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if I transform the time-dependent Schrödinger equation without a potential I get:

$$ - \hbar \omega \psi(\omega,x) = \frac{- \hbar^2}{2m} \frac{\partial^2 \psi(\omega,x)}{\partial x^2}$$

The solutions is clealy: $$\psi(\omega,x)={\it C1}\,{{\rm e}^{{\frac {\sqrt {2\omega m}x}{\sqrt {\hbar}}}} }+{\it C2}\,{{\rm e}^{{-\frac {\sqrt {2\omega m}x}{\sqrt {\hbar}}}} } $$

I don't really understand this result. The problem is, that if I want to transform back, the Fourier-transform will be divergent, so what does this mean regarding my solution? Is there a work-around to get rid of this divergence? Why did this Fourier-transform fail?

(Should I have used the Laplace-transform?)

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    $\begingroup$ How did you get that first equation? $\endgroup$ – Kyle Kanos Jun 22 '14 at 2:36
  • $\begingroup$ Just like @KyleKanos, how come you still have derivative w.r.t. $x$ in the transformed equation? $\endgroup$ – 299792458 Jun 22 '14 at 5:41
  • $\begingroup$ Your solutions do not satisfy your equation! Pay attention to signs... $\endgroup$ – Valter Moretti Jun 22 '14 at 7:19
  • $\begingroup$ @KyleKanos I did the Fourier transform with respect to $t$ to the equation $i \hbar \partial_t \psi(x,t) = -\frac{\hbar^2}{2m} \partial_x^2 \psi(x,t)$ $\endgroup$ – Xin Wang Jun 22 '14 at 9:17
  • $\begingroup$ @V.Moretti sorry, had a minus sign too much in the original equation. $\endgroup$ – Xin Wang Jun 22 '14 at 9:18
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----edited, as suggested in the comments----

The function ψ(t,x) is not square integrable with respect to time. So the Fourier transform of it may have meaning only in the distributional sense. You are treating a distribution (with respect to ω) like a function, and then solving an ODE in x. This in general cannot be done, since distributions does not behave like functions (e.g. you cannot define product of distributions).

To find the solution you proceed in a very standard way as follows. Consider the operator $-\Delta$: it is self-adjoint on a dense domain of $L^2(\mathbb{R}^d)$ (assuming you are in $d$ dimensions), and thus to it can be associated an unitary one-parameter group $\exp(it\Delta)$ (I am assuming here $m$ and $\hbar$ to be one-half and one respectively, for simplicity). This unitary one parameter group is defined for all functions of $L^2$, so given $\psi(t_0,x)=\psi_0(x)\in L^2(\mathbb{R}^d)$, the solution of your Cauchy problem is $$\psi(t,x)=e^{i(t-t_0)\Delta}\psi_0(x)\; .$$ If you take the spatial fourier transform (in $x$), that is another unitary transformation on $L^2$, you obtain the maybe more explicit formula $$\hat{\psi}(t,k)=e^{-i(t-t_0)k^2}\hat{\psi}_0(k)$$ where $\hat{\psi}$ is the Fourier transform of $\psi$.

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  • $\begingroup$ I am aware of this approach, but I was wondering: why did my approach fail? could you try to rather comment on the question? $\endgroup$ – Xin Wang Jun 22 '14 at 10:12
  • $\begingroup$ I mean, if anything is unclear about my question I will try to explain what I did, but your answer is really not what I am looking for. $\endgroup$ – Xin Wang Jun 22 '14 at 10:39
  • $\begingroup$ @yuggib: I think you should put this comment in your answer, since that's really the answer to the question. $\endgroup$ – Adam Jun 22 '14 at 21:54

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