Question

Invert these relations

\[\left|S_{n+}\right\rangle=\cos{\frac{\theta}{2}}\left|S_{z+}\right\rangle+\sin{\frac{\theta}{2}}\left|S_{z-}\right\rangle\]

\[\left|S_{n-}\right\rangle=\sin{\frac{\theta}{2}}\left|S_{z+}\right\rangle-\cos{\frac{\theta}{2}}\left|S_{z-}\right\rangle\]

I need to find \(\left|S_{z+}\right\rangle\) and \(\left|S_{z-}\right\rangle\). I know this should be really simple but it is throwing me off.

Answer 1

Combine both and/or use the multiple angle formula.

Answer 2

ok so I tried to combine both, but I still get lost.

\[
\left|S_{n+}\right\rangle+\left|S_{n-}\right\rangle=
\cos{\frac{\theta}{2}}\left|S_{z+}\right\rangle
+\sin{\frac{\theta}{2}}\left|S_{z-}\right\rangle
+\sin{\frac{\theta}{2}}\left|S_{z+}\right\rangle
-\cos{\frac{\theta}{2}}\left|S_{z-}\right\rangle
\]
\[
\cos{\frac{\theta}{2}}\left|S_{z+}\right\rangle
+\sin{\frac{\theta}{2}}\left|S_{z+}\right\rangle
=\cos{\frac{\theta}{2}}\left|S_{z-}\right\rangle
-\sin{\frac{\theta}{2}}\left|S_{z-}\right\rangle
-\left|S_{n+}\right\rangle-\left|S_{n-}\right\rangle

\]

Answer 3

\[\left(\cos{\frac{\theta}{2}}
+\sin{\frac{\theta}{2}}\right)\left|S_{z+}\right\rangle
=\left(\cos{\frac{\theta}{2}}
-\sin{\frac{\theta}{2}}\right)\left|S_{z-}\right\rangle
-\left|S_{n+}\right\rangle-\left|S_{n-}\right\rangle\]

Answer 4

yeah I am super lost. here. I could divide both sides by \((\cos\frac{\theta}{2}+\sin{\frac{\theta}{2}})\) but it seems that will get me nowhere.

Answer 5

http://www.mathsrevision.net/double-angle-formulae

Answer 6

I'm a little too busy to help now. How good is your trig?

Answer 7

basically my solutions he is just flipping the eigenkets. I just would like to see how this is done.

Answer 8

Is this Quantum or just math?

Answer 9

he does it again here.

Answer 10

it's quantum mechanics, but it's the math that I don't understand. my trig is OK. Like I could follow any math using trig identities. sometimes I have trouble "Seeing" which identities to use. I can derive them all though.

Answer 11

Well, if it's Quantum, you'll need more information about the kets.

Answer 12

what information is that?

Answer 13

Hold on, post the full question.

Answer 14

ok. I can figure out the answer without ever "inverting the relations". but in my solutions he did that and I would like to know how he did it.

Here is the question

Answer 15

and here is the solution I was given, with the part I am unclear of highlighted. This same method comes up a lot and I do not understand how he can just "switch" the eigenkets like that.

Answer 16

also the "results from question 2" was just writing \(\left|S_{n+}\right\rangle\) and \(\left|S_{n-}\right\rangle\) in the basis of \(\left|S_{z+}\right\rangle\) and \(\left|S_{z-}\right\rangle\) which I posted initially.

Answer 17

Can I come back to this later? I'm busy now.
Also try posting it in the Physics section. And BUMP it now if you can.

Answer 18

ok thanks a lot. I would really like to know this but for now I can just assume it's always gonna be true on my test! thanks for considering it though!

Answer 19

There's no quantum mechanics here.

Answer 20

There's no trigonometry needed for this -- multiply the first equation by cos(theta / 2) and the second by sin(theta / 2). Adding them together will give you your expression for z+.

Then multiply the first equation by sin(theta / 2) and the second by cos(theta / 2). Subtracting them gives you the expression for z-

Answer 21

ah. clever. thanks!