Probability of measuring the ground state of a particle (quantum mechanics)

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SUMMARY

The discussion focuses on calculating the probability of measuring the ground-state energy of a particle in an infinite potential well defined by the wave function \(\psi(x,0) = Ax(a-x)\). The eigenvalues of the Hamiltonian for this potential are given by \(\phi_n = \sqrt{\frac{2}{a}} \sin\left(\frac{n \pi x}{a}\right)\). The correct method to find the probability involves calculating \(\left| \langle \psi | \phi_n \rangle \right|^2\) for \(n=1\), which simplifies to \(\left[ \int^a_0 \psi \phi_1 dx \right]^2\). The initial calculation yielded an incorrect probability greater than 1, indicating an algebraic error in the computation.

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BenR-999
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Have to find the probability of measureing the ground-state energy of a particle.
-in infinite potential well 0<x< a
has wave-function \psi (x,0) = Ax(a-x)
where a is the (known) length of the well, and the norm. const. A has already been found.

The eigenvalues of the hamiltonian in this potential are;
\phi_n = \sqrt{\frac{2}{a}} \sin(\frac{n \pi x}{a}})

I think that to do this i should take
\left| \langle \psi | \phi_n \rangle \right|^2
for n=1.

which becomes
\left[ \int^a_0 \psi \phi_1 dx \right] ^2 (as all are real-valued)

I'm not sure if this is correct..it just seems a little to simple.

(with A=\frac{ \sqrt{30} }{a^{5/2} }
i got 60/ \pi ^2 which is obviously incorrect..as it is greater than 1.
But, if the method is correct and i have just made algebra error?

Thanks
 
Physics news on Phys.org
Yes, your method is correct, and you just made an algebra mistake.
 
beauty,
thanks mate
 

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