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