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turnerre
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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.
turnerre said: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.
turnerre said: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.
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) for all physical systems relativistic or not. This is because we believe it is a law of nature.
It is the form of the Hamiltonian that determines the domain of application of Schrodinger equation.
regards
sam
samalkhaiat said:turnerre said: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) for all physical systems relativistic or not. This is because we believe it is a law of nature.
It is the form of the Hamiltonian that determines the domain of application of Schrodinger equation.
regards
sam
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?
blechman said: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?
kdv said: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
[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 said: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...
blechman said:What's the form of the "KG Hamiltonian"? I've never seen this before.
Unfortunately, the second order quantization of the real/complex scalar fieldskdv said:It's simply the "second-quantized" KG Hamiltonian, no?
blechman said:samalkhaiat said: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
samalkhaiat said:[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]
samalkhaiat said:blechman said:samalkhaiat said:Slowdown and readme correctly!
alright, I think I was just being silly. sorry about the confusion.
jostpuur said: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.
blechman said:I'm not quite sure what you mean by non-regular
jostpuur said:You don't :rofl:I don't think this is right.
Open any textbook on field theory and see how easy it is to derive the Hamiltonian densityThe Lagrange's function of Dirac field is non-regular, and the usual Hamiltonian formulation doesn't exist for it.
[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
Hans de Vries said: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]
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!
..., So, if one transforms correctly, then
the other one doesn't.
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].
regards
sam
jostpuur said:[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.
I'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.
The Wiki page on Dirac brackets is far too brief when it says
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.
By not elaborating on this statement for the fermion case, it can cause confusion.
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.
samalkhaiat said:You don't :rofl:
Open any textbook 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
This is just a numerical normalization with [itex]E_o[/itex], the rest mass which is a Lorentz scalar.samalkhaiat said: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!
samalkhaiat said: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].
strangerep said:I'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.
I knew what he was getting at! He needs to learn to count correctly the physical degrees of freedom. I worked with the non-hermitian Dirac 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 the hermitian Dirac 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
jostpuur said: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]
samalkhaiat said:jostpuur said:You don't :rofl:
Open any textbook on field theory and see how easy it is to derive the Hamiltonian density
sam
Except that it isn't. In order to properly do it, you need to use Dirac brackets and Dirac's treatment of constraints, since, as was said earlier, the map from Lagrangian to Hamiltonian is singular for fermions.
lbrits said:Except that it isn't. In order to properly do it, you need to use Dirac brackets and Dirac's treatment of constraints, since, as was said earlier, the map from Lagrangian to Hamiltonian is singular for fermions.
Wow, thank you for this valuable piece of information! Do read my posts to learn more about the use of Dirac brackets in constraint systems.
regards
sam
Always welcome! (unless my sarcasm detector is malfunctioning)samalkhaiat said:Wow, thank you for this valuable piece of information! Do read my posts to learn more about the use of Dirac brackets in constraint systems.
regards
sam