Why is there an "i" in the Schrodinger equation?

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e^x
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On other forums I've read its got something to do with the conservation of probability.

But I can't seem to find a proper explanation?

Thanks
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atsruser
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(Original post by e^x)
On other forums I've read its got something to do with the conservation of probability.

But I can't seem to find a proper explanation?

Thanks
The Schroedinger equation is a postulate of non-relativistic QM, and can't in fact be derived, but the hand waving explanation of how Schroedinger motivated it goes like this:

1. Bohr et al suggested that all quantum effects could be explained by supposing that there was some kind of wave effect going on at the atomic level.

2. De Broglie and Einstein at different times contributed to this quantum wave idea by proposing that

a) all particles have associated waves where the momentum and wave number are related by:

p =\frac{h}{2\pi}k (De Broglie)

b) photons have energy related to angular frequency by

E=\frac{h}{2\pi}\omega (Einstein)

where h is Planck's constant.

3. Schroedinger suggested that a particle travelling along a straight line in empty space would naturally have a plane wave representation of the form:

\psi = A e^{i( \bold{k} \cdot \bold{r} - \omega t)}

where A is a complex amplitude. This is standard stuff apart from complex A, I think. Now we can rewrite this due to Messrs De Broglie and Einstein to get:

\psi = A e^{\frac{2\pi i}{h}(\bold{p} \cdot \bold{r} - E t)} =A e^{\frac{i}{\hbar}(\bold{p} \cdot \bold{r} - E t)}

where \hbar=h/2\pi

Now note that:

\frac{\partial \psi}{\partial t} = -\frac{iE}{\hbar} \psi \Rightarrow E\psi = i \hbar \frac{\partial \psi}{\partial t}

where E is the energy of the particle.

4. Heisenberg had developed a theory of wave mechanics by assuming that observable quantities (e.g. energy, momentum, etc) are the real eigenvalues of certain matrices representing the physical system. The matrices had to be Hermitian. We can translate this idea into operator theory by saying that if we have a representation of a quantum system, \psi, then we operate on it with an Hermitian operator e.g. \hat{E} and we solve an eigenvalue equation to find the possible values of the observable e.g. for energy we have an operator eqn:

\hat{E}\psi = E\psi

Now compare this to Schroedinger's plane wave logic above. We must have an equation:

\hat{E}\psi = E\psi = i\hbar \frac{\partial \psi}{\partial t}

i.e we have an operator eqn

\hat{E}\psi = i\hbar \frac{\partial \psi}{\partial t}

which would appear to relate the time evolution of \psi to the energy operator of the system, and it has the i that you enquired about. Schroedinger then suggested that this applied to *all* quantum systems, and Bob's your uncle, all is done and dusted, Nobel prize in the bag for our dearest Erwin.

The only thing left to do is to figure out a sensible form for \hat{E}, and Schroedinger did that by translating a conservation of energy statement in classical mechanics (E=\frac{p^2}{2m}+V) into a sensible quantum version. And then he solved it for the hydrogen atom to show that it worked, and gave the right discrete spectrum for the the electron energy levels (modulo various relativistic and spin-orbit coupling corrections that no one yet knew about)
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(Original post by atsruser)
The Schroedinger equation is a postulate of non-relativistic QM, and can't in fact be derived, but the hand waving explanation of how Schroedinger motivated it goes like this:

1. Bohr et al suggested that all quantum effects could be explained by supposing that there was some kind of wave effect going on at the atomic level.

2. De Broglie and Einstein at different times contributed to this quantum wave idea by proposing that

a) all particles have associated waves where the momentum and wave number are related by:

p =\frac{k}{2\pi} (De Broglie)

b) photons have energy related to angular frequency by

E=\frac{h}{2\pi}\omega (Einstein)

where h is Planck's constant.

3. Schroedinger suggested that a particle travelling along a straight line in empty space would naturally have a plane wave representation of the form:

\psi = A e^{i( \bold{k} \cdot \bold{r} - \omega t)}

where A is a complex amplitude. This is standard stuff apart from complex A, I think. Now we can rewrite this due to Messrs De Broglie and Einstein to get:

\psi = A e^{\frac{2\pi i}{h}(\bold{p} \cdot \bold{r} - E t)} =A e^{\frac{i}{\hbar}(\bold{p} \cdot \bold{r} - E t)}

where \hbar=h/2\pi

Now note that:

\frac{\partial \psi}{\partial t} = -\frac{iE}{\hbar} \psi \Rightarrow E\psi = i \hbar \frac{\partial \psi}{\partial t}

where E is the energy of the particle.

4. Heisenberg had developed a theory of wave mechanics by assuming that observable quantities (e.g. energy, momentum, etc) are the real eigenvalues of certain matrices representing the physical system. The matrices had to be Hermitian. We can translate this idea into operator theory by saying that if we have a representation of a quantum system, \psi, then we operate on it with an Hermitian operator e.g. \hat{E} and we solve an eigenvalue equation to find the possible values of the observable e.g. for energy we have an operator eqn:

\hat{E}\psi = E\psi

Now compare this to Schroedinger's plane wave logic above. We must have an equation:

\hat{E}\psi = E\psi = i\hbar \frac{\partial \psi}{\partial t}

i.e we have an operator eqn

\hat{E}\psi = i\hbar \frac{\partial \psi}{\partial t}

which would appear to relate the time evolution of \psi to the energy operator of the system, and it has the i that you enquired about. Schroedinger then suggested that this applied to *all* quantum systems, and Bob's your uncle, all is done and dusted, Nobel prize in the bag for our dearest Erwin.

The only thing left to do is to figure out a sensible form for \hat{E}, and Schroedinger did that by translating a conservation of energy statement in classical mechanics (E=\frac{p^2}{2m}+V) into a sensible quantum version. And then he solved it for the hydrogen atom to show that it worked, and gave the right discrete spectrum for the the electron energy levels (modulo various relativistic and spin-orbit coupling corrections that no one yet knew about)
Thanks!
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atsruser
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(Original post by e^x)
Thanks!
My pleasure. I'm also glad to note that you saw fit to repost my response in the complete fullness of its magnificence. It was - and I want to stress this - in no way de trop.

I ought to point out that my description of how the SE came about is not really historically accurate, and Schroedinger actually got it from a variational argument, I believe. So please don't take it too seriously - it was just intended to show how some of the ideas fit together. I think I first saw that argument, or something similar, in a physical chemistry book (Moore?) and if it's simple enough for a chemist to understand then it's probably logically very questionable.

One thing that I didn't really mention is the question of how to turn classical quantities into their equivalent quantum operators, and I can't really remember how all the arguments work here anyway - here is a similar bit of hand-waving that derives a plausible momentum operator though:

https://en.wikipedia.org/wiki/Momentum_operator
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e^x
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(Original post by atsruser)
My pleasure. I'm also glad to note that you saw fit to repost my response in the complete fullness of its magnificence. It was - and I want to stress this - in no way de trop.

I ought to point out that my description of how the SE came about is not really historically accurate, and Schroedinger actually got it from a variational argument, I believe. So please don't take it too seriously - it was just intended to show how some of the ideas fit together. I think I first saw that argument, or something similar, in a physical chemistry book (Moore?) and if it's simple enough for a chemist to understand then it's probably logically very questionable.

One thing that I didn't really mention is the question of how to turn classical quantities into their equivalent quantum operators, and I can't really remember how all the arguments work here anyway - here is a similar bit of hand-waving that derives a plausible momentum operator though:

https://en.wikipedia.org/wiki/Momentum_operator
Can you explain why in this document derivation is written under " " in the first line for the Schrodinger equation? Is not a proper derivation if not what is it?
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atsruser
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(Original post by e^x)
Can you explain why in this document derivation is written under " " in the first line for the Schrodinger equation? Is not a proper derivation if not what is it?
The argument in that document is essentially the same as the one I gave above. However, it's not a derivation since it's not a conclusion of some logical argument. It is merely a plausible result that you can get if you assume some physics - namely that particles in free space can be represented by a plane wave of the form given.

It's also not a derivation since it assumes without proof that you can start with a plane wave, which presumably applies only in free space (i.e. one with no potential fields), invoke the Einstein/De Broglie relations, and then stick the result into an equation with a potential function V(x). This is a leap of blind faith, and the result can only be checked by having it predict experimental results (such as the spectrum of hydrogen).

It's not possible to derive the SE without guessing some physics at some point. You can derive a Schrodinger-like equation, though, if you start with some plausible assumptions about the properties of a time evolution operator for the wave function, but you then have to guess that an operator that appears along the way is in fact the Hamiltonian. This can be done by appealing to Noether's theorem.

For more details, see this paper:

http://www.uio.no/studier/emner/matn...evelopment.pdf

and this thread on physics.stackexchange:

http://physics.stackexchange.com/que...ay/83458#83458

and scroll down to Rod Vance's answer.
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e^x
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(Original post by atsruser)
The argument in that document is essentially the same as the one I gave above. However, it's not a derivation since it's not a conclusion of some logical argument. It is merely a plausible result that you can get if you assume some physics - namely that particles in free space can be represented by a plane wave of the form given.

It's also not a derivation since it assumes without proof that you can start with a plane wave, which presumably applies only in free space (i.e. one with no potential fields), invoke the Einstein/De Broglie relations, and then stick the result into an equation with a potential function V(x). This is a leap of blind faith, and the result can only be checked by having it predict experimental results (such as the spectrum of hydrogen).

It's not possible to derive the SE without guessing some physics at some point. You can derive a Schrodinger-like equation, though, if you start with some plausible assumptions about the properties of a time evolution operator for the wave function, but you then have to guess that an operator that appears along the way is in fact the Hamiltonian. This can be done by appealing to Noether's theorem.

For more details, see this paper:

http://www.uio.no/studier/emner/matn...evelopment.pdf

and this thread on physics.stackexchange:

http://physics.stackexchange.com/que...ay/83458#83458

and scroll down to Rod Vance's answer.
Thanks, I'll check that out.
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Pessimisterious
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In short: The "i" is there because particles are described as waves, and wave equations take the form A e^{i (kx-\omega t)} .

Solving differential equations always requires the inclusion of constants or coefficients to allow for the most general possibilities for solutions (even if those constants turn out to be 1 or 0). In the case of solving the Schrodinger wave equation, the constant A needs to contain the value i (amongst other things) for the solution to be correct.

It's essentially the result of solving an equation that was constructed with the specific intention of describing particles with a wave-like nature.

An interesting test would be to take the Schrodinger equation and remove the 'i' from it, then run a generic wave function through it, i.e. try:

 \hbar \frac{\partial }{\partial t} e^{i (kx-\omega t)} = - \frac{\hbar^2}{2m}\frac{\partial^2 }{\partial x^2} e^{i (kx-\omega t)}

 - \hbar i \omega e^{i (kx-\omega t)} = - \frac{\hbar^2}{2m} (i k)^2 e^{i (kx-\omega t)}

 - \hbar i \omega = \frac{\hbar^2}{2m} k^2}

 - i  = \frac{\hbar k^2}{2m\omega}}

It just doesn't make sense because the equation is supposed to describe something real, not imaginary. So the equation clearly needs to start out with another "i" in there. If you put the "i" back in you get:

 - i*i = \frac{\hbar k^2}{2m\omega}}

 1 = \frac{\hbar k^2}{2m\omega}}

 k^2 = \frac{2m \omega}{\hbar}}

and since k = \frac{\sqrt{2mE}}{\hbar}

you end up with:

E = \hbar \omega

which is a well known result in physics. So the "i" was necessary!

Physics is often just the act of solving known differential equations and finding out what the constants are. The difficult part is knowing how to model whatever system you're looking at.

(3rd year physics student)
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e^x
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(Original post by atsruser)
The argument in that document is essentially the same as the one I gave above. However, it's not a derivation since it's not a conclusion of some logical argument. It is merely a plausible result that you can get if you assume some physics - namely that particles in free space can be represented by a plane wave of the form given.

It's also not a derivation since it assumes without proof that you can start with a plane wave, which presumably applies only in free space (i.e. one with no potential fields), invoke the Einstein/De Broglie relations, and then stick the result into an equation with a potential function V(x). This is a leap of blind faith, and the result can only be checked by having it predict experimental results (such as the spectrum of hydrogen).

It's not possible to derive the SE without guessing some physics at some point. You can derive a Schrodinger-like equation, though, if you start with some plausible assumptions about the properties of a time evolution operator for the wave function, but you then have to guess that an operator that appears along the way is in fact the Hamiltonian. This can be done by appealing to Noether's theorem.

For more details, see this paper:

http://www.uio.no/studier/emner/matn...evelopment.pdf

and this thread on physics.stackexchange:

http://physics.stackexchange.com/que...ay/83458#83458

and scroll down to Rod Vance's answer.
In my attachment does the r mean |(x,y,z)| for the bit explaining about 3-D?
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Pessimisterious
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(Original post by e^x)
In my attachment does the r mean |(x,y,z)| for the bit explaining about 3-D?
It means a 3D co-ordinate position

The basic 1-D wave is \Psi(x,t) = e^{i(kx - \omega t)}

if it's to be extended to general 3D space, it becomes

\Psi(r,t) = e^{i(k \cdot r  - \omega t)}

where k \cdot r = k_x x + k_y y + k_z z

giving \Psi(r,t) = \Psi(x,y,z,t) = e^{i(k_x x + k_y y + k_z z - \omega t)}

It's the same as the 1D version but it has y and z directions included too.
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(Original post by Pessimisterious)
It means a 3D co-ordinate position

The basic 1-D wave is \Psi(x,t) = e^{i(kx - \omega t)}

if it's to be extended to general 3D space, it becomes

\Psi(r,t) = e^{i(k \cdot r  - \omega t)}

where k \cdot r = k_x x + k_y y + k_z z

giving \Psi(r,t) = \Psi(x,y,z,t) = e^{i(k_x x + k_y y + k_z z - \omega t)}

It's the same as the 1D version but it has y and z directions included too.
so its just equal to (x,y,z)?
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(Original post by e^x)
so its just equal to (x,y,z)?
Yep, pretty much. It's a way of neatly throwing full co-ordinate systems into an equation.
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(Original post by Pessimisterious)
Yep, pretty much. It's a way of neatly throwing full co-ordinate systems into an equation.
Thanks
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