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posion117
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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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