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Two time dependent wavefunctions:

$\Psi _1(t)= \psi_1*exp(\frac{-i * E_1}{\hbar}*t)$

$\Psi _2(t)= \psi_2*exp(\frac{-i * E_2}{\hbar}*t)$

Both a solution to the timeindependent (note "in") Schrödinger eq. with the same H. We know they are solutions. Furthermore $E_1$ and $E_2$ are different.

$\mid \psi_1exp(\frac{-i E_1}{\hbar}t) + \psi_2exp(\frac{-i E_2}{\hbar}t)\mid ^2 $

$= \mid \psi_1exp(\frac{-i E_1}{\hbar}*t)\mid^2 + \mid\psi_2exp(\frac{-i E_2}{\hbar}t)\mid^2 + 2 \mid\psi_1exp(\frac{-i E_1}{\hbar}t) \psi_2exp(\frac{-i E_2}{\hbar}t) \mid$

$= \mid \psi_1exp(\frac{-i E_1}{\hbar}*t)\mid^2 + \mid\psi_2exp(\frac{-i E_2}{\hbar}t)\mid^2 + 2 \mid\psi_1 \psi_2exp(\frac{-i (E_2-E_1)}{\hbar}t) \mid$

Is the following correct?:

$\mid \psi_1exp(\frac{-i E_1}{\hbar}*t)\mid^2 = \mid\psi_1^{(*)}\psi_1\mid = \mid\psi_1\mid^2* exp(\frac{-i E_1}{\hbar}t) * exp(\frac{i E_1}{\hbar}t) = \mid\psi_1\mid^2$

leading to:

$\mid \psi_1exp(\frac{-i E_1}{\hbar}t) + \psi_2exp(\frac{-i E_2}{\hbar}t)\mid ^2 = \mid\psi_1\mid^2 + \mid\psi_2\mid^2 + 2 \mid\psi_1 \psi_2exp(\frac{-i (E_2-E_1)}{\hbar}t) \mid$

meaning that: $\mid\Psi _1(t)+\Psi _2(t)\mid $ ocilliates with $\frac{\hbar}{(E_2-E_2)}$ ?

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  • $\begingroup$ How can those wave functions be solutions of the time independent Schrodinger equation, if they do not depend on the spatial coordinates? $\endgroup$
    – Dante
    May 20 '17 at 10:14
  • $\begingroup$ Sorry I was just to lazy to write $\Psi(r,t)$, but you are right, it should have been said that the wavefunctions depended on the spartial coordinates. Any how it does not change the problem. $\endgroup$
    – DUDEofDK
    May 20 '17 at 10:17
  • $\begingroup$ I think what you wrote is right. What's the question? $\endgroup$
    – Martino
    May 20 '17 at 10:43
  • $\begingroup$ If what I wrote was correct, which seems to be the case, which answers my question. thx! $\endgroup$
    – DUDEofDK
    May 20 '17 at 10:46
  • $\begingroup$ Related: Is there oscillating charge in a hydrogen atom? $\endgroup$ May 20 '17 at 13:11
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You obviously mean that $$\begin{aligned} H\psi_1 &= E_1\psi_1,\\ H\psi_2 &= E_2\psi_2. \end{aligned}$$ Then, the solution of the Schrödinger equation (I will use units such that $\hbar=1$ throughout: old habit of a former theoretical physicist!!) $$H\Psi_k = i\dfrac{\partial \Psi_k}{\partial t}$$ with the initial condition $$\Psi_k(t=0) = \psi_k$$ is indeed your $$\Psi_k = \psi_k \exp(-iE_k t).$$ Then you consider a superposition of these two states, $\Psi = \Psi_1 + \Psi_2$, which is also solution of the Schrödinger equation for the initial condition $$\Psi(t=0) = \psi_1 + \psi_2,$$ i.e. a superposition of a state of energy $E_1$ and a state of energy $E_2$.

Your calculation of $|\Psi_1 + \Psi_2|^2$ is a wee bit incorrect: $$|\Psi_1 + \Psi_2|^2 = |\Psi_1|^2 + |\Psi_2|^2 + \underbrace{\Psi_1 {\Psi_2}^{\!*} + {\Psi_1}^{\!*}\Psi_2}_{2\Re\Psi_1{\Psi_2}^{\!*}}$$ where $\Re$ denotes the real part. That is to say $$|\Psi_1 + \Psi_2|^2 = |\psi_1|^2 + |\psi_2|^2+2\Re \psi_1{\psi_2}^{\!*}\exp[ i(E_2-E_1)t]$$ This does not change your conclusion though: oscillations with a pulsation $E_2 - E_1$.

Right, now that this is all clear and correct, could you elaborate on your question? I can't guess what you want to know!

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  • $\begingroup$ The question was: was my calculation correct. Did I properly undersand the calculation behind the superposition squared. Your correction of the $\mid\Psi_1+\Psi_2\mid^2$ answered any doubt I had $\endgroup$
    – DUDEofDK
    May 20 '17 at 11:55
  • $\begingroup$ Ok! Happy to help! $\endgroup$
    – user154997
    May 20 '17 at 12:06

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