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Quantum mechanics challenge


Active member
Nov 3, 2013
A tester of basic quantum mechanics:

1) Let the state of a quantum particle be represented by \(\displaystyle \phi\). Show that if \(\displaystyle \phi\) satisfies Schrodinger's equation, then its norm is constant.

2) Now consider a quantum particle with state \(\displaystyle \phi_{t}\) defined on [-a,a] subject to potential V=0.

State the differential equation that \(\displaystyle \phi_{t}\) solves in terms of partial derivatives of x and t.


Indicium Physicus
Staff member
Jan 26, 2012
Partial solution to #1:

We assume $\phi$ satisfies the Schrodinger equation
$$i \hbar \phi_{t}=(- ( \hbar^{2}/2m) \nabla^{2}+V)\phi. \qquad (1)$$
Since $|\phi|^{2}= \phi^{*} \phi$, we seek to show that
$$ \frac{ \partial}{ \partial t}( \phi^{*} \phi)=0.$$
This is tantamount to showing that
$$ \phi \frac{ \partial \phi^{*}}{ \partial t}+ \phi^{*} \frac{ \partial \phi}{ \partial t}=0.$$
Take the complex conjugation of the Schrodinger equation thus:
$$-i \hbar \phi^{*}_{t}=(- ( \hbar^{2}/2m) \nabla^{2}+V)\phi^{*}. \qquad (2)$$
We have assumed that $V$ is real. Multiply $(1)$ by $\phi^{*}$ and $(2)$ by $-\phi$ to obtain:
i \hbar \phi^{*} \phi_{t}&=(- ( \hbar^{2}/2m) \phi^{*} \nabla^{2}+ \phi^{*}V)\phi \\
i \hbar \phi \phi^{*}_{t}&=(( \hbar^{2}/2m) \phi \nabla^{2}- \phi V)\phi^{*}.
Adding these equations together yields
$$i \hbar (\phi^{*} \phi_{t}+ \phi \phi^{*}_{t})=-( \hbar^{2}/2m) \phi^{*} \nabla^{2} \phi+( \hbar^{2}/2m) \phi \nabla^{2} \phi^{*}
= \frac{ \hbar^{2}}{2m}\left( \phi \nabla^{2} \phi^{*} - \phi^{*} \nabla^{2} \phi \right).$$
We have now reduced the problem down to showing that
$$\phi \nabla^{2} \phi^{*} - \phi^{*} \nabla^{2} \phi=0.$$
Let us examine one of these. I claim that $\phi \nabla^{2} \phi^{*}=- \nabla \phi \cdot \nabla \phi^{*}$; moreover, an analogous line of reasoning will show that $ \phi^{*} \nabla^{2} \phi=- \nabla \phi^{*} \cdot \nabla \phi$. Consider the integral
$$\iiint_{U} \phi^{*} \nabla^{2} \phi \, dV.$$ By Green's First Identity, we know that
$$\iiint_{U} \phi^{*} \nabla^{2} \phi \, dV=- \iiint_{U} \nabla \phi^{*} \cdot \nabla \phi \, dV+ \iint_{ \partial U} \phi^{*}( \nabla \phi \cdot \mathbf{n}) \, dS.$$
This holds for all regions $U$. What I want to say now is that the surface integral has to be zero. However, I'm not sure I can conclude that.

I am at home now; I can post more when I get back to school on Monday.


Indicium Physicus
Staff member
Jan 26, 2012
Question on part 2: are you using the notation
$$\phi_{t}= \frac{ \partial \phi}{ \partial t}?$$