Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am always dubious when I need write Schrödinger equation: do I write $\partial / \partial t$ or $d/dt$ ?

I suppose it depends on the space in which it is considered. How?

share|improve this question
add comment

1 Answer

The most general Schrödinger equation has total derivatives $$ i\hbar \frac{d}{dt}|\psi\rangle = \hat H |\psi\rangle $$ because the state vector $|\psi\rangle$ only depends on one variable, $t$. It's a complicated object that knows about the probability of anything in the given state, but this is hidden "inside" the state vector.

However, if you rewrite the state vector in a given representation, e.g. as $\psi(t,x,y,z,X,Y,Z)$ for the wave function of two particles, then the dependence on $x,y,z,X,Y,Z$, the coordinates of two particles, is put on equal footing with the $t$-dependence, and therefore the $t$-derivatives have to be written as partial ones, $\partial/\partial t$, to emphasize that $x,y,z,X,Y,Z$ are kept fixed during the differentiation. $$ i\hbar \frac{\partial}{\partial t}\psi(t,x,y,z,X,Y,Z) = \hat H \psi(t,x,y,z,X,Y,Z) $$ where the Hamiltonian contains things like the kinetic energy of the first particle $$ \hat H = \dots -\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right)+\dots $$ and similarly the kinetic energy of the second particle $$ \hat H = \dots -\frac{\hbar^2}{2M} \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} \right)+\dots $$ Note that there are partial derivatives everywhere because $\psi$ is now not a "general state vector" whose information is compactified; it is a complex-valued function of many variables.

share|improve this answer
    
Well, that is what I would have said. But both in my course and in Oxford quantum course $\partial$ is used instead of $d$ even in "the most general Schrödinger equation" ... So I am still not convinced. –  Isaac Jan 14 '12 at 10:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.