Schrodingers Equation in 3 space Dimension

  • Context: Graduate 
  • Thread starter Thread starter hell18
  • Start date Start date
  • Tags Tags
    Dimension Space
Click For Summary

Discussion Overview

The discussion revolves around the interpretation of Schrödinger's equation in the context of three-dimensional space and the implications of adding a time dimension. Participants explore the nature of vectors in three-dimensional space and the conceptual challenges associated with visualizing higher dimensions.

Discussion Character

  • Exploratory
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the meaning of a "vector of x" and emphasize that a position vector must have three components, typically labeled x, y, z.
  • There is a suggestion that the term "vector of x" may have been misused, and a more appropriate notation would be to denote a vector as x = (x1, x2, x3).
  • One participant expresses confusion about the dimensionality of space and time, noting that while space is three-dimensional, time is considered one-dimensional, leading to the concept of four-dimensional spacetime.
  • Another participant asserts that a vector has both magnitude and direction, providing an explanation of how vectors are represented in three-dimensional space and how they relate to the origin.
  • There is a mention that while the discussion involves three dimensions, the implications of relativity introduce a fourth dimension, which may complicate the understanding of the Schrödinger equation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of vectors and the dimensionality of space and time. There is no consensus on the clarity of the initial statements regarding vectors, and confusion remains about the visualization of four-dimensional spacetime.

Contextual Notes

Some participants indicate that further study of geometric concepts may be necessary to fully grasp the discussion, particularly in relation to the Schrödinger equation and the nature of vectors.

hell18
Messages
21
Reaction score
0
i read a book on quantium theory, it states that a vector of x = x,y,z. When i think about 3 dimensional space, we cannot see that, is that correct?
 
Astronomy news on Phys.org
We live in 3d space, addition of a time coordinate makes it 4d.

I am not sure what you are asking.
 
A large part of our confusion is "vector of x = x,y,z". What do YOU mean by a "vector of x"?

Certainly it is true that a "position vector" must have 3 components: typically labeled x, y, z: that's what "3 dimensional" MEANS. It would not be very good practice to give a vector the same name as one of its components!
 
Maybe he meant to bold the x for the vector? Still not goood usage, since one expects x = (x1,x2,x3).
 
i thought a vector was a small section of the whole. but i don't think that is the case. In the book it says a 3 dimensional space vector of x = x,y,z. So space is 3d? so time is a 1 dimensional element? add those together get 4d? but i thought we cannot see the 4th dimension? or am i missing something out?
 
I think you need to go back and study some precalculus. You need to spec up on some geometric concepts before tackling the Schroedinger equation.

A vector has two qualities, magnitude and direction. (Bear with me vector space fans, sufficient unto the day is the rigor thereof). For example a force has its magnitude (so many Newtons) and the direction in which it is applied. If you set up coordinates with an origin, any point is determined by a vector whose magnitude is the distance from the origin and whose direction is the direction from the origin to the point. This particular vector is called the radius vector.

Vectors have components. If you set up x y and z axes at right angles to each other, then a given radius vector will have projections on those axes and the length of the projections will give the components. So if the vector v has compnents a, b, and c we write v = (a,b,c).

This works for three dimensions, but relativity requires four, and that is another story entirely. Don't worry about it yet; the Schroedinger equation at the beginning level is not relativistic.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
20
Views
3K
  • · Replies 61 ·
3
Replies
61
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K