What are the dimensions of abs(b_{n})^2 and abs(b(k))^2 in particle functions?

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SUMMARY

The discussion focuses on determining the dimensions of the coefficients abs(b_{n})^2 and abs(b(k))^2 in the context of quantum mechanics, specifically for a particle in a box and a free particle. It is established that these coefficients relate to the normalization condition of the wave function, ψ(x), which must satisfy the integral ∫ ψ*(x)ψ(x) dx = 1. The dimensions of ψ(x) are derived to be [L^(-3/2)], leading to the conclusion that abs(b_{n})^2 has dimensions of [L] and abs(b(k))^2 has dimensions of [L^(-1)].

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Homework Statement


If an arbitrary intial state function for a particle in a box is expanded in the discrete series of eigenstates of the Hamiltonian relevant to the box configuration, one obtains:

\psi(x,0) = \Sigma^{\infty}_{n=1}b_{n}(0)\varphi_{n}(x)

If the particle is free, we obtain:

\psi(x,0) = \int^{\infty}_{-\infty}b(k)\varphi(k)dk

(a)
What are the dimensions of:
abs(b_{n})^2
and abs(b(k))^2?

I don't understand what they mean by dimensions! any hints?
 
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By dimensions they mean, well, dimensions as in dimensional analysis in terms of the standard three, Mass, Length and Time. Start by considering

\int \psi^{*}(x)\psi(x)\: dx = 1

What are the dimensions of ψ(x)? Sort it out from there.
 

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