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Given the normalised ground-state wave-function: $$\Psi(x, t)=\begin{cases} \sqrt\frac{2}{d}\cos(\frac{\pi x}{d})e^\frac{-i\hbar\pi^2t}{2md^2} & \ \lvert x\rvert<\frac{d}{2}, \\ 0 & \text{otherwise.} \end{cases}$$ for a particle of mass $m$ confined to potential well of form: $$V(x)=\begin{cases} 0 & \ \lvert x\rvert<\frac{d}{2}, \\ \infty & \text{otherwise.} \end{cases}$$ show that $\langle x\rangle =0$

I'm told that the correct answer is $$\langle x\rangle=\int_{-\frac{d}{2}}^\frac{d}{2} x\frac{2}{d}\cos^2\biggl(\frac{\pi x}{d}\biggr)\mathrm{d}x=0$$ as the integrand is an odd function. But I don't understand why the integrand takes this form. Could some please explain it to me? I would like to know where the $x$ comes from? Thanks.

Expectation of $x$ is the position of the particle and by my logic this is $\langle x\rangle = \int\psi^*(x)\psi(x)dx=\int_{-\frac{d}{2}}^\frac{d}{2} \frac{2}{d}\cos^2\biggl(\frac{\pi x}{d}\biggr)\mathrm{d}x$ basically without the factor $x$ as shown above which was my reason for asking about it.

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    $\begingroup$ Do you know what the definition of $\langle x \rangle$ is? $\endgroup$ – FenderLesPaul Jul 19 '15 at 8:55
  • $\begingroup$ @FenderLesPaul Thanks for your reply. Yes expectation of $x$ is the position of the particle and by my logic this is $\langle x\rangle = \int\psi^*(x)\psi(x)dx=\int_{-\frac{d}{2}}^\frac{d}{2} \frac{2}{d}\cos^2\biggl(\frac{\pi x}{d}\biggr)\mathrm{d}x$ basically without the factor $x$ as shown above which was my reason for asking about it. $\endgroup$ – BLAZE Jul 19 '15 at 9:45
  • $\begingroup$ Methinks you need a review of expectation values. $\endgroup$ – Kyle Kanos Jul 19 '15 at 14:41
  • $\begingroup$ I'm curious as to why people are still voting to close this. It looks like a perfectly fine homework question to me. $\endgroup$ – David Z Jul 19 '15 at 15:57
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The definition of the expectation value of an operator A is \begin{equation} \langle A\rangle=\int{\psi^* (x) A(x) \psi (x) dx} \end{equation} (because it represents "the value of the variable" $A(x)$ times "the probability of being in that configuration" $P(x)=\psi^* (x) \psi (x)$) and for the particular case of the expectation value of the position operator \begin{equation} \langle x\rangle=\int{x \psi^* (x) \psi (x) dx} \end{equation} that is the origin of your extra $x$.

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$$\int_{-\frac{d}{2}}^{\frac{d}{2}}\psi^*(x)\psi(x)dx$$ is the probability of finding particle between $-\frac{d}{2}$ and $\frac{d}{2}$.

Expectation value is : $$\langle x\rangle=\langle \psi|\hat{x}|\psi\rangle$$

$|\psi\rangle$ is the summation of probability amplitude times given basis kets

$$|\psi\rangle = \sum_i c_i |{e_i}\rangle$$ $$c_i=\langle{e_i}|\psi\rangle$$ probability is ${c_i}^2=({c_i}^*)c_i=(\langle{e_i}|\psi\rangle^*)\langle{e_i}|\psi\rangle=\langle\psi|{e_i}\rangle\langle{e_i}|\psi\rangle$.

if you want expectation value, you need "value times probability". An example is casting a dice, expectation value is 1*1/6+2*1/6+3*1/6+4*1/6+5*1/6+6*1/6.

back to the particle in a box, you need $x_i \times \langle \psi|x_i\rangle\langle x_i|\psi\rangle$, $|x_i\rangle$ is the state of the particle located at $x_i$. you need to add up all "x * prob of the particle located at x" from -d/2 to d/2. since position is continuous, you need to do integration.

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