Verifying Equality: \mathcal{Im}[A+B+Te^{2ip}]=0

[AtoZ]

I have an expression
$\mathcal{Im}[RT^*e^{-2ip}]=|T|^2\sin p$, where $R=Ae^{ip}+Be^{-ip}$ and $p$ is a real number.

This ultimately should lead to $\mathcal{Im}[A+B+Te^{2ip}]=0$ upto a sign (perhaps if I didn't do a mistake).
There is a condition on $R$ that it is real, i.e., $R^*=R$, but $A$ and $B$ are not in general real. Further, $T$ depends on $A$ and $B$ in such a way that if $A=0$, $B=0$ then $T=0$, and $A\neq B$ so the (desired) equality holds. Here is what I do to achieve the desired result:

$\mathcal{Im}[(Ae^{ip}+Be^{-ip})T^*e^{-2ip}-|T|^2e^{ip}]=0$

Then I take common $T^*e^{-ip}$ from the above expression and it leads me to
$\mathcal{Im}[\{(Ae^{2ip}+B)e^{-ip}-Te^{2ip}\}]=0$

This leads to $\mathcal{Im}[Ae^{ip}+Be^{-ip}-Te^{2ip}]=0$

This I rewrite as (since $R=Ae^{ip}+Be^{-ip}$ is real)

$\mathcal{Im}[A+B-Te^{2ip}]=0$, This is the result which is correct upto a sign.

I want to know whether I made a mistake? or there is a mistake in what I want to achieve (regarding the plus sign in front of $T$ expression in the desired versus achieved)? Thanks.

 

[PeroK]

https://www.physicsforums.com/help/latexhelp/

 

[AtoZ]

https://www.physicsforums.com/help/latexhelp/

Thanks, I updated the post with Latex now.

 

[mfb]

$\mathcal{Im}[(Ae^{ip}+Be^{-ip})T^*e^{-2ip}-|T|^2e^{ip}]=0$

Then I take common $T^*e^{-ip}$ from the above expression and it leads me to
$\mathcal{Im}[\{(Ae^{2ip}+B)e^{-ip}-Te^{2ip}\}]=0$

Something went wrong with the exponentials here. In addition you can't just divide by a complex number, the imaginary part might change. As a simple example, $Im(1)=0$ but $Im(\frac{1}{i}) \neq 0$.

This leads to $\mathcal{Im}[Ae^{ip}+Be^{-ip}-Te^{2ip}]=0$

This I rewrite as (since $R=Ae^{ip}+Be^{-ip}$ is real)

$\mathcal{Im}[A+B-Te^{2ip}]=0$

That doesn't look like a correct mathematical operation. If $R=Ae^{ip}+Be^{-ip}$ is real and you take the imaginary part of your expression then you should just remove it from the sum, without leaving in A and B.

 

[AtoZ]

Something went wrong with the exponentials here. In addition you can't just divide by a complex number, the imaginary part might change. As a simple example, $Im(1)=0$ but $Im(\frac{1}{i}) \neq 0$.
That doesn't look like a correct mathematical operation. If $R=Ae^{ip}+Be^{-ip}$ is real and you take the imaginary part of your expression then you should just remove it from the sum, without leaving in A and B.

@mfb Thanks. I will recheck the first part of your answer. Regarding the second part I want to clarify that even if $R$ is real, but $A$ and $B$ need not be real, so I kept $A$ and $B$ inside the $\mathcal{Im}[..]$ is it not allowed then?

 

[mfb]

Regarding the second part I want to clarify that even if $R$ is real, but $A$ and $B$ need not be real, so I kept $A$ and $B$ inside the $\mathcal{Im}[..]$ is it not allowed then?

You can't just randomly decide to "keep something in" in some modified version. As an example, consider $A=B=i$, $p=\frac \pi 2$. Then $R=Ae^{ip}+Be^{-ip} = i^2-i^2 = 0$ but $\mathcal{Im}(A+B)=2$.

 

[AtoZ]

You can't just randomly decide to "keep something in" in some modified version. As an example, consider $A=B=i$, $p=\frac \pi 2$. Then $R=Ae^{ip}+Be^{-ip} = i^2-i^2 = 0$ but $\mathcal{Im}(A+B)=2$.

@mfb Thank you. In a physics perspective, if $p$ is a phase, can we choose an overall phase so that $R$ becomes real? Consequently yielding the required equality?

 

[mfb]

I thought p was some unknown constant. Since when can we choose it? For given A,B there will always be value of p where R is real. For p=0 and p=pi you get opposite imaginary parts and the imaginary part is continuous in p, therefore there must be a zero crossing.

 

[AtoZ]

I thought p was some unknown constant. Since when can we choose it? For given A,B there will always be value of p where R is real. For p=0 and p=pi you get opposite imaginary parts and the imaginary part is continuous in p, therefore there must be a zero crossing.

Okay let me reproduce here, which I got.

Since $Im[z]=\frac{z-\bar{z}}{2i}$ (definition), our $z$ is defined as $z=RT^*e^{-2ip}$ since $R$ is real, so we can write
$\frac{R(T^*e^{-2ip}-Te^{2ip})}{2i}-|T|^2sinp=0$
since $T$ is in general complex, we can replace $T=re^{i\phi}$, with arbitrary $\phi$ and that $|T|=r$
This leads to
$-\frac{R|T|\left[e^{i(\phi+2p)}-e^{-i(\phi+2p)}\right]}{2i}-|T|^2sinp=0$ becase $r=|T|$,
$Rsin(\phi+2p)+|T|sinp=0$
which can we rewrite as?
$Im[Re^{i(\phi+2p)}+Te^{ip}]=0$, now here the assumption of overall phase could matter that "we choose an overall phase so that $R$ is real, and that leads to the expression which is desired i.e., $Im[A+B+Te^{2ip}]=0$.

One more thing: We can make $R$ real only in the following case
if we make the replacement $A=c_1e^{-ip}$ and $B=c_2e^{ip}$ then $R$ is real. which essentially means that $R=c_1+c_2$, but we can rename $c_1$ and $c_2$ as $A$ and $B$ later to match the result with the required equality.
So choosing and overall phase so that $R$ is real, makes sense with this argument? My intuition regarding these phase and stuff is bad, sorry.

I suppose I am quite close but still not exactly there.

 

[mfb]

which can we rewrite as?

How?

If R is real, then $Im[R+Te^{ip}]=Im[Te^{ip}]$.

 

[AtoZ]

How?

If R is real, then $Im[R+Te^{ip}]=Im[Te^{ip}]$.

because there is a $sin(\phi+2p)$ being multiplied by $R$. oh wait. am I correct there? yes you are correct then there should be $Re^{i(\phi+2p)}$ not just $R$

 

[mfb]

That is not an explanation how you got the following expression. Can you break it down step by step?

 

[AtoZ]

That is not an explanation how you got the following expression. Can you break it down step by step?

Okay, I have modified my above answer with more steps. Can you please see whether I made a mistake?

 

[AtoZ]

That is not an explanation how you got the following expression. Can you break it down step by step?

Essentially now the question boils down to whether
$Im[Ae^{i(\phi+2p)}+Be^{i(\phi+2p)}+Te^{ip}]=0$ is equal to $Im[A+B+Te^{2ip}]=0$ or not, if equal, then what should be the condition on $\phi$ or $p$ I think.

 

[mfb]

They are not equal in general, see my example a few posts ago.

They are identical if ϕ+2p is a multiple of 2pi (trivial). They can be identical for other values, that depends on A and B.

Okay, I have modified my above answer with more steps.

I don't see more steps at the point I asked about.

Where does all that come from? Who gave you the final expression you want to get? I feel there is some context missing, because in the way you describe it the answer you want to get is simply wrong.