Schrodingers Equation in 3 space Dimension

[hell18]

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?

 

[Integral]

We live in 3d space, addition of a time coordinate makes it 4d.

I am not sure what you are asking.

 

[HallsofIvy]

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!

 

[selfAdjoint]

Maybe he meant to bold the x for the vector? Still not goood usage, since one expects x = (x₁,x₂,x₃).

 

[hell18]

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?

 

[selfAdjoint]

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.