Confused about some notation used by Griffiths

  • #1
kmm
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I worked out the expectation values of the components of a 1/2 spin particle. However, I'm confused about Griffiths notation for the x and y components.

For the x component I got, ## \left< S_x \right> = \frac \hbar 2 (b^*a+a^*b)## which is correct, but Griffiths equates this to ## \hbar~Re(ab^*) ##.

For the y component I got, ## \left< S_y \right> = \frac \hbar 2 i (ab^*-a^*b)##, and Griffiths equates this to ## - \hbar~Im(ab^*)##.

All I know is that Re and Im refer to the real and imaginary parts of a complex number. Since we only have a factor of ##i## for the y component of spin, it makes sense why we use Im here, but I don't understand how ##Re(ab^*) ## or ## Im(ab^*)## absorbed the factor of 1/2. I also don't understand why for both, we have ## ab^*## only inside the indices, when for the x component we have the factor ##(b^*a+a^*b)## and for the y component after factoring out the minus sign, we have the factor ##(a^*b-ab^*)##. Thanks for any help with this.
 

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  • #2
fresh_42
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The factor ##2## turned ##\dfrac{\hbar}{2}## into ##\hbar##.
 
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  • #3
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These relations are true for any complex number......if you add the complex conjugate, the parts with the i in front cancel!! And vice-versa
 
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  • #4
kmm
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OK thanks. Well it looks like I need to just review complex numbers, because the factor of 2 and these relations aren't obvious to me.
 
  • #5
fresh_42
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Well, you add/subtract basically ##a+b## and ##a-b##, and then ##2## of the same are left.
 
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  • #6
PeroK
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I worked out the expectation values of the components of a 1/2 spin particle. However, I'm confused about Griffiths notation for the x and y components.

For the x component I got, ## \left< S_x \right> = \frac \hbar 2 (b^*a+a^*b)## which is correct, but Griffiths equates this to ## \hbar~Re(ab^*) ##.

For the y component I got, ## \left< S_y \right> = \frac \hbar 2 i (ab^*-a^*b)##, and Griffiths equates this to ## - \hbar~Im(ab^*)##.

All I know is that Re and Im refer to the real and imaginary parts of a complex number. Since we only have a factor of ##i## for the y component of spin, it makes sense why we use Im here, but I don't understand how ##Re(ab^*) ## or ## Im(ab^*)## absorbed the factor of 1/2. I also don't understand why for both, we have ## ab^*## only inside the indices, when for the x component we have the factor ##(b^*a+a^*b)## and for the y component after factoring out the minus sign, we have the factor ##(a^*b-ab^*)##. Thanks for any help with this.
What exactly was stopping you calculating ##Re(ab^*) ## and ##\frac 1 2 (b^*a+a^*b)##?

Even if you didn't immediately recognise this, what stops you checking these are equal?
 
  • #7
kmm
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What exactly was stopping you calculating ##Re(ab^*) ## and ##\frac 1 2 (b^*a+a^*b)##?

Even if you didn't immediately recognise this, what stops you checking these are equal?
I’m not sure what lead you to this assumption that I didn’t try. I did try, and then came here when it was clear I was missing something. The previous comments gave me some clues of what I need to review, so I will be continuing to try.
 
  • #8
kmm
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It turns out my problem was in making an embarrassingly simple mistake. I often have erroneously thought of numbers like ##a## or ##b^*## as merely real numbers or a real number with a factor of ##i## attached, and not like the complex number, ##z=x+iy##. With this in mind I was then able to take ##Re(ab^*)##, and have ##a=w+ix## and ##b^*=y-iz##. Working out ##b^*a+a^*b## and ##ab^*-a^*b##, I was able to recover the correct relations. Thanks for the help everyone!
 
  • #9
vanhees71
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Isn't this simply due to
$$\text{Re} z=\frac{z+z^*}{2}, \quad \text{Im} z=\frac{z-z^*}{2\mathrm{i}}$$
for any ##z \in \mathbb{C}##?
 
  • #10
PeroK
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Isn't this simply due to
$$\text{Re} z=\frac{z+z^*}{2}, \quad \text{Im} z=\frac{z-z^*}{2\mathrm{i}}$$
for any ##z \in \mathbb{C}##?
I think the OP has already admitted that he misunderstood what is meant by ##b^*##. I think he assumed ##a, b## were real and that ##b^* = bi##.
 
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