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

• I
• Sparky_
In summary, the conversation discusses calculating the expectation of <Sx>, equation 4.164, in Griffiths Intro to Quantum (second edition). The conversation includes a question about the presence of "2cos(yB0t)" terms after using Euler's for the e^i terms, and the conclusion is that the mistake was due to a missing factor of 2 in the denominator of the sin terms. The conversation also mentions that using symbols for complex expressions and substituting at the end can help avoid or troubleshoot mistakes.
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_

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?

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_

Last edited:
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
$$\begin{pmatrix} a & b\end{pmatrix} \begin{pmatrix} 0 &1\\1 & 0 \end{pmatrix} \begin{pmatrix}a^* \\b^* \end{pmatrix} =\begin{pmatrix} a & b\end{pmatrix}\begin{pmatrix}b^* \\a^* \end{pmatrix}=ab^*+a^*b=2 Re(ab^*)\\=2\cos(\alpha/2)\sin(\alpha/2)\cos(\gamma B_0 t).$$

## 1. What is the purpose of Example 4.3 in Griffiths' "Intro To Quantum" textbook?

The purpose of Example 4.3 is to illustrate the concept of expectation values in quantum mechanics. It uses the specific example of measuring the x-component of the spin of a particle to demonstrate how the expectation value is calculated.

## 2. How does Example 4.3 clarify the concept of expectation values?

Example 4.3 clarifies the concept of expectation values by providing a step-by-step calculation of the expectation value for the x-component of spin. It also explains the significance of the expectation value and how it relates to the actual measurements of the particle's spin.

## 3. What is the significance of the notation in Example 4.3?

The notation represents the expectation value of the x-component of spin. It is a mathematical representation of the average value that would be obtained if multiple measurements of the spin were made on identical particles.

## 4. How is the expectation value calculated in Example 4.3?

In Example 4.3, the expectation value is calculated by taking the inner product of the state vector with the operator for the x-component of spin. This is followed by multiplying the result by the complex conjugate of the state vector and integrating over all possible values of the spin.

## 5. What does Example 4.3 tell us about the measurement of spin in quantum mechanics?

Example 4.3 tells us that the measurement of spin in quantum mechanics is probabilistic. The expectation value gives us the most probable value for the spin measurement, but the actual measurement may vary each time it is performed. This is due to the inherent uncertainty in quantum systems.

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