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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.

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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