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In the theory of relativistic wave equations, we derive the Dirac equation and Klein-Gordon equation by using representation theory of Poincare algebra.

For example, in this paper

http://arxiv.org/abs/0809.4942

the Dirac equation in momentum space (equation [52], [57] and [58]) can be derived from the 1-particle state of irreducible unitary representation of the Poincare algebra (equation [18] and [19]). The ordinary wave function in position space is its Fourier transform (equation [53], [62] and [65]).

Note at this stage, this Dirac equation is simply a classical wave equation. i.e. its solutions are classical Dirac 4-spinors, which take values in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$.

If we regard the Dirac waves $\psi(x)$ and $\bar{\psi}(x)$ as a 'classical fields', then the quantized Dirac fields are obtained by promoting them into fermionic harmonic oscillators.

What I do not understand is that when we are doing the path-integral quantization of Dirac fields, we are, in fact, treating $\psi$ and $\bar{\psi}$ as Grassmann numbers, which are counter-intuitive for me. As far as I understand, we do path-integral by summing over all 'classical fields'. While the 'classical Dirac wave $\psi(x)$' we derived in the beginning are simply 4-spinors living in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$. How can they be treated as Grassmann numbers instead?

As I see it, physicists are trying to construct a 'classical analogue' of Fermions that are purely quantum objects. For instance, if we start from a quantum anti-commutators

$$[\psi,\psi^{\dagger}]_{+}=i\hbar1 \quad\text{and}\quad [\psi,\psi]_{+}=[\psi^{\dagger},\psi^{\dagger}]_{+}=0, $$

then we can obtain the Grassmann numbers in the classical limit $\hbar\rightarrow0$. This is how I used to understand the Grassmann numbers. The problem is that if the Grassmann numbers are indeed a sort of classical limit of anticommuting operators in Hilbert space, then the limit $\hbar\rightarrow0$ itself does not make any sense from a physical point of view since in this limit $\hbar\rightarrow0$, the spin observables vanish totally and what we obtain then would be a $0$, which is a trivial theory.

Please tell me how exactly the quantum Fermions are related to Grassmann numbers.

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  • $\begingroup$ Some minor comments: the components or $\psi$ are Grassmann numbers at the classical level. In QM, they are no longer Grassmann numbers, but operators instead. We choose $\psi_\alpha(x)$ to be Grassmann odd at the classical level so that the CCR (as induced by the Poisson bracket algebra) are anticommutators instead of commutators, thus giving rise to fermionic states. $\endgroup$ – AccidentalFourierTransform Jan 28 '16 at 19:30
  • $\begingroup$ @ AccidentalFourierTransform There is no direct evidence that convinces me that the Dirac fields we derived from representation of the Poincare algebra should be grassmann-valued. Instead, I find that people tend to believe that classical Dirac fields are grassmann numbers because the quantum fields of electrons are Fermionic. Neglecting the fact that electrons are indeed fermions, if we started from representation of the Poincare algebra and derive the relativistic wave equations, it is clear that they are complex valued spinors. $\endgroup$ – Xiaoyi Jing Jan 28 '16 at 19:39
  • $\begingroup$ the classical Dirac field is irrelevant: we can define it whatever we want it to be. There is no use of $(-i\partial+m)\psi=0$ for a classical field $\psi$. We define it to have the properties that work better when quantised: we do know what $\hat\psi$ has to be, so we define $\psi$ so that everything works fine. Remember that relativistic wave equation are useless: $\psi$ is not a wave function. The important object is $\hat \psi$ (there is no classical limit of fermion fields, because they don't exist at the classical level) $\endgroup$ – AccidentalFourierTransform Jan 28 '16 at 19:45
  • $\begingroup$ @ AccidentalFourierTransform. $\psi$ is a wave function. What is important is that the wave we obtained from representation theory of the Poincare algebra is not the quantum mechanical probabilistic wave. Instead, it is the wave of the classical fields. It is well-known that some classical fields satisfy the Schrodinger equation. But it is not the quantum mechanical wave equation since the classical wave here does not have probabilistic interpretation. $\endgroup$ – Xiaoyi Jing Jan 28 '16 at 19:49
  • $\begingroup$ @ AccidentalFourierTransform For example, you can quantize the classical Schrodinger field, which is the non-relativistic limit of classical Klein-Gordon field. It is also the classical limit of the quantum Schrodinger field. This quantum Schrodinger field can also be regarded as a continuum limit of quantum mechanics of many body identical particles. We simply sandwich the classical Hamiltonian by creation and annihilation operators. Note that the classical Hamiltonian takes the same form as the quantum Hamiltonian of the ordinary 1-particle quantum mechanics. $\endgroup$ – Xiaoyi Jing Jan 28 '16 at 19:55
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  1. First of all, recall that a super-Lie bracket $[\cdot,\cdot]_{LB}$ (such as, e.g., a super-Poisson bracket $\{\cdot,\cdot\}$ & the super-commutator $[\cdot,\cdot]$), satisfies super-antisymmetry $$ [f,g]_{LB} ~=~ -(-1)^{|f||g|}[g,f]_{LB},\tag{1} $$ and the super-Jacobi identity $$\sum_{\text{cycl. }f,g,h} (-1)^{|f||h|}[[f,g]_{LB},h]_{LB}~=~0.\tag{2}$$ Here $|f|$ denotes the Grassmann-parity of the super-Lie algebra element $f$. Concerning supernumbers, see also e.g. this Phys.SE post and links therein.

  2. In order to ensure that the Hilbert space has no negative norm states and that the vacuum state has no negative-energy excitations, the Dirac field should be quantized with anticommutation relations $$ [\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)]_{+} ~=~ \hbar\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1} ~=~[\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)]_{+}, $$ $$ [\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)]_{+} ~=~ 0, \qquad [\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)]_{+}~=~ 0, \tag{3} $$ rather than with commutation relations, cf. e.g. Ref. 1 and this Phys.SE post.

  3. According to the correspondence principle between quantum and classical physics, the supercommutator is $i\hbar$ times the super-Poisson bracket (up to possible higher $\hbar$-corrections), cf. e.g. this Phys.SE post. Therefore the corresponding fundamental super-Poisson brackets read$^1$
    $$ \{\psi_{\alpha}({\bf x},t), \psi^{\ast}_{\beta}({\bf y},t)\} ~=~ -i\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y}) ~=~\{\psi^{\ast}_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\}, $$ $$ \{\psi_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\} ~=~ 0, \qquad \{\psi^{\ast}_{\alpha}({\bf x},t), \psi^{\ast}_{\beta}({\bf y},t)\}~=~ 0. \tag{4} $$

  4. Comparing eqs. (1), (3) & (4), it becomes clear that the Dirac field is Grassmann-odd, both as an operator-valued quantum field $\hat{\psi}_{\alpha}$ and as a supernumber-valued classical field $\psi_{\alpha}$.

  5. It is interesting that the free Dirac Lagrangian density$^2$ $$ {\cal L}~=~\bar{\psi}(\frac{i}{2}\stackrel{\leftrightarrow}{\cancel{\partial}} -m)\psi \tag{5} $$ is (i) real, and (ii) its Euler-Lagrange (EL) equation is the Dirac equation$^3$
    $$(i\cancel{\partial} -m)\psi~\approx~0,\tag{6}$$ irrespectively of the Grassmann-parity of $\psi$!

  6. The Dirac equation (6) itself is linear in $\psi$, and hence agnostic to the Grassmann-parity of $\psi$.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 3.5.

  2. H. Arodz & L. Hadasz, Lectures on Classical and Quantum Theory of Fields, Section 6.2.

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$^1$ In this answer, we are for simplicity just considering dequantization, i.e. going from a quantum system to a classical system. Normally in physics, one is faced with the opposite problem: quantization. Given the Lagrangian density (5), one could (as a first step in quantization) find the Hamiltonian formulation via the Dirac-Bergmann recipe or the Faddeev-Jackiw method. The Dirac-Bergmann procedure leads to second class constraints. The resulting Dirac bracket becomes eq. (4). The Faddeev-Jackiw method leads to the same result (4). For more details, see also this Phys.SE post and links therein.

$^2$ The variables $\psi^{\ast}_{\alpha}$ and $\bar{\psi}_{\alpha}$ are not independent of $\psi_{\alpha}$, cf. this Phys.SE post and links therein. We disagree with the sentence "Let us stress that $\psi_{\alpha}$, $\bar{\psi}_{\alpha}$ are independent generating elements of a complex Grassmann algebra" in Ref. 2 on p. 130.

$^3$ Conventions. In this answer, we will use $(+,-,-,-)$ Minkowski sign convention, and Clifford algebra

$$\{\gamma^{\mu}, \gamma^{\nu}\}_{+}~=~2\eta^{\mu\nu}{\bf 1}_{4\times 4}.\tag{7}$$ Moreover, $$\bar{\psi}~=~\psi^{\dagger}\gamma^0, \qquad (\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{8} $$ The Hermitian adjoint of a product $\hat{A}\hat{B}$ of two operators $\hat{A}$ and $\hat{B}$ reverses the order, i.e. $$(\hat{A}\hat{B})^{\dagger}~=~\hat{B}^{\dagger}\hat{A}^{\dagger}.\tag{9} $$ The complex conjugation of a product $zw$ of two supernumbers $z$ and $w$ reverses the order, i.e. $$(zw)^{\ast}~=~w^{\ast}z^{\ast}.\tag{10} $$

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  • $\begingroup$ Thanks for pointing out the super-Poisson bracket, which I've been ignored since the beginning. But please allow me to point it out that there is still a tiny deficit in the above answer. $\endgroup$ – Xiaoyi Jing Sep 25 '16 at 18:39
  • $\begingroup$ I recently learned from the book "Lectures on Classical and Quantum Theory of Fields" by Henryk Arodz and Leszek Hadasz, in section 6.2 'The Dirac Field' of Chapter 6 ‘The Quantum Theory of Free Fields’ page 131-132 that the conjugate variables of $\psi$ should be $\bar{\psi}$, instead of $\psi^{\dagger}$ since there are Dirac constraints for first order systems. $\endgroup$ – Xiaoyi Jing Sep 25 '16 at 18:45
  • $\begingroup$ @ Qmechanic You didn't correct it. The vonjugate variables of $\psi$ is not $\psi^{\dagger}$, but $\bar{\psi}$. $\endgroup$ – Xiaoyi Jing Sep 27 '16 at 2:36
  • $\begingroup$ Thank you. But I still don't see that $\psi^{\dagger}$ is replaced by $\bar{\psi}$. $\endgroup$ – Xiaoyi Jing Sep 29 '16 at 2:18
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Oct 1 '16 at 11:07

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