Say that we have a wave function $$\Psi(x,t)=c_1\Psi_1(x,t)+c_2\Psi_2(x,t)$$ where the particle is confined to $[0,a]$, $\Psi_i$ corresponds to the normalised wave function for energy $E_i$, and $c_i$ comes from an initial condition.

How do we find the probability that it is an interval say $[0,\frac{1}{2}a]$ at time $t$?

Can we condition on the probability by saying that $\lvert c_i\rvert^2$ is the probability it is of energy $E_i$, and find $$\int_0^{\frac{1}{2}a}\lvert \Psi_i\rvert^2dx$$ for each $i$ or must we calculate the integral for $\lvert \Psi\rvert^2$ over the half-interval?

  • $\begingroup$ The latter is what your book should say. What makes you consider the integral you wrote before "or"? $\endgroup$ Commented Sep 14, 2021 at 19:35
  • $\begingroup$ I have just seen that given an initial condition, the values $\lvert c_i \rvert^2$ correspond to the probability that the particle is of energy $E_i$, so I was wondering if we could condition i.e. $$\mathbb{P}(\text{in }[0,\frac{1}{2}a])=\mathbb{P}(\text{in }[0,\frac{1}{2}a] \lvert E=E_1)\mathbb{P}(E=E_1)+\mathbb{P}(\text{in }[0,\frac{1}{2}a] \lvert E=E_2)\mathbb{P}(E=E_2)$$ $\endgroup$
    – jaycneek
    Commented Sep 14, 2021 at 19:40
  • $\begingroup$ You invented an unwarranted algorithm. In QM interference one adds amps, not probabilities. $\endgroup$ Commented Sep 14, 2021 at 19:54
  • $\begingroup$ I'm sorry, I dont follow. I am a maths student after all... I suppose I'm just trying to clarify if the basic law of total probability applies here $\endgroup$
    – jaycneek
    Commented Sep 14, 2021 at 19:57
  • $\begingroup$ There is no "basic law of total probability", for dependent, interfering alternatives in QM. That's what you are meant to learn in your course! You may contrast your conjectural pseudoanswer to the correct one, also provided in the answer provided. $\endgroup$ Commented Sep 14, 2021 at 20:00

1 Answer 1


The probability that $\psi(x,t)$ is measured to lay in the interval $[0, a/2]$ at the time $t$ is given by $$ \int_0^{a/2} |\psi(x,t)|^2 \mathrm{d}x = \int_0^{a/2} |c_1|^2 |\psi_1|^2 + c_1^*\psi_1^*c_2\psi_2 + c_1\psi_1 c_2^* \psi_2^* + |c_2|^2|\psi_2|^2 \,\mathrm{d}x $$ since the whole wavefunction $\psi$ should be normalised. If you assume that the energy eigenstates $\psi_i$ are normalised then that'll give you a restriction on the possible values that $c_1$ and $c_2$ can take on.


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