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Physics
Quantum Physics
Griffiths: Intro To Quantum Example 4.3 clarification , <Sx> =
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[QUOTE="Sparky_, post: 6166934, member: 69836"] 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_ [/QUOTE]
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Quantum Physics
Griffiths: Intro To Quantum Example 4.3 clarification , <Sx> =
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