# Schrodinger's Equation validity for relativistic particles

1. Mar 25, 2008

### turnerre

Hey guys, I wanted to know if there is any other (more physical math) reason why schrodinger's equation is not valid for relativistic particles besides that it is not an invariant under lorentz.

2. Mar 26, 2008

### smallphi

If you represent the wave function of the particle with a plane wave ~ exp(-Et+kx) and insert that in Schrodinger, you will get E = k^2 / 2m which is the non-relativistic relation.

3. Mar 26, 2008

### blechman

isn't that enough? not being Lorentz-invariant (up to phase) means that physics looks different in different reference frames, which violates the special principle of relativity. therefore it cannot be the correct description, unless of course you want to claim that the principle of relativity is wrong...

4. Mar 26, 2008

### samalkhaiat

Last edited: Mar 26, 2008
5. Mar 27, 2008

### blechman

6. Mar 27, 2008

### kdv

It's simply the "second-quantized" KG Hamiltonian, no?
In QFT the starting point is to assume that the time evolution of a state is given by
$$|\psi(t)> = e^{-iHt} |\psi(0)>$$ which is essentially Schrodinger's equation. The point, I think, is that now H is not simply a differential operator acting on a wavefunction in a Hilbert space but an operator acting in Fock space. Maybe I am mistaken.

7. Mar 27, 2008

### blechman

What's the form of the "KG Hamiltonian"? I've never seen this before.

This sounds like putting the cart before the horse. You "assume" the Schrodinger equation, which has a solution in terms of your formula (when H is time independent; but generalizations are straight-forward).

There is such a thing as the "Dirac Hamiltonian" - it is a differential operator, first order in space-derivatives (linear in momentum), and its action on a spinor wavefunction is proportional to the (first) time derivative of the said wavefcn. However, this is because Dirac's equation is already first order in time, and the condition for such a thing to happen was the introduction of spin. How does one do the same trick for KG, which is second-order in time and has no spin? In other words, the propagator for KG is 1/p^2, not 1/p. Therefore it does not obey the time-evolution equation you wrote down with H a well-defined linear operator, as the Schrodinger Eqn expects it to be.

Well, I might be wrong, but that's what it looks like to me...

8. Mar 27, 2008

### smallphi

The KG equation is re-interpreted in QFT as an equation for the wave operator, not the states (wave function). So KG in QFT does not correspond to Schrodinger in non-relativistic QM. The Hamiltonian corresponding to KG, that must be put in the analogue of Schrodinger equation can be found in any text quantizing the free scalar field. Note, they usually don't write the Schrodinger equation in QFT because they work in Heisenberg picture for free fields - the operators evolve with time, the states don't.

9. Mar 27, 2008

### kdv

I thought that the Hamiltonian density was imply
$$H =\frac{1}{2} \Pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2$$

where phi and Pi are the usual expressions in terms of creation/annihilation operators.

Maybe I am missing something but I thought that the whole canonical quantization procedure was the following:

-> Define the S matrix elements as the transition amplitudes between free states at t= minus infinity to free states at t = plus infinity. This is where one makes use of the expression I wrote, Psi(t) = e^{-iHt} Psi(0) .

-> One then relates the expectation values of annihilation/creation operators in terms of expectation values of the field itself. This is the LSZ reduction fromula.

-> Then one introduces the interaction picture, one separates the time evolution due to the free part of the hamiltonian from the interaction part.

-> One then Taylor expands the exponential of the interaction part. One introduces Wick's theorem, Feynman rules, etc.

10. Mar 27, 2008

### Hans de Vries

The most realistic would be this one:

$$H~~=~~\int dx^3 {\cal H}\ =\ \int dx^3\ \frac{1}{2E_o} \bigg[~ +\dot{\phi}^*\dot{\phi}\ +\ \nabla\phi^* \nabla\phi\ +\ m^2 \phi^*\phi\ \bigg]$$

The Hamiltonian density of the wave function corresponds with the correct
Hamiltonian in the case of a free particle (H=E). The Hamiltonian transforms
like $\gamma$ and the Hamiltonian density transforms like $\gamma^2$, (an extra $\gamma$ factor due
to Lorentz contraction.)

However, it contains the derivative in time of $\phi$....

Unfortunately, the second order quantization of the real/complex scalar fields
is inherently non-relativistic due to “replacements” like $\frac{1}{2}mv^2 \rightarrow \frac{1}{2}\dot{\phi}^2$. The latter
is interpreted as the kinetic term used to obtain the Hamiltonian (density) from
the Lagrangian (density) and visa versa, So, if one transforms correctly, then
the other one doesn't.

for instance, it gives the pair:

$${\cal L}\ =\ \frac{1}{2E_o}\ \bigg[\ +(\dot{\phi})^2\ -\ (\nabla\phi)^2\ -\ m^2 \phi^2\ \bigg]$$

$${\cal H}\ =\ \frac{1}{2E_o}\ \bigg[\ +(\dot{\phi})^2\ +\ (\nabla\phi)^2\ +\ m^2 \phi^2\ \bigg]$$

The Lagrangian density transforms correctly as a Lorentz scalar as it comes from
the Euler Lagrange formalism. The Hamiltonian density however transforms like $\beta\gamma^2$.
That is, the Hamiltonian density of a particle at rest $\phi=e^{-iEt}$, with $E=m$ would be
zero!

$${\cal H}\ =\ \frac{1}{2E_o}\ \bigg[\ -m^2 \phi^2\ -\ 0\ +\ m^2 \phi^2\ \bigg]\ =\ 0$$

I'm a proponent of leaving relativistic QED separated from subjects like second
order quantization, instead of what you see in the Coleman and P&S style lectures:

http://www.damtp.cam.ac.uk/user/dt281/qft/qft.pdf

A freely intermixing of all kinds of stuff like The Schrödinger picture, The Heisenberg
picture, Second order quantization, QED etcetera. I prefer a more separated approach
like for instance in Ryder.

Regards, Hans

PS: to see how a quantity transforms one can use the free particle expression:
$$\phi\ =\ e^{-iEt=ipx}\ = e^{-i\gamma t+i\beta\gamma x}, \ \ \ (m=1)$$
and insert this in the expressions.

11. Mar 27, 2008

### samalkhaiat

Last edited: Mar 27, 2008
12. Mar 28, 2008

### jostpuur

I don't think this is right. The Lagrange's function of Dirac field is non-regular, and the usual Hamiltonian formulation doesn't exist for it.

13. Mar 28, 2008

### blechman

14. Mar 28, 2008

### blechman

I'm not quite sure what you mean by non-regular (it's been a while since I've studied analysis seriously), but you can write down a Hamiltonian for a Dirac "wavefunction", remembering that it acts in spin-space as well ($\alpha,\beta$ are the Dirac matrices, or products of them, and $\psi$ is a spinor). Check out, for example, Liboff's "Intro to QM" - not one of my favorite texts, but the one I used as an undergrad. Although I've never used it myself, I'm sure Bjorken and Drell have stuff on this as well, or Sakurai's "Advanced QM" text... in fact, pretty much ANY text on "Relativistic QM".

15. Mar 28, 2008

### jostpuur

I mean that when you solve $(\psi, \Pi)$ as a function of $(\psi, \partial_0\psi)$, you cannot reverse this mapping. Thus, the Hamilton's equations don't give the equivalent dynamics as the Euler-Lagrange equations.

16. Mar 28, 2008

### samalkhaiat

17. Mar 28, 2008

### samalkhaiat

18. Mar 28, 2008

### strangerep

19. Mar 29, 2008

### jostpuur

I see this.

No.

The problem is that the Hamilton's density is a function of these real valued fields

$$\textrm{Re}(\psi_1), \textrm{Re}(\psi_2), \textrm{Re}(\psi_3), \textrm{Re}(\psi_4), \textrm{Im}(\psi_1), \textrm{Im}(\psi_2), \textrm{Im}(\psi_3), \textrm{Im}(\psi_4),$$

and that there is no dependence on time derivatives. There is no obvious way to decide where you want to substitute

$$\textrm{Re}(\pi_1), \textrm{Re}(\pi_2), \textrm{Re}(\pi_3), \textrm{Re}(\pi_4), \textrm{Im}(\pi_1), \textrm{Im}(\pi_2), \textrm{Im}(\pi_3), \textrm{Im}(\pi_4).$$

The problem is the same as it was with the Lagrangian

$$L=\dot{x}y - x\dot{y} - x^2 - y^2$$

that gave the equation of motion

$$\dot{x}(t)=y(t)$$
$$\dot{y}(t)=-x(t).$$

The canonical momenta are

$$p_x = y,\quad p_y=-x.$$

and the Hamiltonian is

$$H=x^2 + y^2.$$

How do you quantize this? Some dumb attempts would be

$$H=x^2 + y^2\quad\implies\quad i\partial_t\Psi = (x^2 + y^2)\Psi$$

or

$$H=y p_x - x p_y\quad\implies\quad i\partial_t\Psi = -i(y\partial_x - x\partial_y)\Psi$$

or

$$H= p_x^2 + p_y^2\quad\implies\quad i\partial_t\Psi = -(\partial_x^2 + \partial_y^2)\Psi$$

but since this cannot be done uniquely, surely none of these attempts is going to be right? Now at this point you explained that the Lagrangian I have written down doesn't describe a physical system. However, when you are given the Dirac field Lagrangian

$$\mathcal{L} = -\textrm{Im}(\overline{\psi}\gamma^{\mu}\partial_{\mu}\psi) - m\overline{\psi}\psi$$

which has precisely the same pathology, and also leads into a equation of motion with only first order time derivatives, and cannot be given the usual Hamiltonian formulation, now you are happily computing the Hamiltonian and substituting canonical momenta operators into those $\psi$-variables where the notation makes it seem most plausible. That doesn't look right to me.

Last edited: Mar 29, 2008
20. Mar 29, 2008

### Hans de Vries

This is just a numerical normalization with $E_o$, the rest mass which is a Lorentz scalar.
$E_o$ does not transform like $cm^{-1}$.

Thus, the given Hamiltonian does transform like $cm^{-1}$, like energy, like it should do.

Without doing this you get expressions like:

$$\int m^2 \phi^*\phi\ \ dx^3\ =\ m$$

Maybe you can understand why I sometimes prefer to numerically normalize this like:

$$\frac{1}{m}\int m^2 \phi^*\phi\ \ dx^3\ =\ m$$

Regards, Hans

Last edited: Mar 29, 2008