Confused about some notation used by Griffiths

[kmm]

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 I am 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 I am 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.

 

[fresh_42]

The factor $2$ turned $\dfrac{\hbar}{2}$ into $\hbar$.

 

[hutchphd]

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

 

[kmm]

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.

 

[fresh_42]

Well, you add/subtract basically $a+b$ and $a-b$, and then $2$ of the same are left.

 

[PeroK]

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 I am 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 I am 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?

 

[kmm]

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.

 

[kmm]

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!

 

[vanhees71]

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 $z \in \mathbb{C}$?

 

[PeroK]

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 $b^* = bi$.