Schrodinger equation/Madelung equation phase transformation

[binbagsss]

When one does the phase transformation $\psi(\vec{x},t)=R(\vec{x},t)\exp^{iS(\vec{x},t)/\bar{h}}$

For this transformation to be valid doesn't it need to have the same asymptotic behaviour $x \to \pm \infty$, and also at $x=0$ as the original wave-function $\psi$ does? How come this is the never mentioned?


Also on the Wikipedia page, it assumes $R(\vec{x},t) \geq 0$ and both $R,S \in \Re$, how is this possible without loss of generality? Also, obviously the w.f squared needs to be $\geq 0$ but, by assuming $R(\vec{x},t) \geq 0$ the amplitude of both the real and imaginary part of the w.f also now need to be positive. This surely restricts what the wave-function is able to describe, or is this true for a lot of problems?

(Although I have seen other derivations which transform the Lagrangian of the Schrodinger Lagrangian instead by $\psi=\sqrt{rho}\exp{i\theta/\bar{h}}$, s.t. the relevant equations are given by the E-L equations,: variation w.r.t $\theta$ yields the cty, and variation wrt $\rho$ yields the quantum Hamilton-Jacobi. so taking the Real and Imaginary part is not required..)


Thanks.

 

[Jonomyster]

For this transformation to be valid doesn't it need to have the same asymptotic behaviour...

The equation $\psi(\mathbf{x}, t) = R(\mathbf{x}, t) \exp \left( i S(\mathbf{x}, t)/\hbar \right)$ is taken to be true for all $\mathbf{x}$ and $t$, by the very definition of $R$ and $S$. Because they are the same for all $\mathbf x$, then it trivially follows that they are the same for $|\mathbf x| \rightarrow \infty$ and $\mathbf x = 0$. In other words, if two functions have the same value for all inputs, it is unnecessary to mention that their limits are also the same.

how is this possible without loss of generality?

Consider an arbitrary complex number $z$. It is always possible to write $z$ as a nonnegative real number multiplied by an exponential phase factor. That is,
$$ z = A \cdot \exp(i \theta) ,\,\,\, \text{where} \,\,\, A \geq 0 $$Note that in cases where $z$ is a negative real number, we can take $\theta = \pi$, for example $-1 = \exp(i \pi)$. More generally, the real and/or imaginary parts of $z$ can be made negative for the appropriate phase angle $\theta$, so there is no restriction on $z$.

Since the value of $\psi$ at a given $\mathbf x$ and $t$ is always a complex number, we can always do this phase transformation. That is what the equation$\psi(\mathbf{x}, t) = R(\mathbf{x}, t) \exp \left( i S(\mathbf{x}, t)/\hbar \right)$ is expressing. It's just that now the magnitude $R$ and phase $S$ depend on the position and time, because the wavefunction $\psi$ does.

 

[binbagsss]

The equation $\psi(\mathbf{x}, t) = R(\mathbf{x}, t) \exp \left( i S(\mathbf{x}, t)/\hbar \right)$ is taken to be true for all $\mathbf{x}$ and $t$, by the very definition of $R$ and $S$. Because they are the same for all $\mathbf x$, then it trivially follows that they are the same for $|\mathbf x| \rightarrow \infty$ and $\mathbf x = 0$. In other words, if two functions have the same value for all inputs, it is unnecessary to mention that their limits are also the same.


Consider an arbitrary complex number $z$. It is always possible to write $z$ as a nonnegative real number multiplied by an exponential phase factor. That is,
$$ z = A \cdot \exp(i \theta) ,\,\,\, \text{where} \,\,\, A \geq 0 $$Note that in cases where $z$ is a negative real number, we can take $\theta = \pi$, for example $-1 = \exp(i \pi)$. More generally, the real and/or imaginary parts of $z$ can be made negative for the appropriate phase angle $\theta$, so there is no restriction on $z$.

Since the value of $\psi$ at a given $\mathbf x$ and $t$ is always a complex number, we can always do this phase transformation. That is what the equation$\psi(\mathbf{x}, t) = R(\mathbf{x}, t) \exp \left( i S(\mathbf{x}, t)/\hbar \right)$ is expressing. It's just that now the magnitude $R$ and phase $S$ depend on the position and time, because the wavefunction $\psi$ does.

Okay thank you. So if I am looking at the Proca action and similarly want to perform a similar phase transformation of the fields $B$ and $B*$, this is equally "allowed"?

$\ell=-\frac{1}{2}(\partial_{\mu}B_{\nu}*-\partial_{\nu}B_{\mu}*)(\partial_{\mu}B_{\nu}-\partial_{\nu}B_{\mu})+\frac{m^2c^2}{\bar{h}^2}B_{\nu}*B^{\nu}$

 

[Jonomyster]

Yes, in general each of the four components of $B^\mu$ is a complex valued function of position and time, and therefore can be written as a magnitude part multiplied by a phase part.

 

[binbagsss]

This may be a stupid question, but in the Proca action , or anything similar, when one tends to do a polar decomposition if

$B*= e^{i\phi} \implies B=e^{-i\phi}$

but, would it also be valid to write , e.g. :

$B*= e^{i\phi_1}$ and $B=e^{-i\phi}$,

since, in the Lagrangian when varying w.r.t B and B* they are treated independently?

Thanks.

 

[binbagsss]

anyone? thank you.

 

[renormalize]

This may be a stupid question, but in the Proca action , or anything similar, when one tends to do a polar decomposition if
$B*= e^{i\phi} \implies B=e^{-i\phi}$
but, would it also be valid to write , e.g. :
$B*= e^{i\phi_1}$ and $B=e^{-i\phi}$,
since, in the Lagrangian when varying w.r.t B and B* they are treated independently?

Without a reference to look at I can't comment on the polar decomposition of a Proca (vector) field. But for a complex scalar field $B$ where $B^*$ is treated as independent, the polar decomposition would be $B=\left|B\right|e^{i\phi}$ where the real quantities $\left|B\right|$ and $\phi$ are now treated as two independent fields.

 

[binbagsss]

Thank for the reply. No, but what I mean is if someone was write $B$ as above. Then $B*$ tends to be written as its complex conjugate. Is one able to do a polar decomposition using two different scalar fields instead, in the exponent? Thanks. (Of course I understand this is introducing more fields so seems counter-productive..)

 

[renormalize]

OK, you wrote:

$B*= e^{i\phi} \implies B=e^{-i\phi}$
$B*= e^{i\phi_1}$ and $B=e^{-i\phi}$

You stated that $B$ and $B^*$ are complex and are conjugate to each other. But can you please clarify: are $\phi$ and $\phi_1$ each real, imaginary, or complex quantities?

 

[binbagsss]

OK, you wrote:

You stated that $B$ and $B^*$ are complex and are conjugate to each other. But can you please clarify: are $\phi$ and $\phi_1$ each real, imaginary, or complex quantities?

well, in line with the second reply here, I don't believe it matters. so to simplify things one should take them as real, and include a real amplitude for each B and B* as welll... then they can span over all complex numbers.

 

[renormalize]

well, in line with the second reply here, I don't believe it matters. so to simplify things one should take them as real, and include a real amplitude for each B and B* as welll... then they can span over all complex numbers.

Sorry but I'm not following. If I decompose complex $B$ as $B=B_\text{real}+i B_\text{imag}$, where $B_\text{real},B_\text{imag}$ are real, and then write that $B=e^{i\phi}$ for $\phi$ real, am I not declaring (nonsensically) that two arbitrary real quantities can be made equal to the exponential of one arbitrary real quantity?

 

[binbagsss]

how are B_real and B_imag real? they are the same as the schrodigner polar decomposition discussed above. they can span all complex numbers?

 

[renormalize]

how are B_real and B_imag real? they are the same as the schrodigner polar decomposition discussed above. they can span all complex numbers?

Any complex number $z$ can by definition be written as $z=x+i\,y$ where $x$ and $y$ are individually real quantities and $i\equiv\sqrt{-1}$. As such, $z$ defines a point in the 2D complex-plane. But $e^{i\phi}$ with $\phi$ a real number defines a point on the 1D unit-circle $\left|z\right|=1$. You cannot equate a 2D space to a 1D space. How much familiarity with complex analysis do you have?

 

[binbagsss]

Any complex number $z$ can by definition be written as $z=x+i\,y$ where $x$ and $y$ are individually real quantities and $i\equiv\sqrt{-1}$. As such, $z$ defines a point in the 2D complex-plane. But $e^{i\phi}$ with $\phi$ a real number defines a point on the 1D unit-circle $\left|z\right|=1$. You cannot equate a 2D space to a 1D space. How much familiarity with complex analysis do you have?

i said to include amplitudes as well, post 10.

 

[renormalize]

i said to include amplitudes as well, post 10.

If $B=B_\text{real}+i\, B_\text{imag}$ in cartesian form, then $B=\left|B\right|e^{i\varphi}$ in polar form, where the amplitude $\left|B\right|=\sqrt{B^2_\text{real}+B^2_\text{imag}}$ and the phase $\varphi=\text{tan}^{-1}\left(\frac{B_\text{imag}}{B_\text{real}}\right)$. $B$ cannot equal $e^{i\phi}$ for $\phi$ real except in the special case that $B^2_\text{real}+B^2_\text{imag}=1\,$; i.e., only if $B$ happens to lie on the unit-circle in the complex plane. Again I ask you: what is your knowledge of complex variables?

 

[binbagsss]

If $B=B_\text{real}+i\, B_\text{imag}$ in cartesian form, then $B=\left|B\right|e^{i\varphi}$ in polar form, where the amplitude $\left|B\right|=\sqrt{B^2_\text{real}+B^2_\text{imag}}$ and the phase $\varphi=\text{tan}^{-1}\left(\frac{B_\text{imag}}{B_\text{real}}\right)$. $B$ cannot equal $e^{i\phi}$ for $\phi$ real except in the special case that $B^2_\text{real}+B^2_\text{imag}=1\,$; i.e., only if $B$ happens to lie on the unit-circle in the complex plane. Again I ask you: what is your knowledge of complex variables?

sorry where does $B=B_\text{real}+i\, B_\text{imag}$ come from? Aren't B and B* treated as independent fields in the Proca action?

 

[renormalize]

sorry where does $B=B_\text{real}+i\, B_\text{imag}$ come from? Aren't B and B* treated as independent fields in the Proca action?

OK, this is for the scalar action, not Proca (vector) action: let $B,B^*$ be your independent complex scalar fields; i.e., 2 degrees of freedom. Then define the field combinations: $$B_{\text{real}}\equiv\frac{1}{2}\left(B+B^{*}\right),\;B_{\text{imag}}\equiv\frac{1}{2i}\left(B-B^{*}\right)\tag{1a,b}$$It's easy to verify that $B_{\text{real}},B_{\text{imag}}$ are each real quantities by definition and carry the same 2 degrees of freedom as the original fields (it's just a change of basis). Insert the inverse of (1):$$B=B_{\text{real}}+i\,B_{\text{imag}},\;B^{*}=B_{\text{real}}-i\,B_{\text{imag}}\tag{2a,b}$$into your scalar action so that it now depends only on the 2 real independent fields $B_{\text{real}},B_{\text{imag}}$ and you're done.
But why are you repeatedly avoiding answering my question about your understanding of complex variables?

 

[binbagsss]

OK, this is for the scalar action, not Proca (vector) action: let $B,B^*$ be your independent complex scalar fields; i.e., 2 degrees of freedom. Then define the field combinations: $$B_{\text{real}}\equiv\frac{1}{2}\left(B+B^{*}\right),\;B_{\text{imag}}\equiv\frac{1}{2i}\left(B-B^{*}\right)\tag{1a,b}$$It's easy to verify that $B_{\text{real}},B_{\text{imag}}$ are each real quantities by definition and carry the same 2 degrees of freedom as the original fields (it's just a change of basis). Insert the inverse of (1):$$B=B_{\text{real}}+i\,B_{\text{imag}},\;B^{*}=B_{\text{real}}-i\,B_{\text{imag}}\tag{2a,b}$$into your scalar action so that it now depends only on the 2 real independent fields $B_{\text{real}},B_{\text{imag}}$ and you're done.
But why are you repeatedly avoiding answering my question about your understanding of complex variables?

because its patronising. its not how i would be helping out someone on these forums. i have definitely done multiple complex analysis coursers during my undergrad and master's dont panic ;). admittedly, i was in the middle of something else as well so only skim read your replies.

 

[binbagsss]

because its patronising. its not how i would be helping out someone on these forums. i have definitely done multiple complex analysis coursers during my undergrad and master's dont panic ;). admittedly, i was in the middle of something else as well so only skim read your replies.

OK, this is for the scalar action, not Proca (vector) action:

which is?This is unclear before you then go on to state textbook complex knowledge again... I am only interested in the Proca action

 

[PeterDonis]

i have definitely done multiple complex analysis coursers

Then I am very, very confused about why you are not thoroughly familiar with the notations $z = x + i y$ with $x$ and $y$ real, and $z = r e^{i \theta}$ with $r$ and $\theta$ real, since those notations are covered as basic knowledge in every complex analysis course I've ever heard of. I am even more confused because this is an "A" level thread and the notations I have just written are "I" level--undergraduates are taught them routinely. In fact I even encountered them in high school, which would be "B" level. Anyone who starts an "A" level thread should have thorough knowledge of "B" and "I" level concepts in the subject area being discussed. When you make posts that make it appear that you don't have that knowledge, it's not "patronizing" to ask.

 

[PeterDonis]

to simplify things one should take them as real, and include a real amplitude for each B and B* as well

You can't have a different real amplitude for $B$ and $B*$ if they are complex conjugates. If $B = r e^{i \theta}$, then $B* = r e^{- i \theta}$. There's no freedom to introduce another amplitude for $B*$ different from $r$.

 

[PeterDonis]

it assumes $R(\vec{x},t) \geq 0$ and both $R,S \in \Re$, how is this possible without loss of generality?

Suppose I have a complex number $z = r e^{i \theta}$, but I decide to let $r$ and $\theta$ be complex themselves. Then I can always write:

$$
z = \left( R e^{i \Theta} \right) e^{i \left( X + i Y \right)}
$$

with $R$, $\Theta$, $X$, and $Y$ all real, and it is then a simple matter to rewrite this as

$$
z = \bar{r} e^{i \bar{\theta}}
$$

where $\bar{r} = R e^{- Y}$ and $\bar{\theta} = e^{i \left( \Theta + X \right)}$ are both real again. So there is no loss of generality in restricting $r$ and $\theta$ in $z = r e^{i \theta}$ to be real numbers. (And by a similar argument one can show that there is also no loss of generality in restricting $r \ge 0$.)

 

[binbagsss]

Then I am very, very confused about why you are not thoroughly familiar with the notations $z = x + i y$ with $x$ and $y$ real, and $z = r e^{i \theta}$ with $r$ and $\theta$ real, since those notations are covered as basic knowledge in every complex analysis course I've ever heard of. I am even more confused because this is an "A" level thread and the notations I have just written are "I" level--undergraduates are taught them routinely. In fact I even encountered them in high school, which would be "B" level. Anyone who starts an "A" level thread should have thorough knowledge of "B" and "I" level concepts in the subject area being discussed. When you make posts that make it appear that you don't have that knowledge, it's not "patronizing" to ask.

becuase i have not done it in a long time !i could have recapped it in 5 minutes.

 

[binbagsss]

Suppose I have a complex number $z = r e^{i \theta}$, but I decide to let $r$ and $\theta$ be complex themselves. Then I can always write:

$$
z = \left( R e^{i \Theta} \right) e^{i \left( X + i Y \right)}
$$

with $R$, $\Theta$, $X$, and $Y$ all real, and it is then a simple matter to rewrite this as

$$
z = \bar{r} e^{i \bar{\theta}}
$$

where $\bar{r} = R e^{- Y}$ and $\bar{\theta} = e^{i \left( \Theta + X \right)}$ are both real again. So there is no loss of generality in restricting $r$ and $\theta$ in $z = r e^{i \theta}$ to be real numbers. (And by a similar argument one can show that there is also no loss of generality in restricting $r \ge 0$.)

this was answered by post 2. why answer it again?

 

[binbagsss]

You can't have a different real amplitude for $B$ and $B*$ if they are complex conjugates. If $B = r e^{i \theta}$, then $B* = r e^{- i \theta}$. There's no freedom to introduce another amplitude for $B*$ different from $r$.

I mentioned if you can have different arguments in the exponent $\theta$ and $\theta_1$... . Do they have to be complex conjugates in the Proca action , I guess is part of my question. Since they are treated as independent when varying to get the Proca equations from it anyway.

 

[PeterDonis]

I mentioned if you can have different arguments in the exponent $\theta$ and $\theta_1$

And my post makes obvious that if they are complex conjugates, you can't.

Do they have to be complex conjugates in the Proca action , I guess is part of my question.

Yes, they do. @renormalize has already addressed this. You have two independent real quantities that can be varied. For various reasons, it is preferred to write those two independent real quantities as two complex numbers that are complex conjugates of each other, which amounts to the same thing. But if you write down two complex numbers that are not complex conjugates of each other, now you have four independent real quantities, not two.

 

[PeterDonis]

this was answered by post 2. why answer it again?

Because a number of your posts in this thread after post #2 made me doubt that you actually understood the answer.

 

[binbagsss]

And my post makes obvious that if they are complex conjugates, you can't.


Yes, they do. @renormalize has already addressed this. You have two independent real quantities that can be varied. For various reasons, it is preferred to write those two independent real quantities as two complex numbers that are complex conjugates of each other, which amounts to the same thing. But if you write down two complex numbers that are not complex conjugates of each other, now you have four independent real quantities, not two.

in post 21 you used the same theta. and then said with that the amps can't be the same

 

[binbagsss]

And my post makes obvious that if they are complex conjugates, you can't.


Yes, they do. @renormalize has already addressed this. You have two independent real quantities that can be varied. For various reasons, it is preferred to write those two independent real quantities as two complex numbers that are complex conjugates of each other, which amounts to the same thing. But if you write down two complex numbers that are not complex conjugates of each other, now you have four independent real quantities, not two.

i understood this. to which I said I understand it may seem illogical as you would then be introducing two more real quantities.

 

[PeterDonis]

in post 21 you used the same theta. and then said with that the amps can't be the same

No, I said the opposite, that the amplitudes must be the same. Go read that post again.

 

[PeterDonis]

to which I said I understand it may seem illogical as you would then be introducing two more real quantities.

You did mention in post #8 that it would introduce more fields. However, once again, a number of other posts you have made in this thread made me wonder exactly what you understood.

 

[binbagsss]

No, I said the opposite, that the amplitudes must be the same. Go read that post again.

i said, with that, as in with theta being the same, the amps must be the same. whereas i asked about differing theta. so basically if i did a polar decomposition using different amplitudes and phases for both B and B*, then I would end up with two more equations when varying the action. Ignoring the fact that that I would now have four independent variables, if I were to write $B=B_1 e^{i\phi}$ and $B*=B_2 e^{i\phi_2}$ , then, from them being complex conjugates of each other, I would obtain constraints, such as $B_1 \exp^{-i\theta}=B_2 \exp^{i\phi_2}$, and $B_1 \exp^{i\theta}=B_2 \exp^{-i\phi_2}$but because of the differing amplitudes/arguments, neither the amplitudes or arguments in the exponents need to be the same?

 

[PeterDonis]

Ignoring the fact that that I would now have four independent variables

In other words, ignoring the fact that in the case you were asking about, the Proca equation, you don't. You only have two. Or more precisely, two for each 4-vector component (since the Proca equation is for a spin-1 vector field instead of a spin-0 scalar field); but all that means is that you stick a 4-vector index $\mu$ on $B$.

You need to make up your mind what you are asking about. Are you asking about the actual Proca equation? Or are you asking about some other case, which AFAIK has no physical relevance and is not discussed in the literature, where you somehow have four independent variables instead of two? You said before that you were asking about the Proca equation. Are you now changing the discussion to be about something else?