Transformation to a different base

[Krischi]

1. A transformation from base S to base T. The three combinations
C(x)=(-1/sqrt(2))(C(+) - C(-)), C(y)=(-i/sqrt(2))(C(+) - C(-)) and C(z)=C(0)
transform to C'(x), C'(y) and C'(z) just the way that x,y,z transform to x',y' and z'. You can check that this is so by using the transformation laws (see below).

Homework Equations

From the point of view of S the state Φ (can be +S, 0S and -S) is described by the three numbers:
C(+)=〈+S ⎢ Φ〉, C(0)=〈0S ⎢ Φ〉and C(-)=〈-S ⎢ Φ〉.
From the point of view of T the state Φis described by the three numbers:
C'(+)=〈+T ⎢ Φ〉, C'(0)=〈0T ⎢ Φ〉and C'(-)=〈-T ⎢ Φ〉.
Transformation laws 〈+T ⎢ +S〉= exp(+ib), 〈0T ⎢ 0S〉=1 and 〈-T ⎢ -S〉=exp(-ib), all other possible amplitudes are equal to zero.

The Attempt at a Solution

I tried the following: C'(x)=(-1/sqrt(2))(C'(+) - C'(-))
with C'(+)=〈+T ⎢ Φ〉=〈+T ⎢ +S〉= exp(+ib) and C'(-)=〈-T ⎢ Φ〉=〈-T ⎢ +S〉=0.
It follows: C'(x)=(-1/sqrt(2))(exp(+ib))

Am I on the right track and if so what does the upper result tell me?

Thanks in advance!
Krischi

 

[Jhero]

Hello Krischi,

Yes, you are on the right track! Your result shows that the transformation from base S to base T for the combination C(x) is just like the transformation from x to x'. This means that the transformation laws are consistent and valid for both the physical quantities (x, y, z) and the abstract quantities (C(+), C(0), C(-)).

In other words, the transformation laws hold true for both the position and the state of the system, which is a fundamental concept in quantum mechanics. This result also confirms that the transformation from base S to base T is a valid and accurate representation of the system.

I hope this helps clarify your understanding. Keep up the good work in your studies!