Modern Physics: Chapter 8 – Normalizing a Wavefunction

I discussed in the previous chapter you the probability of a particle being detected in the continous range (x_{0}, x_{1}) can be calculated from its wavefunction using,

  • \int_{x_{0}}^{x_{1}} \psi^{*} \psi dx

Now, assuming the particle actually exists, if we essentially look for it at every point in the entire universe we’re bound the detect it somewhere right? Hence the probability of finding the particle in the range (-\infty, \infty) should be 1. In other words,

  • \int_{-\infty}^{\infty} \psi^{*} \psi dx must equal 1.

So, what if we get a wavefunction, say. g(x) where \int_{-\infty}^{\infty} g^{*} g dx doesn’t equal 1 and equals some other number k instead? Then we can make the function’s probability equal 1 bu multiplying it with a constant A, such that the corrected wavefunction is:

  • \psi(x) = A \times g(x)

This process is called normalizing the wavefunction. But how do we figure out what the value of A is? Notice that now that \psi(x) = A \times g(x),

  • \int_{-\infty}^{\infty} \psi^{*} \psi dx
  • \Rightarrow \int_{-\infty}^{\infty} |A|^2 g^{*}(x) g(x) dx
  • \Rightarrow |A|^2 \int_{-\infty}^{\infty} g^{*}(x) g(x) dx

Now we already know that \int_{-\infty}^{\infty} g^{*} g dx = k, so the above expression simplifies to:

  • \Rightarrow |A|^2 \int_{-\infty}^{\infty} g^{*}(x) g(x) dx \Rightarrow |A|^2 \times k

Now to ensure \int_{-\infty}^{\infty} \psi^{*} \psi dx = 1, we need a value for A such that

  • |A|^2 \times k = 1  so
  • |A|^2 = \frac{1}{k}
  • \Rightarrow |A| = \frac{1}{\sqrt{k}}

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s