- #1

Hany_Draidi

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ψ*ψ= A^2 (x^2)exp(-2a(mx^2/h)) ; where m,a,A,and h are constants and I want the graph in terms of them.

I tried: Plot[A^2 *(x^2)*exp(-2a(mx^2/h)),{x,-1000,1000}]

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- Thread starter Hany_Draidi
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- #1

Hany_Draidi

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ψ*ψ= A^2 (x^2)exp(-2a(mx^2/h)) ; where m,a,A,and h are constants and I want the graph in terms of them.

I tried: Plot[A^2 *(x^2)*exp(-2a(mx^2/h)),{x,-1000,1000}]

- #2

vela

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You can't.

- #3

NasuSama

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ψ*ψ= A^2 (x^2)exp(-2a(mx^2/h)) ; where m,a,A,and h are constants and I want the graph in terms of them.

I tried: Plot[A^2 *(x^2)*exp(-2a(mx^2/h)),{x,-1000,1000}]

I would say you can't do that since Wolfram Alpha assumes that these terms are variables. Hence, there is no approach for this. You could just substitute the constants with any values and then, check the graph. You just have a graph; that doesn't guaranteed the exact graph you should expect. You will need to substitute random values!

I may be wrong. I am just assuming that is true.

- #4

vela

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What you want to do is come up with some combination of the constants that is a characteristic length in the problem. In this case, your wave function is

$$\psi(x) = A x e^{-\frac{ma}{\hbar}x^2}.$$ The argument of the exponential has to be unitless, and if you check the units, you'll indeed find that the combination ##ma/\hbar## has units of 1/length

- #5

phyzguy

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