I'm having some trouble finding an expression for the probability to find the particle outside the classical area in the harmonic oscillator.

I have a wavefunction defined as:

$\psi \left( x,\,t \right)=\frac{1}{2}\left( \sqrt{3}i{{\phi }_{1}}\left( x \right){{e}^{-i{{E}_{1}}t/\hbar }}+{{\phi }_{3}}\left( x \right){{e}^{-i{{E}_{3}}t/\hbar }} \right)$

I'm supposed to give the expression by $P(x,t)$, but not explicitly calculated.

My thought was to use:

${{\int_{a}^{b}{\left| \psi \left( x,t \right) \right|}}^{2}}dx$,

but then I pretty ran into the wall.

So anyone who could give me a hint of what to do ?

Thanks in advance.

  • $\begingroup$ You don't need to take the integral : you are at a situation where $a=x$, $b=x+dx$. $\endgroup$ Jun 12, 2013 at 10:51
  • $\begingroup$ Hmmm, why does that imply that I don't have to do the integral ? $\endgroup$ Jun 12, 2013 at 11:58
  • $\begingroup$ The integral you wrote is the probability of being betwwen $a$ and $b$, i.e. $\int_a^bP(x,t) dx$, what you are looking for is $P(x,t)$, which is under the integral. Another way to see it is to look at the probability to be in a small interval close to $x$. In this case, the integral reduces to the simple product $P(x,t)\times dx$, if $x$ is small enough. $\endgroup$ Jun 12, 2013 at 12:50
  • $\begingroup$ Sorry, I misunderstood the question. Forget my comments, and read @Nivalth's answer $\endgroup$ Jun 12, 2013 at 13:11

1 Answer 1


The classically forbidden region coresponds to the region in which

$$ T(x,t)=E(t)-V(x) <0$$

in this case, you know the potential energy $V(x)=\displaystyle\frac{1}{2}m\omega^2x^2$ and the energy of the system is a superposition of $E_{1}$ and $E_{3}$. This should be enough to allow you to sketch the forbidden region, we call it $\Omega$, and with $\displaystyle\int_{\Omega} dx \psi^{*}(x,t)\psi(x,t) $ probability you're asked for,

  • $\begingroup$ Ok. Kind of strange question, but I think I know what you mean :) Thank you very much. $\endgroup$ Jun 12, 2013 at 18:37

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