Understanding Vector Components in Quantum Mechanics

phrygian
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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?
 
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phrygian said:
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.
 
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?
 
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?
 
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.
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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