Understanding Vector Components in Quantum Mechanics

[phrygian]

Homework Statement

I read how to solve a problem I am working on, and part of it deals with adding vector components. A is the vector, Ax is the x component, Ay the y component, and theta is the angle A makes from the y axis.

Homework Equations

The Attempt at a Solution

The solution involves using Ax Sin(theta) + Ay Cos(theta) = A.

I know it seems easy but I can't seem to figure out why this would be true?

 

[╔(σ_σ)╝]

The solution involves using Ax Sin(theta) + Ay Cos(theta) = A.

I know it seems easy but I can't seem to figure out why this would be true?

This actually doesn't make sense if A is a scalar.

When you add vectors you get vectors not scalars.

If you were referring to $$\left\|A_x Sin(\theta) + A_y Cos(\theta) \right\| =A$$ then that would make more sense.

 

[phrygian]

What I mean was Ax Sin(theta) + Ay Cos(theta) = A where A, Ax and Ay are vectors, I don't see where this relation comes from?

 

[╔(σ_σ)╝]

In what direction is A,if it is a vector ? The only way this would make sense is if A has the direction of a_rho in cylindrical coordinates.

So I have to ask... in what direction is the unit vector of A?

 

[HallsofIvy]

Actually, your original question doesn't make sense if A is a vector- for exactly the opposite reason! If Ax and Ay are the x and y components of vector A, then "Ax cos(theta)+ Ay sin(theta)" is a scalar and cannot be equal to the vector A.

Assuming that this is in two dimensions, and vector A makes angle theta with the x-axis, then what is true is that Ax= |A|cos(theta) and Ay= |A| sin(theta) where |A| is the length of the vector A. You could also write that as "|A|cos(theta) i+ |A|sin(theta) j= A" where i and j are the unit vectors in the directions of the x and y axes respectively.

 

[phrygian]

It's problem 4.50 in griffiths quantum mechanics and here is a quote from the solution manual:

We may as well choose axes so that a lies along the z axis and b is in the xz plane. Then S(1)a= S(1)z , and S(2)b = cosθ S(2)z + sinθ S(2)x .

S(1)a means the spin operator of particle 1 in the direction a.