ – Robert L. May 19 '18 at 1:02
  • $\begingroup$ @TheLedge $\mathrm{d}v=\psi^*(x)\psi(x)x\mathrm{d}x$ and $v=1$ are contradictory unless $\psi=0$. $\endgroup$ – user110373 May 19 '18 at 1:08
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    $\begingroup$ Your integration by parts is certainly not correct. Presumably you mean $dv=\psi^*\psi dx$, not with the extra $x$ factor. To proceed you need to find the anti derivative of $\psi^*\psi$, and it’s certainly not $1$ since the derivative of $1$ is not $\psi^*\psi$. Note that $\psi^*\psi \ne 1$ at the boundaries either. $\endgroup$ – ZeroTheHero May 19 '18 at 21:43
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I figured it out. Thanks to those who commented for attempting to guide me. I appreciate your effort.

$P(x)$ is a probability function, so $\begin{gather*} \int_{-\infty}^{\infty}P(x)\mathrm{d}x = 1 \end{gather*}$ is true by definition. This is a definite integral. It has the value $1$ only for the bounds $-\infty$ to $\infty$. In general, $\begin{gather*} \int_{a}^{b}P(x)\mathrm{d}x \leq 1 \end{gather*}$.

In integration by parts,

$\begin{gather*} \frac{\mathrm{d}}{\mathrm{d}x}uv = u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x} \implies \int_{a}^{b}u\,\mathrm{d}v = (uv-\int v\,\mathrm{d}u)|_{a}^{b} \end{gather*}$.

The integral $\begin{gather*} \int v\,\mathrm{d}u \end{gather*}$ is indefinite. Applied to my problem, $\begin{gather*} u=x,\, \mathrm{d}u=\mathrm{d}x,\,\mathrm{d}v=P(x)\,\mathrm{d}x \end{gather*}$, and $\begin{gather*} v=\int P(x)\,\mathrm{d}x \end{gather*}$, also an indefinite integral.

Then

$\begin{gather*} \langle x\rangle = \int_{-\infty}^{\infty}P(x)x\,\mathrm{d}x = (x\int P(x)\,\mathrm{d}x-\int\int P(x)\,\mathrm{d}x\,\mathrm{d}x)|_{-\infty}^{\infty} \end{gather*}$.

In each part of this final difference the integral $\begin{gather*} \int P(x)\,\mathrm{d}x \end{gather*}$ is evaluated and applied to another function ($x$ and $\begin{gather*} \int\mathrm{d}x \end{gather*}$ respectively) before the bounds are applied to the final expression. Because this integral is defined as $1$ only for the case where the bounds $-\infty$ to $\infty$ are applied during integration, the expression cannot be evaluated until $\psi(x)$ is known and $P(x)$ is replaced by $\psi^*(x)\psi(x)$.

Herein lies my error. I attempted to apply the bounds to the integral $\begin{gather*} \int P(x)\,\mathrm{d}x \end{gather*}$ before I multiplied it to the other functions.

If anyone sees any further errors in my work, I'd appreciate any additional corrections.

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