- 2

- 0

## Main Question or Discussion Point

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.

- Thread starter turnerre
- Start date

- 2

- 0

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.

- 441

- 2

blechman

Science Advisor

- 779

- 8

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...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.

samalkhaiat

Science Advisor

- 1,637

- 833

Your question is meaningless if by "Schrodinger's equation" you meant the nonrelativistic Schrodinger equation!

The general form of Schrodinger equation

[tex]i \frac{\partial}{\partial t} \Psi (...) = H \Psi (...)[/tex]

is valid (or we assume to be valid) forall physical systemsrelativistic or not. This is because we believe it is alaw of nature.

It is theform of the Hamiltonianthat determines thedomain of applicationof Schrodinger equation.

regards

sam

Last edited:

blechman

Science Advisor

- 779

- 8

This is true for the Dirac equation, but how about the Klein-Gordon equation? It's second-order in time, and therefore not obviously related to Schrodinger. What is the "Hamiltonian" for such a beast?Your question is meaningless if by "Schrodinger's equation" you meant the nonrelativistic Schrodinger equation!

The general form of Schrodinger equation [...] is valid (or we assume to be valid) forall physical systemsrelativistic or not. This is because we believe it is alaw of nature.

It is theform of the Hamiltonianthat determines thedomain of applicationof Schrodinger equation.

regards

sam

- 336

- 1

It's simply the "second-quantized" KG Hamiltonian, no?This is true for the Dirac equation, but how about the Klein-Gordon equation? It's second-order in time, and therefore not obviously related to Schrodinger. What is the "Hamiltonian" for such a beast?

In QFT the starting point is to assume that the time evolution of a state is given by

[tex] |\psi(t)> = e^{-iHt} |\psi(0)> [/tex] 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.

blechman

Science Advisor

- 779

- 8

What's the form of the "KG Hamiltonian"? I've never seen this before.It's simply the "second-quantized" KG Hamiltonian, no?

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).In QFT the starting point is to assume that the time evolution of a state is given by

[tex] |\psi(t)> = e^{-iHt} |\psi(0)> [/tex] 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.

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...

- 441

- 2

- 336

- 1

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...

I thought that the Hamiltonian density was imply

[tex] H =\frac{1}{2} \Pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2 [/tex]

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.

Hans de Vries

Science Advisor

Gold Member

- 1,085

- 21

The most realistic would be this one:What's the form of the "KG Hamiltonian"? I've never seen this before.

[tex] 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][/tex]

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 [itex]\gamma[/itex] and the Hamiltonian density transforms like [itex]\gamma^2[/itex], (an extra [itex]\gamma[/itex] factor due

to Lorentz contraction.)

However, it contains the derivative in time of [itex]\phi[/itex]....

It's simply the "second-quantized" KG Hamiltonian, no?

Unfortunately, the second order quantization of the real/complex scalar fields

is inherently non-relativistic due to “replacements” like [itex]\frac{1}{2}mv^2 \rightarrow \frac{1}{2}\dot{\phi}^2[/itex]. 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:

[tex]

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

[/tex]

[tex]

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

[/tex]

The Lagrangian density transforms correctly as a Lorentz scalar as it comes from

the Euler Lagrange formalism. The Hamiltonian density however transforms like [itex]\beta\gamma^2[/itex].

That is, the Hamiltonian density of a particle at rest [itex]\phi=e^{-iEt} [/itex], with [itex]E=m[/itex] would be

zero!

[tex]

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

[/tex]

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:

[tex]\phi\ =\ e^{-iEt=ipx}\ = e^{-i\gamma t+i\beta\gamma x}, \ \ \ (m=1) [/tex]

and insert this in the expressions.

samalkhaiat

Science Advisor

- 1,637

- 833

Slowdown and readme correctly!This is true for the Dirac equation, but how about the Klein-Gordon equation? It's second-order in time, and therefore not obviously related to Schrodinger.

You will find it below.What is the "Hamiltonian" for such a beast?

In the relativistic domain, QM is a field theory. So we can still work with the Scrodinger picture where the operators are independent of time. Why do you think I wrote [itex]\Psi(...)[/itex] and not [itex]\Psi(x,t)[/itex]? In field theory, my friend, we call it a wavefunctional, [itex]\Psi [u( \vec{x}),t][/itex], with [itex]u( \vec{x})[/itex] is the relevant field operator in the Schrodinger picture. It satisfies the functional differential equation of Schrodinger

[tex]

i \partial_{t} \Psi[u( \vec{x}),t] = \int d^{3}x \mathcal{H}\left( \pi ( \vec{x}), u( \vec{x}\right)) \Psi[ u( \vec{x}),t]

[/tex]

with

[tex]

\mathcal{H} = \frac{1}{2} \left( - \frac{ \delta^{2}}{\delta \phi^{2}( \vec{x})} + | \nabla \phi |^{2} + m^{2}\phi^{2} \right)

[/tex]

for the K-G field [itex]u( \vec{x}) \equiv \phi(\vec{x})[/itex],

[tex]

\mathcal{H} = \frac{\delta}{ \delta \psi ( \vec{x})} \left( -i \alpha \ . \ \nabla + m \beta \right) \psi( \vec{x})

[/tex]

for Dirac field [itex] u \equiv \psi[/itex] and

[tex]

\mathcal{H} = \frac{1}{2} \left( - \frac{\delta}{\delta \vec{A}(\vec{x})} . \frac{\delta}{\delta \vec{A}(\vec{x})} + \vec{B}(\vec{x}) . \vec{B}(\vec{x}) \right)

[/tex]

for photon field [itex]u \equiv \vec{A}(\vec{x})[/itex] in the temporal gauge [itex]A_{0} = 0[/itex].

If you require more relativistic Schrodinger equations, don't hesitate to ask

regards

sam

Last edited:

- 2,099

- 16

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.[tex]

\mathcal{H} = \frac{\delta}{ \delta \psi ( \vec{x})} \left( -i \alpha \ . \ \nabla + m \beta \right) \psi( \vec{x})

[/tex]

for Dirac field [itex] u \equiv \psi[/itex]

blechman

Science Advisor

- 779

- 8

alright, I think I was just being silly. sorry about the confusion.Slowdown and readme correctly!

blechman

Science Advisor

- 779

- 8

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 ([itex]\alpha,\beta[/itex] are the Dirac matrices, or products of them, and [itex]\psi[/itex] 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".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.

- 2,099

- 16

I mean that when you solve [itex](\psi, \Pi)[/itex] as a function of [itex](\psi, \partial_0\psi)[/itex], you cannot reverse this mapping. Thus, the Hamilton's equations don't give the equivalent dynamics as the Euler-Lagrange equations.I'm not quite sure what you mean by non-regular

samalkhaiat

Science Advisor

- 1,637

- 833

You don't :rofl:I don't think this is right.

OpenThe Lagrange's function of Dirac field is non-regular, and the usual Hamiltonian formulation doesn't exist for it.anytextbook on field theory and see how easy it is to derive the Hamiltonian density

[tex]\mathcal{H} = \psi^{\dagger} \left( -i \alpha . \nabla + m \beta \right) \psi[/tex]

In the coordinate representation of Schrodinger picture, we put;

[tex]\psi \rightarrow \psi[/tex]

[tex]\pi = i \psi^{\dagger} \rightarrow i \frac{\delta}{\delta \psi}[/tex]

just like

[tex]x \rightarrow x[/tex]

[tex]p \rightarrow - i \frac{\partial}{\partial x}[/tex]

in ordinary QM. OK?

Now, you be clever and tell me why is it (unlike ordinary QM) we don't put;

[tex]\pi \rightarrow - i \frac{\delta}{\delta \psi}[/tex]

sam

samalkhaiat

Science Advisor

- 1,637

- 833

Do your units correctly! The Hamiltonian is not a dimensionless quantity! Hamiltonian and Lagrangian have the same dimension as energy. In the natural units, it is [itex]cm^{-1}[/itex]. Including the DODGY factor [itex]\frac{1}{E_{0}}[/itex] makes your expression for H dimensionless!The most realistic would be this one:

[tex] 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][/tex]

The Hamiltonian density is not Lorentz scalar and it does not need to be Lorentz scalar! It is the (0,0)-component of the energy-momentum tensor [itex]T^{\mu \nu}[/itex]........, So, if one transforms correctly, then

the other one doesn't.

regards

sam

strangerep

Science Advisor

- 3,006

- 808

I'm guessing that Sam gave a ridiculing reply because he didn't realize you were[tex]

\mathcal{H} = \psi^{\dagger} \left( -i \alpha . \nabla + m \beta \right) \psi

[/tex]

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.

trying to understand this stuff in the context of Dirac-Bergman (constraint) quantization.

The Wiki page on Dirac brackets is far too brief when it says

By not elaborating on this statement for the fermion case, it can cause confusion.Wiki: When the Lagrangian is at most linear in the velocity of at least one coordinate;

in which case, the definition of the canonical momentum leads to a constraint. This is the

most frequent reason to resort to Dirac brackets. For instance, the Lagrangian (density) for

any fermion is of this form.

I'll try to work through it in a bit more detail...

First, let's take a arbitrary Lagrangian [itex]L(q,\dot q)[/itex] and the usual Euler-Lagrange

equations:

[tex]

\frac {d}{d t} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) = \frac {\partial L}{\partial q_j}

[/tex]

which can be re-written as

[tex]

\frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j} ~ d^2 q_j/dt^2

~=~ - \frac{\partial^2 L}{\partial \dot{q}_i \partial q_j}\dot{q}_j

+ \frac {\partial L}{\partial q_j}

[/tex]

This can only be inverted to find the accelerations [itex]d^2 q_j/dt^2[/itex] if we

have a non-singular Hessian matrix

[tex]

W_{ij}(q,\dot q) ~:=~ \frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j}

[/tex]

I.e., the accelerations at a given time are uniquely determined by [itex](q,\dot q)[/itex]

at that time only if [itex]\det W \ne 0[/itex].

In the case of the Dirac field, we start assuming that [itex]\Psi,\Psi^\dagger[/itex]and

their time derivatives are independent in the Lagrangian, but if you calculate

the Hessian above for that case you'll find the entire Hessian matrix is 0. It is thus

a trivial case. All the Dirac-Bergman constraint quantization stuff is intended for less

trivial cases where det W = 0, but W is not all-zeros.

For the free Dirac field, we just find that there are only 2 independent "variables",

i.e., [itex]\Psi[/tex] and [itex]\pi_\Psi = i\Psi^\dagger[/itex]. It is trivial to handle

the constraints in this case, and arrive at the correct Hamiltonian (i.e., the

one above which is quoted in textbooks).

Rigorous quantization for the case of interacting Dirac fields

(i.e., quantization of gauge fields) requires the full machinery.

I hope that helps.

- 2,099

- 16

I see this.You don't :rofl:

Openanytextbook on field theory and see how easy it is to derive the Hamiltonian density

[tex]\mathcal{H} = \psi^{\dagger} \left( -i \alpha . \nabla + m \beta \right) \psi[/tex]

No.In the coordinate representation of Schrodinger picture, we put;

[tex]\psi \rightarrow \psi[/tex]

[tex]\pi = i \psi^{\dagger} \rightarrow i \frac{\delta}{\delta \psi}[/tex]

just like

[tex]x \rightarrow x[/tex]

[tex]p \rightarrow - i \frac{\partial}{\partial x}[/tex]

in ordinary QM. OK?

The problem is that the Hamilton's density is a function of these real valued fieldsNow, you be clever and tell me why is it (unlike ordinary QM) we don't put;

[tex]\pi \rightarrow - i \frac{\delta}{\delta \psi}[/tex]

sam

[tex]

\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),

[/tex]

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

[tex]

\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).

[/tex]

The problem is the same as it was with the Lagrangian

[tex]

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

[/tex]

that gave the equation of motion

[tex]

\dot{x}(t)=y(t)

[/tex]

[tex]

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

[/tex]

The canonical momenta are

[tex]

p_x = y,\quad p_y=-x.

[/tex]

and the Hamiltonian is

[tex]

H=x^2 + y^2.

[/tex]

How do you quantize this? Some dumb attempts would be

[tex]

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

[/tex]

or

[tex]

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

[/tex]

or

[tex]

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

[/tex]

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

[tex]

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

[/tex]

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 [itex]\psi[/itex]-variables where the notation makes it seem most plausible. That doesn't look right to me.

Last edited:

Hans de Vries

Science Advisor

Gold Member

- 1,085

- 21

Do your units correctly! The Hamiltonian is not a dimensionless quantity! Hamiltonian and Lagrangian have the same dimension as energy. In the natural units, it is [itex]cm^{-1}[/itex]. Including the DODGY factor [itex]\frac{1}{E_{0}}[/itex] makes your expression for H dimensionless!

This is just a numerical normalization with [itex]E_o[/itex], the rest mass which is a

[itex]E_o[/itex] does not transform like [itex]cm^{-1}[/itex].

Thus, the given Hamiltonian does transform like [itex]cm^{-1}[/itex], like energy, like it should do.

Without doing this you get expressions like:

[tex]

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

[/tex]

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

[tex]

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

[/tex]

Regards, Hans

Last edited:

Hans de Vries

Science Advisor

Gold Member

- 1,085

- 21

Of course not. I never said so .The Hamiltonian density is not Lorentz scalar and it does not need to be Lorentz scalar! It is the (0,0)-component of the energy-momentum tensor [itex]T^{\mu \nu}[/itex].

The densities transform like:

[tex]

\begin{array}{lll}

{\cal H} & \mbox{transforms as} & \gamma^2 \\

{\cal L} & \mbox{transforms as} & "1" \\

\end{array}

[/tex]

The integrated versions transform like:

[tex]

\begin{array}{lll}

H & \mbox{transforms as} & \gamma \\

L & \mbox{transforms as} & 1/\gamma \\

p\dot{q} & \mbox{transforms as} & \beta^2\gamma \\

\end{array}

[/tex]

Regards, Hans

samalkhaiat

Science Advisor

- 1,637

- 833

I knew what he was getting at! He needs to learn to count correctly the physical degrees of freedom. I worked with theI'm guessing that Sam gave a ridiculing reply because he didn't realize you were

trying to understand this stuff in the context of Dirac-Bergman (constraint) quantization.non-hermitianDirac Lagrangian

[tex]\mathcal{L} = \bar{\psi}\left( i \gamma . \partial - m \right) \psi \ \ (1)[/tex]

This Lagrangian describes non-degenerate, 2-dimensional phase-space. Therefore, the whole issue of constraints become meaningless.

Dirac method becomes necessary (waste of time) when the formalism is based on thehermitianDirac Lagrangian

[tex]

\mathcal{L} = \bar{\psi} \left( \frac{i}{2} \gamma . \partial^{\leftrightarrow} - m \right) \psi \ \ (2)

[/tex]

The large (degenerate) phase-space described by eq(2) is, in my case, a total waste of time because the two Lagrangians are identical up to a

4-divergence. Therefore, they lead to the same equation of motion and to the same conserved quantities. So, one does not need to bother oneself with

eq(2) when the equivalent Lagrangian eq(1) obviates the need to work with constraints.

regards

sam

Last edited:

samalkhaiat

Science Advisor

- 1,637

- 833

The problem is the same as it was with the Lagrangian

[tex]

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

[/tex]

Your "Lagrangian" is of the form

[tex]L = P_{x}\dot{x} + P_{y}\dot{y} - H[/tex]

So, if you identify [itex](x, P_{x}=y)[/itex] and [itex](y, P_{y}=-x)[/itex] as two canonical conjugate pairs, you end up with degenerate (4-dimensional) phase-space. This is because; the fundamental (Poisson) brackets;

[tex]\{x , P_{x} \}_{P} \equiv \{x , y \}_{P} = 1[/tex]

[tex]\{y , P_{y}\}_{P} \equiv \{y , -x\}_{P} = 1[/tex]

are identical. This means that the "physical" phase-space is 2-dimensional. Phase-space reduction can be achieved by choosing an equivalent ( gauge-fixed) Lagrangian. In english, we use the fact that Lagrangians which differ by a total time derivative are equivalent, i.e., they lead to the same E-L equations and the same conserved quantities (the Hamiltonian is one of these quantities).

Now, your Lagrangian can be written as

[tex]L = \bar{L} - \frac{d}{dt}(xy)[/tex]

where

[tex]\bar{L} = 2 x \dot{y} - V(x,y)[/tex]

is now of the form

[tex]\bar{L} = P_{y}\dot{y} - H[/tex]

and immediately identifies [itex](y,2x)[/itex] as a canonical pair. In this 2-dimensional phase space, the fundamental Poisson bracket describes non-commutative coordinates

[tex]\{y , x \}_{P} = \frac{1}{2}[/tex]

Notice that Dirac bracket is not needed in this "physical" phase space, i.e., you do not need to worry about constraints.

Now, let us FORMALLY quantize the thing by the usual rule

[tex]

\{x , y \}_{P} \rightarrow - i \left[ \hat{x},\hat{y} \right]_{-} = - \frac{1}{2}

[/tex]

This means that we can not diagonalize [itex]\hat{x}[/itex] and [itex]\hat{y}[/itex] simultaneously, in better words, these operators posses no common eigenstates, or even better, it means that the wavefunction can not depend on both variables x and y. So, in the coordinate y-representation:

[tex]\hat{y}\rightarrow y[/tex]

[tex]\hat{x} \rightarrow -\frac{i}{2} \partial_{y}[/tex]

Schrodinger equation becomes

[tex]

i\partial_{t}\Psi (y,t) = H( y ,\hat{x}) \Psi (y,t) = V( y , -\frac{i}{2}\partial_{y}) \Psi (y,t)

[/tex]

Last edited:

samalkhaiat

Science Advisor

- 1,637

- 833

Planar motion in a constant magnetic field [itex]\vec{B}= \nabla \times \vec{A}[/itex]:

1) Gauge-fixed Lagrangian formulation (= no worry about constraints)

In the gauge [itex]A = (0, Bx)[/itex], the Lagrangian for such motion, with additional potential, is

[tex]

L = \frac{m}{2} ( \dot{x}^{2} + \dot{y}^{2} ) + eBx \dot{y} - V(x,y) \ \ (1)

[/tex]

The equation for the 2-vector r = (x,y) is

[tex]m\ddot{x}_{i} = eB \epsilon_{ij}\dot{x}_{j} - \partial_{i}V \ \ (2)[/tex]

In the limit [itex]m \rightarrow 0[/itex], the Lagrangian becomes

[tex]

L = (eBx)\dot{y} - V(x,y) \equiv P_{y}\dot{y} - H(P_{y},y)

[/tex]

Thus, [itex]y[/itex] is the one and only physical degree of freedom, and [itex]P_{y} = eBx[/itex] is the conjugate momentum. Hence

[tex]\{x_{i},x_{j}\}_{P} = \frac{1}{eB}\epsilon_{ij}[/tex]

is the fundamental Poisson bracket and

[tex] H(P_{y},y) \equiv V( \frac{P_{y}}{eB},y)[/tex]

is the Hamiltonian. Therefore, the whole dynamics is given in terms of the usual Poisson bracket:

[tex]

\dot{x}_{i} = \{ V , x_{i} \}_{P} = \{ x_{i},x_{j}\}_{P} \partial_{j}V= \frac{1}{eB}\epsilon_{ij}\partial_{j}V

[/tex]

which is exactly the equation of motion, eq(2), in the limit [itex]m \rightarrow 0[/itex].

samalkhaiat

Science Advisor

- 1,637

- 833

2) Hamiltonian formulation in the limit [itex]m \rightarrow 0[/itex] (= do worry about constraints)

To give a canonical derivation of the Lagrangian results, we start with the Hamiltonian

[tex]H = \frac{\pi^{2}}{2m} + V \ \ (3)[/tex]

where [itex]\pi_{i} = m\dot{x}_{i}[/itex] is the kinematical momentum related to the canonical momentum [itex]P_{i} = \frac{\partial L}{\partial \dot{x}_{i}}[/itex] by

[tex]\pi_{i} = P_{i} - eA_{i}[/tex]

Upon bracketing with [itex]x_{i}[/itex], the above Hamiltonian gives the equation of motion eq(2) (in post#24) provided the following Poisson brackets hold:

[tex]\{x_{i},x_{j}\}_{P}=0 \ \ (4a)[/tex]

[tex]\{x_{i}, \pi_{j}\}_{P} = \delta_{ij} \ \ (4b)[/tex]

[tex]\{\pi_{i},\pi_{j}\}_{P} = -eB \epsilon_{ij} \ \ (4c)[/tex]

Now we would like to set m to zero in H. This can only be done provided [itex]\pi_{i}[/itex] vanishes. So, we impose [itex]\pi_{i} \approx 0[/itex] as a constraint surface. But according to eq(4c), the Poisson bracket of the constraints (i.e., the matrix [itex]C_{ij} = -eB\epsilon_{ij}[/itex]) does not vanish on the constraint surface, and the constraints are recognized to be "second-class" in Dirac's terminology. To proceed with Hamiltonian formalism, we must introduce Dirac bracket (like poisson's, Dirac bracket is classical object):

[tex]

\{a,b\}_{D} = \{a,b\}_{P} - \{a,\pi_{k}\}_{P} \left( C^{-1}\right)_{kl} \{\pi_{l},b\}_{P}

[/tex]

Since

[tex]\{a , \pi_{i} \}_{D} = 0 \ \ \mbox{for any} \ \ a[/tex]

the 2nd-class constraints can indeed be set EQUAL to zero. But the Dirac bracket of two coordinates is now non-vanishing;

[tex]

\{x_{i},x_{j}\}_{D} = - \{x_{i}, \pi_{k}\}_{P} \left(\frac{1}{eB}\epsilon_{kl}\right) \{\pi_{l},x_{j}\}_{P} = \frac{1}{eB}\epsilon_{ij}

[/tex]

So, generally speaking, in this approach the original PB is discarded after having served its purpose of identifying the second-class constraints. All the equations are now written in terms of the Dirac bracket, and the 2nd-class constraints become identities (i.e., strong equalities).

In the presence of constraints, quantization is carried out by the rule

[tex]\{a,b\}_{D} \rightarrow -i [\hat{a},\hat{b}]_{-}[/tex]

for all the classically-commuting (Bose) numbers ab = ba, and by the rule

[tex]\{a,b\}_{D} \rightarrow -i [\hat{a},\hat{b}]_{+}[/tex]

for all classically-anticommuting (Grassman-valued) (Fermi) numbers ab = - ba.

sam

To give a canonical derivation of the Lagrangian results, we start with the Hamiltonian

[tex]H = \frac{\pi^{2}}{2m} + V \ \ (3)[/tex]

where [itex]\pi_{i} = m\dot{x}_{i}[/itex] is the kinematical momentum related to the canonical momentum [itex]P_{i} = \frac{\partial L}{\partial \dot{x}_{i}}[/itex] by

[tex]\pi_{i} = P_{i} - eA_{i}[/tex]

Upon bracketing with [itex]x_{i}[/itex], the above Hamiltonian gives the equation of motion eq(2) (in post#24) provided the following Poisson brackets hold:

[tex]\{x_{i},x_{j}\}_{P}=0 \ \ (4a)[/tex]

[tex]\{x_{i}, \pi_{j}\}_{P} = \delta_{ij} \ \ (4b)[/tex]

[tex]\{\pi_{i},\pi_{j}\}_{P} = -eB \epsilon_{ij} \ \ (4c)[/tex]

Now we would like to set m to zero in H. This can only be done provided [itex]\pi_{i}[/itex] vanishes. So, we impose [itex]\pi_{i} \approx 0[/itex] as a constraint surface. But according to eq(4c), the Poisson bracket of the constraints (i.e., the matrix [itex]C_{ij} = -eB\epsilon_{ij}[/itex]) does not vanish on the constraint surface, and the constraints are recognized to be "second-class" in Dirac's terminology. To proceed with Hamiltonian formalism, we must introduce Dirac bracket (like poisson's, Dirac bracket is classical object):

[tex]

\{a,b\}_{D} = \{a,b\}_{P} - \{a,\pi_{k}\}_{P} \left( C^{-1}\right)_{kl} \{\pi_{l},b\}_{P}

[/tex]

Since

[tex]\{a , \pi_{i} \}_{D} = 0 \ \ \mbox{for any} \ \ a[/tex]

the 2nd-class constraints can indeed be set EQUAL to zero. But the Dirac bracket of two coordinates is now non-vanishing;

[tex]

\{x_{i},x_{j}\}_{D} = - \{x_{i}, \pi_{k}\}_{P} \left(\frac{1}{eB}\epsilon_{kl}\right) \{\pi_{l},x_{j}\}_{P} = \frac{1}{eB}\epsilon_{ij}

[/tex]

So, generally speaking, in this approach the original PB is discarded after having served its purpose of identifying the second-class constraints. All the equations are now written in terms of the Dirac bracket, and the 2nd-class constraints become identities (i.e., strong equalities).

In the presence of constraints, quantization is carried out by the rule

[tex]\{a,b\}_{D} \rightarrow -i [\hat{a},\hat{b}]_{-}[/tex]

for all the classically-commuting (Bose) numbers ab = ba, and by the rule

[tex]\{a,b\}_{D} \rightarrow -i [\hat{a},\hat{b}]_{+}[/tex]

for all classically-anticommuting (Grassman-valued) (Fermi) numbers ab = - ba.

sam

Last edited:

- Replies
- 13

- Views
- 2K

- Replies
- 3

- Views
- 2K

- Replies
- 6

- Views
- 1K

- Replies
- 13

- Views
- 4K

- Replies
- 12

- Views
- 2K

- Last Post

- Replies
- 12

- Views
- 1K

- Replies
- 2

- Views
- 1K

- Replies
- 1

- Views
- 1K