Is the wave function normalized?

  • Thread starter jg370
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  • #1
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Homework Statement



The ground state wave function for the electron in a hydrogen atom is:

[tex]\psi(r) = \frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o}[/tex]

where r is the radial coordinate of the electron and a_o is the Bohr radius.

Show that the wave function as given is normalized.


Homework Equations



Any wave function satisfying the following equation is said to be normalized:

[tex]\int_{-\infty}^{+\infty} |\psi|^2\dx = 1[/tex]


The Attempt at a Solution



Because the sum of all probabilities over all values of r must be 1,

[tex]\int_{-\infty}^{+\infty} (\frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o})^2 dr = \frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = 1[/tex]

Since the integral can be expressed as the sum of two integrals, we have,

[tex]\frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = \frac{2}{\pi a_o^3} \int_0^{+\infty} e^\frac{-2r}{a_o^2} dr = 1 [/tex]

After integrating, I obtain,

[tex]\frac{1}{\pi a_o^3} = 1[/tex]

which is definitely incorrect. However, I do not see any other way to proceed. could someone give some assitance.

Thanks for your kind assistance

jg370
 

Answers and Replies

  • #2
1,444
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you integrated wrong -
[itex]\int_0^{\infty} e^{-\frac{2r}{a_0^2}} dr = [-\frac{a_0^2}{2} e^{-\frac{2r}{a_0^2}]_0^{\infty}=\frac{a_0^2}{2} [/itex]

which gives [itex]\frac{1}{2 \pi a_0}=1[/itex]
 
  • #3
441
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the a_0 in your exponent shouldn't be squared, also the normalization condition says you should integrate over the bounds of your function. if r is the radial coordinate then how can it have a value of negative infinity? similarly your no longer normalizing in 1-d but 3D I recommend using spherical coordinates in which case your integral inherits a r^2 sin theta
 
  • #4
18
0
Tks. I wil try again using spherical coordinates
 

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