Expectation value of momentum

Ψ(x)=(2/a)^(1/2) [csin(nπx/a)]

The Expectation value of momentum <P>=∫Ψ＊(x)[-ih d/dx ] Ψ(x) dx = 0
the average momentum is zero.It means the particle is moving equally in the +x and -x.

And
if Ψ(x)=(2/a)^(1/2) {csin(πx/a)+dsin(2πx/a)}

I calculate the average momentum is also zero,too,
Is this means the same?
My teacher didn't tell me the answer,and he say it is not zero "intuitively".
But i think it is zero,isn't?

Because Ψ(x)=(2/a)^(1/2) [csin(πx/a)]+ (2/a)^(1/2) [dsin(2πx/a)]
according to the Ψ(x)=(2/a)^(1/2) [csin(nπx/a)] ---> <p>=0
so <p>=0+0=0

thanks.

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It means the particle is moving equally in the +x and -x.

That's not quite what it means. It means that upon measurement, the particle has an equal likelihood of being found moving in either the +x or -x direction (I assume we're talking about a particle in a box).

For any* wavefunction at all (even something weird like $\psi(x) = 3x^x$), the expected momentum will be 0 so long as $\psi(0) = \psi(a) = 0$.

Here's why:

$$\langle p_x \rangle = \int_{0}^{a} \psi^{*}(x) \left[ -i \hbar \frac{d\psi}{dx} \right]\, dx$$

$$\langle p_x \rangle = -i \hbar \int_{0}^{a} \psi(x) \frac{d\psi}{dx}\, dx$$

$$\langle p_x \rangle = \frac{-i \hbar}{2}\left[ \psi(a)^2 - \psi(0)^2 \right]$$

$$\langle p_x \rangle = 0$$

*The above math does not verify if this is true for complex wavefunctions, but my guess is "probably". If someone wants to prove/disprove that, feel free.

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so if the Ψ(x)=(2/a)^(1/2) {csin(πx/a)+dsin(2πx/a)}
do the <P>=0 ??

so if the Ψ(x)=(2/a)^(1/2) {csin(πx/a)+dsin(2πx/a)}
do the <P>=0 ??

Yep -- for a particle in a box. (Although I am assuming you meant to put an "n" in the argument of each sine function).

thanks ^^