Partial Differentiation in free Schrodinger Equation

[karkas]

Hello again! This time I have another calculus question for you, coming straight out of my study of the free Schrodinger equation, since I am not that experienced with that kind of derivative.

It all starts with a given wavefunction (which I think is 2-dimensional,correct me if wrong)
$\psi(x,t)=e^{i(px-Et)}/\hbar$ (1)

My book says|: We derive 3.4 according to t and x so we have:
$\frac{\partial \psi}{\partial t}=-\frac{iE}{\hbar} \psi$
and $\frac{\partial \psi}{\partial x}=\frac{ip}{\hbar}\psi$.

What I don't understand is how this partial deriving works. What I can figure out is that given the (1) equation, you derive the $i(px)/\hbar$ and $i(-Et)/\hbar$ parts differently, one by one not together. Is that it? So you basically remove the e (2.71)(this must be the way to derive from exponential equations, I don't remember clearly though) and derive the power...partially?

Alas, I am confused.. Any understandable link or advice is appreciated!

 

[ice109]

Hello again! This time I have another calculus question for you, coming straight out of my study of the free Schrodinger equation, since I am not that experienced with that kind of derivative.

It all starts with a given wavefunction (which I think is 2-dimensional,correct me if wrong)
$\psi(x,t)=e^{i(px-Et)}/\hbar$ (1)

My book says|: We derive 3.4 according to t and x so we have:
$\frac{\partial \psi}{\partial t}=-\frac{iE}{\hbar} \psi$
and $\frac{\partial \psi}{\partial x}=\frac{ip}{\hbar}\psi$.

What I don't understand is how this partial deriving works. What I can figure out is that given the (1) equation, you derive the $i(px)/\hbar$ and $i(-Et)/\hbar$ parts differently, one by one not together. Is that it? So you basically remove the e (2.71)(this must be the way to derive from exponential equations, I don't remember clearly though) and derive the power...partially?

Alas, I am confused.. Any understandable link or advice is appreciated!

i have no idea what you said but

if

$$\psi(x,t)=e^\frac{i(px-Et)}{\hbar}$$

then you compute the partial derivatives by taking regular derivatives while considering the other variables as constants. hence what the second and third equation say is that

$$\frac{\partial}{\partial t}\psi(x,t) = \frac{-iE}{\hbar}e^\frac{i(px-Et)}{\hbar}=\frac{-iE}{\hbar}\psi(x,t)$$

this means $\psi(x,t)$ is an eigenfunction of the $\frac{\partial}{\partial t}$ operator with eigenvalue $\frac{-iE}{\hbar}$.