I am given the wavefunction $\psi(x,t)$.
How does one determine the probability of a measurement of the energy giving the $n^{th}$ eigenstate?

My guess is it should be something like $P_n = \int_{-\infty}^{\infty}|\langle O_n|\psi(x,t)\rangle|^2dx$, but what is the $O_n$ operator that I should use?


1 Answer 1


You first need to calculate the spectrum of the Hamiltonian, i.e. you need to solve the time-independent Schrodinger equation

$$\hat{H} \vert \varepsilon _n \rangle = E_n \vert \varepsilon _n \rangle$$

to find the eigenvalues $E_n$ and eigenvectors $\vert \varepsilon _n \rangle$.

With that in mind, if the $n^\mathrm{th}$ eigenstate corresponds to a non-degenerate eigenvalue, you can simply write:

$$\mathrm{Pr}(E=E_n , t) = \vert \langle \varepsilon_n \vert \Psi (t) \rangle \vert^2 = \Big\vert \int_{- \infty}^\infty dx \ \varepsilon_n(x)^* \psi(x,t) \Big\vert^2$$

And for the general degenerate case, where the eigenvalue $E_n$ corresponds to the orthonormal set of eigenvectors $\{\vert \varepsilon_{n,k} \rangle \}_{k=1}^{g_n}$ you have:

$$\mathrm{Pr}(E=E_n,t) = \sum_{k=1}^{g_n} \vert \langle \varepsilon_{n,k} \vert \Psi (t) \rangle \vert^2 = \sum_{k=1}^{g_n} \Big\vert \int_{- \infty}^\infty dx \ \varepsilon_{n,k}(x)^* \psi(x,t) \Big\vert^2$$ with $g_n$ being the order of degeneracy for the eigenvalue $E_n$.


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