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Given a normalised wavefunction say $$\psi(x) = A\sin(n\pi x),$$ (where $A$ is a normalisation constant) I can calculate the probability of finding the particle being between a position $x$ and $x + dx$ as $$\int_{x}^{x+dx} \psi(x)^*\psi(x) dx.$$ Likewise I can calculate the expected value of the energy $$\langle E \rangle = \int_{-\infty}^\infty \psi(x)^*\hat{H} \psi(x) dx.$$ However, how can I calculate the probability of a given value of the energy say $$E = \frac{\hbar^2 \pi^2}{2}$$ as my wavefunction $\psi(x)$ is defined in terms of position not energy?

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    $\begingroup$ Find the eigenstate $\lvert E \rangle$ of $H$ that belongs to $E$, calculate $\lvert \langle E \vert \psi \rangle \rvert^2$, same as with any other observables. $\endgroup$ – ACuriousMind May 28 '15 at 14:43
  • $\begingroup$ $sin(n\pi x)$ is not an allowed wavefunction on the whole $L^2(\mathbb{R}^3)$; but only in a box... $\endgroup$ – yuggib May 28 '15 at 14:57
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Elaborating on the answer of zeldredge, I want to say why the following expression works:

$$P(E_0) = \left| \int_{-\infty}^{\infty} \ \Phi^* \psi \d x \right|^2 \tag{1}$$

Notice that the eigenfunctions of the Hamiltonian spans the space. That is you can write any function $\psi$ as a linear combination of eigenfunctions of the Hamiltonian ie.

$$\psi=a_0\Phi_0+a_1\Phi_1+a_2\Phi_2 \dots = \sum_i a_i \Phi_i \tag{2}$$

with $a_i \in \mathbb C$ and $\sum_i \left|a_i \right|^2=1$. Notice that as in all superposition of states the probability of finding particle in a particular state is given by $\left| a_i \right|^2$. Assuming that you have chosen your states cleverly such that they are orthonormal ie they satisfy the following property,

$$\int \Phi_i^*\Phi_j \d x = \delta_{ij}$$

you can see immediately why Eqn. (1) works.

$$\int \Phi_i^* \psi \d x = \int \Phi_i^* \sum_j a_j \Phi_j \d x = \sum_j a_j\int \Phi_i^* \Phi_j \d x = \sum_j \delta_{ij} a_j =a_i $$

Therefore the norm squared of this expression just gives the probability to find the state in the energy eigenfunction $\Phi_i$.

Notice I assumed that the energy levels are discrete to write the equation (2). Furthermore I dropped $x$'s in $\psi(x)$ and $\Phi(x)$ because they were cluttering the equations.

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  • $\begingroup$ Yes they both missed the absolute value squared. For the case of continuous spectrum see my comment below the other answer. That the abs value squared is necessary can be easily seen for the given formula observing that for a discrete (non-degenerate) spectrum such that in $[E_0,E_1]$ only the value $E\in [E_0,E_1]$ belongs to the spectrum with eigenvector $\Phi_E$, the projector $1_{[E_0,E_1]}(H)=\lvert \Phi_E\rangle\langle \Phi_E\rvert$, therefore the formula you wrote is the correct one. $\endgroup$ – yuggib May 28 '15 at 15:29
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You have to know the energy eigenstates. Suppose you know that $\Phi (x)$ is an energy eigenstate with $E = E_0$. Then the probability that your state $\psi(x)$ will be measured to have energy $E_0$ is:

$$ P(E_0) = \int_{-\infty}^{\infty} \mathrm{d} x \ \Phi^*(x) \psi(x) $$

You can do this for the full set of eigenstates $\Phi_i$ to get the wavefunction in an energy basis (instead of the position basis you have it represented in now). Note you can't always do this for an arbitrary $E$, because only a subset of them (the eigenvalues) are going to be energy eigenstates, although for some systems (scattering states) you have a continuous basis for energy.

EDIT: Another thing I forgot to add is that your energy states may be degenerate--there might be multiply $\Phi$ that could give you $E_0$. In that case you need to add up the different probabilities from each one.

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  • $\begingroup$ The energy eigenstates has to be orthonormal for this to work. $\endgroup$ – Gonenc May 28 '15 at 14:54
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    $\begingroup$ Are you missing a norm square in your equation with $P(E_o)$? $\endgroup$ – Gonenc May 28 '15 at 15:14
  • $\begingroup$ I assumed both states were normalized $\endgroup$ – zeldredge May 28 '15 at 15:15
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    $\begingroup$ @zeldredge nonetheless you need a norm squared I am afraid... also, in general for energies with continuous spectrum the maximum you can say is the probability of finding a value of $E$ in a given interval of reals $[E_0,E_1]$. This is given by (assuming the state is normalized) $\langle \psi, 1_{[E_0,E_1]}(H)\psi\rangle$, where $1_{[E_0,E_1]}(H)$ is the suitable spectral projection associated to the operator $H$. $\endgroup$ – yuggib May 28 '15 at 15:19
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First you have to find the eigenstate of the Hamiltonian, in other words, you have to solve the equation: $$\hat H \psi_{E}(x)=E \psi_{E}(x),$$ then take the inner product of $$\psi_{E}(x)\quad and \quad \psi(x)$$,it is the probability you want.

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    $\begingroup$ Am I wrong or both of you miss a norm squared? $\endgroup$ – Gonenc May 28 '15 at 15:18
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    $\begingroup$ we just ignored the norm and think that the wave function is normalized,you are right. $\endgroup$ – xjtan Jun 2 '15 at 13:23
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    $\begingroup$ Can someone explain why this answer got downvoted? It seems helpful (i.e. to point into the right direction), even though not complete/elaborate. $\endgroup$ – Genti Saliu Jun 19 '16 at 17:09

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