- #1
HJ Farnsworth
- 126
- 1
Greetings everyone,
I haven’t done any quantum in a while, and was reviewing my textbook, Griffiths Ed. 1. The form of the Schrödinger equation I’m using is:
i\hbar\partial\Psi/\partialt = -\hbar2/2m * \partial2\Psi/\partialx2 + V\Psi
The book says if V is a function of x only, then the wave function is separable. This makes sense to me, but what I wanted to know is, why is it not generally separable otherwise? My understanding is that any linear homogeneous equation is separable (could someone please verify that?), so that if L(\Psi) = 0, then \Psi should be separable, where L is a linear operator. Referring to the Schrödinger equation, L would be:
i\hbar\partial/\partialt + \hbar2/2m * \partial2/\partialx2 - V.
I’m not going to worry about the first two terms right now, since it seems obvious enough that they're linear. By definition of linearity:
L(c1u1 + c2u2) = c1L(u1) + c2L(u2)
It seems like V(x, t) should be a linear operator, since as far as I can tell it won't behave so much as an operator, but instead as a function that just gets multiplied by given solutions of the wave function:
V(x, t)(c1\Psi1 + c2\Psi2) = c1V(x, t)\Psi1 + c2V(x, t)\Psi2
This seems like it should be true by basic algebra to me – what’s my mistake here? It seems like I'm either applying the definition of linearity wrong, or the potential function can, in fact, be something that will actually "affect" the wavefunction it operates on, like something ending with partials - but this doesn't make sense to me, since the potential function should just give numbers for the potential energy.
A simple example of a potential function for which the above doesn’t work and/or an explanation of why I'm using the definition of linearity wrong would be very much appreciated.
Thanks very much for any help you can give.
-HJ Farnsworth
I haven’t done any quantum in a while, and was reviewing my textbook, Griffiths Ed. 1. The form of the Schrödinger equation I’m using is:
i\hbar\partial\Psi/\partialt = -\hbar2/2m * \partial2\Psi/\partialx2 + V\Psi
The book says if V is a function of x only, then the wave function is separable. This makes sense to me, but what I wanted to know is, why is it not generally separable otherwise? My understanding is that any linear homogeneous equation is separable (could someone please verify that?), so that if L(\Psi) = 0, then \Psi should be separable, where L is a linear operator. Referring to the Schrödinger equation, L would be:
i\hbar\partial/\partialt + \hbar2/2m * \partial2/\partialx2 - V.
I’m not going to worry about the first two terms right now, since it seems obvious enough that they're linear. By definition of linearity:
L(c1u1 + c2u2) = c1L(u1) + c2L(u2)
It seems like V(x, t) should be a linear operator, since as far as I can tell it won't behave so much as an operator, but instead as a function that just gets multiplied by given solutions of the wave function:
V(x, t)(c1\Psi1 + c2\Psi2) = c1V(x, t)\Psi1 + c2V(x, t)\Psi2
This seems like it should be true by basic algebra to me – what’s my mistake here? It seems like I'm either applying the definition of linearity wrong, or the potential function can, in fact, be something that will actually "affect" the wavefunction it operates on, like something ending with partials - but this doesn't make sense to me, since the potential function should just give numbers for the potential energy.
A simple example of a potential function for which the above doesn’t work and/or an explanation of why I'm using the definition of linearity wrong would be very much appreciated.
Thanks very much for any help you can give.
-HJ Farnsworth