- #1

- 1

- 0

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}]

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Hany_Draidi
- Start date

- #1

- 1

- 0

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

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 14,998

- 1,574

You can't.

- #3

- 326

- 3

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

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 14,998

- 1,574

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

Science Advisor

- 4,802

- 1,746

Share: