An example in Griffith's Intro. to Quantum Mechanics is:

Suppose we have two non-interacting particles both of mass $m$ in a infinite square well. The one particle states are $$\phi_n (x) = \sqrt{\frac{2}{a}} \sin \left( \frac{n\pi}{a} x \right), \hspace{3mm} E_n = n^2 K$$

He states that if the particles are distinguishable, the ground state wave function is $$\phi_{1,1} = \frac{2}{a} \sin(\pi x_1/a) \sin(\pi x_2/a)$$

and this makes sense, because $\phi_{n_1, n_2}$ is just the product of respective wavefucntions for distinguishable particles.

However, he then claims that for bosons, the ground state is the same. I am not sure I can verify this.

I know from earlier in the chapter that for a boson, $$\phi_{a, b}(x_1, x_2) = A [\phi_a(x_1) \phi_b(x_2) + \phi_b(x_1) \phi_a(x_2)]$$

Here I don't quite know what $A$ is, but inside the brackets I get: $$\phi_{1, 1} = \left[\sqrt{\frac{2}{a}} \sin \left( \frac{\pi x_1}{a}\right) \sqrt{\frac{2}{a}} \sin \left( \frac{\pi x_2}{a}\right) + \sqrt{\frac{2}{a}} \sin \left( \frac{\pi x_1}{a}\right) \sqrt{\frac{2}{a}} \sin \left( \frac{\pi x_2}{a}\right) \right]$$ $$= \frac{4}{a} \sin \left(\frac{\pi x_1}{a}\right) \sin \left(\frac{\pi x_2}{a}\right)$$

But now I have an extra factor of 2 in the ground state wave function. Is this taken care of by the $A$ or am I doing something wrong?

[Edit: I've also added a picture to the actual exercise in case I'm misinterpreting anything!][enter image description here]1

  • 2
    $\begingroup$ You just have to renormalize. I am sure Griffiths is saying this from the point of view that a constant multiple for the wavefunction doesn't modify the actual state. In general when you do math like this with normalized states you end up with non-normalized states. You either have to renormalize or modify how you determine the probabilities/normalizations (which is just a difference of constant factors out front of course) $\endgroup$ – Aaron Stevens Apr 15 at 22:25

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