Chapter 3: 3.10P (page 106)

Is the ground state of the infinite square well an eigenfunction of momentum? If so, what is its momentum? If not, why not?

Short Answer

Expert verified

The wavefunction $ψ_{n}$ is not a momentum eigenfunction.

The magnitude of the momentum is a constant, and it is given by,

$| p | = \sqrt{2 m E} = \frac{n π ℏ}{a}$

Step by step solution

Concept used

The wave function of the infinite square well's stationary states is calculated as follows:

$ψ_{n} (x) = \sqrt{\frac{2}{a}} s i n (\frac{nπx}{a})$

Where $a$ is the width of the well.

Calculate the momentum

The wave function of the infinite square well's stationary states is calculated as follows:

$\hat{p} = - i ℏ \frac{d}{d x}$

We can check that directly as:

$\hat{p} ψ_{n} = - i ℏ \frac{d}{d x} \sqrt{\frac{2}{a}} \sin (\frac{n π x}{a}) = - i ℏ \sqrt{\frac{2}{a}} \frac{n π}{a} \cos (\frac{n π x}{a}) = - i \frac{ℏ n π}{a} \cot (\frac{n π x}{a}) ψ_{n} (x)$

Since the momentum operator doesn't yield the original wave function multiplied by a constant, then the wavefunction $ψ_{n}$ is not an eigenfunction of momentum. The mean momentum is:

$⟨ p ⟩ = \int_{0}^{a} ψ_{n} \hat{p} ψ_{n} d x = - i \frac{2 n π ℏ}{a^{2}} \int_{0}^{a} \sin (\frac{n π x}{a}) \cos (\frac{n π x}{a}) d x = 0$

Which means the particle is just as likely to be found traveling to the left as to the right. The magnitude of momentum is a constant that is given by,

$| p | = \sqrt{2 m E} = \frac{n π ℏ}{a}$

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Is the ground state of the infinite square well an eigenfunction of momentum? If so, what is its momentum? If not, why not?

Short Answer

Step by step solution

Concept used

Calculate the momentum

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