# Unwanted $i$ floating around when trying to calculate $\langle p\rangle$

$\def\sp#1{\left\langle#1\right\rangle}$I am given $$\Psi(x,0)=A_0 \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \cdot \exp\left(\frac{i}{\hbar}p_0x\right)\tag1$$

where $A_0=(\pi \sigma_0^2)^{-\frac{1}{4}}$

I am trying to find $\sp p$, which is $$\int\limits_{-\infty}^{+\infty}\Psi^*(x,0)\left(-i\hbar \frac{\partial \Psi(x,0)}{\partial x}\right)\,dx\tag2$$

I have $$\frac{\partial \Psi(x,0)}{\partial x}= A_0 \left[ \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x\right) \frac{i}{\hbar}p_0+ \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x]\right) \frac{2x}{2\sigma_0^2}2 \right]\tag3$$

Collecting terms:$$\frac{\partial \Psi(x,0)}{\partial x}=A_0\cdot \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x\right)\left( \frac{i}{\hbar}p_0+\frac{2x}{\sigma_0^2}\right)\tag4$$

Multiplying by $-i\hbar$: $$-i\hbar \frac{\partial \Psi(x,0)}{\partial x}=A_0\cdot \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x\right)\left(p_0-\frac{2i \hbar x}{\sigma_0^2}\right)\tag5$$

Premultiplying by $\Psi^*(x,0)$ and integrating:$$\int\limits_{-\infty}^{+\infty} \Psi^*(x,0) \left(-i\hbar \frac{\partial \Psi(x,0)}{\partial x}\right)\,dx= \int\limits_{-\infty}^{+\infty} A_0 \exp\left( -\frac{x^2}{2\sigma_0^2}\right)\left(p_0-\frac{2i \hbar x}{\sigma_0^2}\right)\,dx\tag6$$

Now, it looks like I have an $i$ hanging around when I don't want one. $\sp p$ should be a real value, and I don't want any imaginary component. I'd appreciate it if someone could show me what I'm doing wrong. Thanks in advance, and apologies for the inconsistent formatting.

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The integral of the term with the $i$ is zero. –  Raskolnikov Nov 21 '12 at 10:55

You are probably doing everything right. Look at the imaginary part of the integral that you have in the end: $$\int_{-\infty}^{+\infty} A_0\exp\left(-\frac{x^2}{2\sigma_0^2}\right)\left(-\frac{2\hbar x}{\sigma_0^2}\right)dx.$$ The function under the integral is odd and quickly decreases when $x \to \infty$. Therefore this integral converges, and converges to $0$, because values on the positive half-axis "cancel out" with values on the negative half-axis. So your integral is real, you can just drop the imaginary part.