<E> of a Particle in a 1D Box

  • Thread starter Domnu
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  • #1
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Problem
Show that for a particle in a one-dimensional box, in an arbitrary state [tex]\psi(x,0)[/tex], [tex] \langle E \rangle \ge E_1[/tex]. Under what conditions does the equality maintain?

Solution
Note that any arbitrary particle in a one-dimensional box can be expanded in terms of the eigenstates

[tex]\phi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi x}{a}[/tex]​

Now, note that [tex]\phi_1[/tex] corresponds to the [tex]E_1[/tex] energy level. So, if the expansion contains anything excepting the [tex]E_1[/tex] energy level, then the expected energy level will be higher than [tex]E_1[/tex]. Note that energy levels can only be positive integers, so 0 cannot be an energy level (the probability of a particle having an energy level of 0 is 0, because the eigenstate when [tex]n=0[/tex] is just 0... in addition, this would imply the particle had no energy, which would mean that the particle's position is infinitely uncertain... but this isn't true, since we know that the particle is constrained to be within a box). [tex]\blacksquare[/tex]

Are my arguments correct?
 

Answers and Replies

  • #2
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Your qualitative description is OK. You could make your solution more rigorous by finding a mathematical expression for <E> in terms of expansion coefficients and then use it to explicitly prove the statement.
 

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