# What are the backwards 6s in the Schrödinger Equation?

Discussion in 'Physics & Math' started by Neugierig, Oct 22, 2013.

1. ### NeugierigRegistered Member

Messages:
21
en.wikipedia.org/wiki/Schrödinger_equation

It doesn't tell you what the backwards 6s mean!

3. ### SorcererPut a Spell on youRegistered Senior Member

Messages:
856
Why don't you post the link properly? It's a small letter d.

5. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
It's a "d"; it stands for "dead cat"

7. ### SorcererPut a Spell on youRegistered Senior Member

Messages:
856
Hey, you don't know that the cat is dead! Oh wait, you didn't peek did you?

8. ### rpennerFully WiredValued Senior Member

Messages:
4,833
In $i \hbar \frac{\partial \quad}{\partial t} \Psi = \hat{H} \Psi$,
$i = \sqrt{-1}$, $\hbar = \frac{h}{2 \pi}$ is the reduced Planck's constant, $\frac{\partial \quad}{\partial t}$ is the partial derivative operator, $\hat{H}$ is an operator( the definition of which along with boundary conditions gives you what you need to solve for $\Psi$) and $\Psi = \Psi( t, \vec{x} )$ is a complex function of both time and position.

Example:
$\Psi = \Psi( t, x, y, z ) \hat{H} = \frac{1}{2m} \hat{p} ^2 = \frac{1}{2m} ( -i \hbar \nabla )^2 = \frac{- \hbar^2}{2m} \left( \vec{e}_{x} \frac{\partial \quad}{\partial x} + \vec{e}_{y} \frac{\partial \quad}{\partial y} + \vec{e}_{z} \frac{\partial \quad}{\partial z} \right)^2 = \frac{- \hbar^2}{2m} \left( \frac{\partial^2 \quad}{\partial x^2} + \frac{\partial^2 \quad}{\partial y^2} + \frac{\partial^2 \quad}{\partial z^2} \right) = \frac{- \hbar^2}{2m} \nabla^2$

Then we can solve for the ansatz $\Psi( t, x, y, z ) = A e^{i \left(k_x x + k_y y + k_z z - \omega t \right)}$
In which case we can rapidly evaluate the differential operators
because $\frac{\partial \quad}{\partial u} e^{au+b} = u e^{a u +b}$ (// Edit : $\frac{\partial \quad}{\partial u} e^{au+b} = a e^{a u +b}$)
so we have:
$\frac{\partial \quad}{\partial t} \Psi = \frac{\partial \quad}{\partial t} A e^{i \left(k_x x + k_y y + k_z z - \omega t\right)} = - i \omega A e^{i \left(k_x x + k_y y + k_z z - \omega t \right)} = - i \omega \Psi( t, x, y, z )$
and
$\nabla^2 \Psi = i^2 k_x^2 \Psi + i^2 k_y^2 \Psi + i^2 k_z^2 \Psi = - ( k_x^2 + k_y^2 + k_z^2 ) \Psi$
which means the original equation: $i \hbar \frac{\partial \quad}{\partial t} \Psi = \hat{H} \Psi$ is a solution only if
$i \hbar ( - i \omega ) \Psi = \frac{- \hbar^2}{2m} \left( - ( k_x^2 + k_y^2 + k_z^2 ) \right) \Psi$
or $\hbar \omega = \frac { \left( \hbar ( k_x \vec{e}_{x} + k_y \vec{e}_{y} + k_z \vec{e}_{z} ) \right)^2 }{ 2 m}$
or with $\vec{k} = k_x \vec{e}_{x} + k_y \vec{e}_{y} + k_z \vec{e}_{z}$ we have:
$\omega = \frac{ \hbar \vec{k}^2 }{2 m}$
$\omega$ is an angular frequency and $\vec{k}$ is a direction vector with units of inverse length.

The describe a wave with cyclical frequency of $f = \frac{\omega}{2 \pi}$ and wavelength $\lambda = \frac{2 \pi}{| \vec{k} |}$ with well-defined kinetic energy $E = \hbar \omega = h f$ and momentum $\vec{p} = \hbar \vec{k}, \quad | \vec{p} | = \hbar | \vec{k} | = \frac{h}{\lambda}$ but without a well-defined position.

Last edited: Jun 11, 2016
9. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
There ya go, OP. Now you know what the backward sixes are...

10. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
Partial derivative

11. ### iceauraValued Senior Member

Messages:
30,246
Is this OK? (Looks like it should be a in the derivative, chain rule.)

(In case the guy is actually taking advantage of the favor, and reading closely)

12. ### rpennerFully WiredValued Senior Member

Messages:
4,833
You are correct to be suspicious. I was flip-flopping between writing $\frac{\partial \quad}{\partial a}$ and $\frac{\partial \quad}{\partial u}$ so I should have settled on either: $\frac{\partial \quad}{\partial u} e^{au+b} = a e^{a u +b}$ or $\frac{\partial \quad}{\partial a} e^{au+b} = u e^{a u +b}$ but not the mishmash that resulted.

But no fear. It appears Neugierig has started a new thread with a misunderstanding of quantum physics rather than trying to engage on this thread.

13. ### eramSciengineerValued Senior Member

Messages:
1,877
I'm trying to visualize a single diagram that explains the partial derivatives of z(x,y) when x and y are unrelated and when x and y are both related.

14. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Isn't the concept that you are trying to illustrate the analogue of a directional derivative.
$\nabla f(x,y) = \hat{e}_x \frac{\partial f(x,y)}{\partial x} + \hat{e}_y \frac{\partial f(x,y)}{\partial y} \\ \nabla_{\vec{u}} f(x,y) = \left( \nabla f(x,y) \right) \cdot \frac{ \vec{u} }{ \left| \vec{u} \right| } = \frac{\frac{\partial f(x,y)}{\partial x} u_x + \frac{\partial f(x,y)}{\partial y} u_y }{\sqrt{u_x^2 + u_y^2}}$

While if y is a function of x, you have:
$\frac{d f(x,y)}{dx} = \frac{\partial f(x,y)}{\partial x} + \frac{\partial f(x,y)}{\partial y} y' = \frac{\partial f(x,y)}{\partial x} \frac{dx}{dx} + \frac{\partial f(x,y)}{\partial y} \frac{d y(x)}{dx} = \left( \nabla f(x,y) \right) \cdot \left( 1 \hat{e}_x + y'(x) \hat{e}_y \right)$
and if both x and y are functions of u, you have:
$\frac{d f(x,y)}{du} = \frac{\partial f(x,y)}{\partial x} x'(u) + \frac{\partial f(x,y)}{\partial y} y'(u) = \frac{\partial f(x,y)}{\partial x} \frac{dx(u)}{du} + \frac{\partial f(x,y)}{\partial y} \frac{d y(u)}{du} = \left( \nabla f(x,y) \right) \cdot \left( x'(u) \hat{e}_x + y'(u) \hat{e}_y \right)$
and if the relationship between x and y is of the form $g(x,y) = c$ ( a level curve) then it follows that $\frac{\partial g}{\partial x} dx + \frac{\partial g}{\partial y} dy = 0$ such that $\frac{d y}{dx} = - \frac{\partial g}{\partial x} \frac{\partial y}{\partial g}$ is the derivative of y with respect to x that keeps on the curve $g(x,y) = c$.
So the normalized tangent vector to the level curve is $\hat{t} = \pm \frac{ \hat{e}_x \frac{\partial g}{\partial y} - \hat{e}_y \frac{\partial g}{\partial x} }{ \sqrt{ \left(\frac{\partial g}{\partial x} \right)^2 + \left( \frac{\partial g}{\partial y} \right)^2 }}$

And the directional derivative of $f(x,y)$ to the tangent vector of the level curve $g(x,y) = c$ is:
$\nabla_{\hat{t}} f(x,y) = \left( \nabla f(x,y) \right) \cdot \hat{t} = \pm \frac{ \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y} \frac{\partial g}{\partial x} }{ \sqrt{ \left(\frac{\partial g}{\partial x} \right)^2 + \left( \frac{\partial g}{\partial y} \right)^2 }}$

where the sign depends on the choice of which way you go on the level curve.

So the gradient to a function in N dimensions is a vector, while the tangent vector to a 1-dimensional parametric curve in N dimensions is a vector and the tangent to the level set of a function in N dimensions is a hyperplane of N-1 dimensions orthogonal to the gradient vector of the same function. This, to me, suggests an illustration of a gradient field, a curve through that field, and decomposing the gradient at a point on the field in directions orthogonal and parallel to the tangent to the curve and doing the same for a level curve of the same function.

Last edited: Jun 11, 2016
15. Rpenner... do you have the best interests when it comes to posts like this? I bet only a handful actually know about advanced mathematics, as such demonstrated by your posts. The OP for sure will be lost in all those variables. Not even that, but I bet half of those who congratulate you on such posts, don't actually understand half of it either. Is it really a matter of showing off your math talents? I dare say it comes across as that, since half posters here post nothing close to your quality of posts.

16. ### YazataValued Senior Member

Messages:
5,364
The backwards 6's are partial derivatives.

The derivative is a basic concept in calculus.

If the value of x is dependent on the value of y, mathematicians say that x is a function of y. The derivative is a measure of how much x will change if we change y a little bit. It's basically just the slope of the graph of the function at a point.

The values of some functions might be dependent on more than one variable. Partial derivatives address how much each of the variables is contributing to the change of the value of the function.

So if you see little backwards 6's in an equation, it's basically just telling you that the value of some quantity is dependent on more than one variable. The guts of the equation specify the details of that dependence.

17. Yes just tell him how it is. We don't need the likes of Rpenner seriously going over-board.

The ''backwward'' sixes are as many and elaborately told you, is the process of partial derivatives.

Just as much $\frac{dE}{dt}$ tells you, we have a derivative of an energy to a time (an energy flow), are in fact derivatives of the function $E$, we can have partial derivatives, which may have several derivatives based to one of those variables.

This statement, among many which may seem daunting at first but can be easily understood. The value in which you are talking about is the partial derivative in question:

$\partial$

What makes this variable different, is that they can form differences in all the variables given, if you have a normal derivative, the normal procedure is to think of some of these numbers as constants. Of course, partial differentiation promises different reviews.

18. eram asked for an explanation on the partial derivative, you realize that you explained the total derivative, don't you?

19. ### rpennerFully WiredValued Senior Member

Messages:
4,833
That's not how I parsed it. The gradient $\nabla f$ is a vector with partial derivatives as components, and that seems to both satisfy eram's request to "diagram" the partial derivatives when x and y and unrelated. But that was only half of eram's request.

In the case when x & y are related, I thought of three types of relations, y is a function of x, both x and y are functions of u (parametric curve) and x & y are fixed by an equational relationship, thus they are on some sort of level curve. In each of these cases the derivative is some sort of dot product with the gradient.

I also discussed the directional derivative, which just that sort of dot product where the directional vector is normalized to unit length. Since the gradient can be decomposed into directional derivatives in orthogonal directions, I think this is a visual way to relate the partial derivatives (which are the directional derivatives in the X and Y directions) to the directional derivatives tangent to and orthogonal to a particular curve.

20. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Someone who doesn't recognize the language the Schrödinger equation is written in needs more help than explaining what "the backward sixes are." In this case I was the first on the thread to identify them as a partial derivative operator which operates on a function of space and time. The OP was welcome to ask additional questions. Or he could just google the terms not understood.

The best interests of whom? The OP asked a question, perhaps sincerely, so the natural thing to answer the question in sufficient context so that any related misunderstandings would be exposed. Specifically, if you confused H with a number you would be missing the point of the Schrödinger equation.
And yet the forum is labeled "Physics & Math" which implies an expectation that questions and answers will have a degree of physics and/or math. Quantum physics is intrinsically mathy.
And yet, if the OP -- the person who expressed interesting in learning the meaning of the Schrödinger equation -- wants to understand the equation, understanding all the pieces is a prerequisite.
My posts are lauded and even you say they are of high quality, and that is a problem, according to you? What qualifications do you have to judge the math talents of posters, when in [post=3120883]this post[/post] you fail to recognize that numerator and denominator are identical? Why beholdest thou the mote that is in thy brother's eye, but considerest not the beam that is in thine own eye?

Why is Neugierig the only poster who matters and why are you Neugierig's special advocate?
Who are you to judge? Do you know Neugierig's age, education and goals for this thread? All I have are the contents of the OP.
"You" appears to refer to Neugierig while the first use of the term "partial derivative" in this thread is post #5, isn't it?

The rest of Trapped's post seems like an unhelpful "clarification" while Yazata's post is actually useful at the conceptional level. But at least as useful is Pete's link to Wikipedia, http://en.wikipedia.org/wiki/Partial_derivative, where both the concept and mathematical details of partial derivatives are described.

Messages:
12,738
Heh Heh.