Probability of Finding a Particle in the Ground-state after the Size of the Box is Doubled [closed]

A particle of mass $m$ is contained in a one-dimensional infinite well extending from $x=-{L\over2}$ to $x=+{L\over2}$. The particle is in its ground state given by $$|{\psi_0(x)}\rangle=\sqrt{2\over{L}}\cos(\frac{\pi x}{L})$$ The walls of the box are moved suddenly to form a box extending from $x=-L$ to $x=+L$. What is the probability that the particle will be in the ground state?

My solution

Using the sudden expansion approximation the initial state of the particle in the box of length $L$ will not have enough time to respond to the expansion and so the eigenstate of the particle after the change is completed is $$|{\psi_n} (x)\rangle=\sqrt\frac{2}{L}\cos(\frac{n\pi x}{L}); x{\epsilon}[-{L\over2},{L\over2}]$$ $$|{\psi_n(x)}\rangle=0; x{\epsilon}[-L,L]$$ The eigenstate for a particle of mass $m$ in a one-dimensional box extending form $x=-L$ to $x=+L$ is $$|{\phi_n(x)}\rangle=\sqrt\frac{1}{L}\cos(\frac{{\pi}x}{2L})$$ for odd $n$ and $$|{\phi_n(x)}\rangle=\sqrt\frac{1}{L}\sin(\frac{{\pi}x}{2L})$$ for even $n$

The probability of finding that the particle will be in the ground-state is $$P= \int_{-{L\over2}}^{L\over2}(\langle{\phi_1(x)}|{\psi_1(x)}\rangle)^2 dx$$ Which comes out to be $$P=\left(8\over{3{\pi}}\right)^2$$ My question is why when calculating the probability the limit of integration is from $x={-{L\over2}}$ to $x={L\over2}$ and not from $x=-L$ to $x=+L$? Does the question mean by ”The probability of finding that the particle will be in the ground-state” - the probability of finding the particle in the ground-state of the new box?

And secondly do I need to use the argument that the eigenvalues of both the hamiltonians will be the same?

closed as off-topic by John Rennie, Jon Custer, Kyle Kanos, Bill N, AccidentalFourierTransformFeb 24 '17 at 12:11

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• not sure what your second question is asking, but I also think you should integrate from -L to L. – Steven Sagona Feb 20 '17 at 19:20
• If I use the argument that the eigenvalues of the two hamiltonians are the same, say in the first case $$E_n=\frac{n^2{\pi^2}{\hbar^2}}{2mL^2}$$ and the second case $$E'_n=\frac{n'^2{\pi^2}{\hbar^2}}{2m(2L)^2}$$ then $$n'=2n$$ and the probability becomes 0.5. – Sayontön Vöttacharjo Feb 20 '17 at 19:35
• Because outside of +-L/2 the integrand is zero anyway, so it doesn't matter. – LLlAMnYP Feb 21 '17 at 9:35