Griffiths: Intro To Quantum Example 4.3 clarification , <Sx> =

[Sparky_]

TL;DR Summary

Clarification on Example 4.3 Griffiths Larmor Frequency Example

Hello,

In Griffiths Intro To Quantum (second edition) example 4.3 page 180 ...

Calculating the expectation of <Sx>, equation 4.164.

I'm sure I am wrong, but it seems like after using Euler's for the e^i terms there should be 2cos(yB0t)'s terms. I agree the sin's cancel (After using Euler) and I agree with the sin(alpha) term - after trig identity

my question: Shouldn't there be "2cos(yB0t)" ... I don't see why not? adding the 2 cos() terms after using Euler

the expectation of <Sy> and <Sz> agrees in form to <Sx> so I know I am missing something.

Help

Thanks
-Sparky_

 

[kuruman]

In Equation 4.164, setting aside the factor of $\hbar/2$, you have the following
$$
\begin{pmatrix} a & b\end{pmatrix} \begin{pmatrix} 0 &1\\1 & 0 \end{pmatrix} \begin{pmatrix}a^* \\b^* \end{pmatrix}
$$where $a=\cos (\alpha/2)e^{i\gamma B_0t/2}$ and $b=\sin (\alpha/2)e^{-i\gamma B_0t/2}$.
What do you get when you multiply this out?

 

[Sparky_]

I get

$$
\begin{pmatrix} \cos (\alpha/2)e^{-i\gamma B_0t/2} & \sin (\alpha/2)e^{+i\gamma B_0t/2}\end{pmatrix} \begin{pmatrix} 0 &1\\1 & 0 \end{pmatrix} \begin{pmatrix}\cos (\alpha/2)e^{+i\gamma B_0t/2} \\\sin (\alpha/2)e^{-i\gamma B_0t/2}\end{pmatrix}
$$

$$
\begin{pmatrix} \cos (\alpha/2)e^{-i\gamma B_0t/2} & \sin (\alpha/2)e^{+i\gamma B_0t/2}\end{pmatrix} \begin{pmatrix}\sin (\alpha/2)e^{-i\gamma B_0t/2} \\\cos (\alpha/2)e^{+i\gamma B_0t/2}\end{pmatrix}
$$

$$
\cos (\alpha/2)e^{-i\gamma B_0t/2} \sin (\alpha/2)e^{-i\gamma B_0t/2} + \sin (\alpha/2)e^{+i\gamma B_0t/2}\cos (\alpha/2)e^{+i\gamma B_0t/2}
$$

$$
\cos (\alpha/2) \sin (\alpha/2)e^{-i\gamma B_0t} + \sin (\alpha/2)\cos (\alpha/2)e^{+i\gamma B_0t}
$$

$$\frac{\sin (\alpha)}{2} e^{-i\gamma B_0t} + \frac{\sin (\alpha)}{2}e^{+i\gamma B_0t}
$$

It is here that I found my mistake ... in my notes I just dropped the 2 in the denominator of the sin's (out of the blue) expanding the exponential terms using Euler's the sin terms within that will cancel and I do get 2 cosine terms but the "2" on "2cos () will cancel with the 2 in the denominator that is lost somewhere

On my scratch pad notes, it looks as if I might have confused the h/2 as factoring out the 2 in the denominator of the sin's - but who knows, methodically transcribing it to this post caught my stupid mistake.

(unless you see a problem ... I am now getting what the book is getting)
Thanks
Sparky_

 

[kuruman]

I see no problem. I just want to point out that you can avoid and/or troubleshoot mistakes of this kind if you use symbols for complex expressions and then substitute at the end. For example, using post #2
$$\begin{pmatrix} a & b\end{pmatrix} \begin{pmatrix} 0 &1\\1 & 0 \end{pmatrix} \begin{pmatrix}a^* \\b^* \end{pmatrix}
=\begin{pmatrix} a & b\end{pmatrix}\begin{pmatrix}b^* \\a^* \end{pmatrix}=ab^*+a^*b=2 Re(ab^*)\\=2\cos(\alpha/2)\sin(\alpha/2)\cos(\gamma B_0 t).$$