How does a Dirac spinor such as:

$$ \psi = \pmatrix{a_0+ib_0\\a_1+ib_1\\a_2+ib_2\\a_3+ib_3} $$

Connect to a probability?

Can one apply the Born rule of this object?

  • 1
    $\begingroup$ I thought this had already been asked/answered somewhere on this site, but I didn't find it in a quick search. In the meantime, here's a related question that I happen to remember because I answered it. It doesn't answer your question directly, but it gives some perspective. $\endgroup$ Jun 28, 2021 at 4:48
  • $\begingroup$ Related: Interpretation of Dirac equation states $\endgroup$
    – Ruslan
    Jun 28, 2021 at 9:26
  • $\begingroup$ The issue have been treated for the Hydrogen atom in my paper arXiv:2010.06375 $\endgroup$ Jan 8, 2023 at 9:32

6 Answers 6


The following has been taken from Greiner' s Relativistic Quantum Mechanics where he makes it clear the Dirac equation can be thought of as a relativistic wave equation. This answer is a precursor to how does it connect to probability (Chapter $2$).

To shorten some of the arguments presented:

We define

$$ \hat \alpha_i = \begin{pmatrix} 0 & \hat \sigma_i \\ \hat \sigma_i & 0 \end{pmatrix} $$

And $\hat \beta$ as:

$$ \hat \beta= \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix} $$

where $\sigma_i$ is the $i$'th $2 \times 2$ Pauli matrix and $I$ is the identity matrix. We wish to construct a four-current density for the Dirac equation:

$$ i \hbar \frac{\partial \psi}{\partial t}= \frac{\hbar c}{i} \sum_{i =1}^3 \hat \alpha_k \frac{\partial}{\partial x^k} \psi + m_0 c^2 \hat \beta \psi $$

Multiplying the above with a Hermitian conjugate:

$$ i \hbar \psi^\dagger \frac{\partial \psi}{\partial t}= \frac{\hbar c}{i} \sum_{i =1}^3 \hat \psi^\dagger \alpha_k \frac{\partial}{\partial x^k} \psi + m_0 c^2 \hat \psi^\dagger \beta \psi $$

Taking the conjugate of the Dirac equation and multiplying $\psi$. We get $2$ equations:

$$ - i \hbar \frac{\partial \psi^\dagger}{\partial t} \psi = - \frac{\hbar c}{i} \sum_{i =1}^3 \hat \alpha_k \frac{\partial \psi^\dagger }{\partial x^k} \psi + m_0 c^2 \hat \beta \psi^\dagger \psi $$

Substracting the above $2$ equations:

$$ i \hbar \frac{\partial \psi^\dagger \psi }{\partial t} = \frac{\hbar c}{i } \sum_{i=1}^k \frac{\partial}{\partial x^k} (\psi^\dagger \hat \alpha_k \psi)$$

(Note $\alpha^\dagger = \alpha$ and $\beta^\dagger = \beta$)

Hence this is precisely of the form:

$$ \frac{\partial \rho}{\partial t} + \nabla j = 0$$

where $\rho = \psi^\dagger \psi $ and $j = (\psi^\dagger \hat \alpha_k \psi) $

Since $ \rho $ is positive definite we accept $j$ as a probability density current.


As soon as you're dealing with relativistic equations, you're no longer in the realm of 1-particle quantum mechanics.

It is a basic preliminary result of quantum field theory that in order to have relativistic causality preserved, you must include both positive and negative frequencies in the amplitude. There is a classic piece of literature where Feynman explains this concept at length: Elementary Particles and the Laws of Physics: The 1986 Dirac Memorial Lectures.

But the same material is covered in any book or lecture notes on QFT. Look for the first chapters where the authors explain the different propagators. Fundamentally, this is because any relativistic equation worth its salt must satisfy Einstein's energy-momentum identity, $$ E^{2}-c^{2}\boldsymbol{p}^{2}=m^{2}c^{4} $$ which gives you two solutions for the dispersion relation, $$ E=\pm\sqrt{m^{2}c^{4}+c^{2}\boldsymbol{p}^{2}} $$ And Dirac's equation is no exception. It is essentially the "square root" of Klein-Gordon's equation.

As @josephh has said, a 4-spinor is not a wave function. Not even the more fundamental Weyl 2-spinors (of which Dirac 4-spinors are essentially direct sums) are wave functions. If you try to build a quadratic bilinear that carries statistical information about the state, you must give up positivity. This forces you to re-interpret the most natural quadratic bilinear as a number operator, instead of a probability density. This number becomes a charge density operator (which can be either positive or negative), the amplitudes become operators instead of states, and the 1-particle states are more fundamental objects on which these space-time dependent operators act.

  • $\begingroup$ This does not really answer the OP's question. One can perform an experiment in which an electron is detected with a certain probability at a given location in space and time. How does one predict (or calculate) that that probability, given a specific state for the electron? $\endgroup$ Jun 28, 2021 at 12:41
  • $\begingroup$ If a spinor is a vector of 4 complex numbers, then the complex conjugate of this vector with itself should produce a positive-definite measure $\psi^H\psi > 0$ - is this not a probability? $\endgroup$
    – Anon21
    Jun 28, 2021 at 15:13
  • $\begingroup$ You're right, @flippiefanus. I missed the 1st question. I did, though, answer the 2nd one, I hope. There is a formal answer to the 1st question that I'm aware of, although I don't think it's a very oft-trodden path, to say the least. Furthermore, a sensible position operator in relativistic QFT is very problematic to define, if at all possible. Mind you, though, the OP is not talking about position. $\endgroup$
    – joigus
    Jun 28, 2021 at 15:16

Sure you can! This is like the whole point of spinors!

For simplicity sake let's think about particles with spin $1/2$ (like electrons): Spinors are vectors in the Hilbert space of spin, you can write them as column vectors

$$\chi = \begin{bmatrix} a\\b\\\end{bmatrix} \ \ \ \ \ \ a,b \in \mathbb{C}$$

or equivalently you can write them as a linear combination of basis vectors

$$ \chi = a |+\rangle + b |-\rangle $$

And this last object is exactly one for which Born's rule applies! The probabilities of measuring the spin along one direction, up or down, are simply derived with the Born's rule (assuming of course normalized basis vectors): just take the module squared of the component and you are done!

  • $\begingroup$ You're referring to a non-relativistic treatment. I think the OP is concerned with the relativistic setting. $\endgroup$ Jun 28, 2021 at 9:22
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    $\begingroup$ The space of spinors is formally defined as the fundamental representation of the Clifford algebra. They do not represent states in Hilbert space. $\endgroup$
    – joseph h
    Jun 28, 2021 at 10:30

The Born rule, in its simplest form, tells us that the probability (density) of finding a particle at a given point is the square of the particle's wave function at that point. That is, $$\rho = \mid \psi (x,t)\mid^2$$ which is always a positive real number.

As such, the Dirac spinor is not a wave function. They do not even represent traditional states in a Hilbert space. By convention, spinors are taken to be the elements of a representation space of the Dirac algebra.

But anyway, from the fermion 4-current $$j^{\mu}=\bar \psi\gamma^{\mu}\psi$$ and if we take the $\mu=0$ component, we define the probability density as $$j^0=\rho=\bar \psi\gamma^{0}\psi = \psi^{\dagger} \psi$$ and this quantity is not always positive definite.

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    $\begingroup$ Can you explain why $\psi^\dagger \psi$ is not always positive definite? If $\psi$ are four complex numbers, I would think $\psi^\dagger \psi$ is a sum of 4 complex norm which should be positive definite? $\endgroup$
    – Anon21
    Jun 29, 2021 at 10:22

When treating the Dirac equations as a classical equation, one can find plane wave solutions with internal degrees of freedom that are represented as Dirac spinors. Let's denote them as: $$ \psi = u_s \exp(-ipx) - \text{fermion} $$ $$ \phi = v_s \exp(ipx) - \text{anti-fermion} , $$ where $s$ represents the spin index. These solutions are combined with ladder (annihilation and creation) operators in the quantization of the fermion field to define field operators. These ladder operators are often denoted as $\hat{a}_s(p)$ and $\hat{a}_s^{\dagger}(p)$ for the fermions and $\hat{b}_s(p)$ and $\hat{b}_s^{\dagger}(p)$ for the anti-fermions. When the creation operator is applied to the vacuum it produces a state with the properties associated with the Dirac spinor solution $$ |p_s\rangle = \hat{a}_s^{\dagger} | vac\rangle . $$ However, note that $|p_s\rangle$ is not a well-defined state in the infinite dimensional case, because it is not normalizable.

One can now use these ladder operators to define states. An arbitrary single particle fermion state would be created by an operator $$ \hat{a}_F^{\dagger} = \sum_s \int \hat{a}_s^{\dagger}(p) F_s(p)\ \text{d}p , $$ where $F_s$ is an arbitrary normalized spectral coefficient function. The state that is produced by this operator is simply produced by $$ |\psi\rangle = \hat{a}_F^{\dagger} | vac\rangle . $$

Although the concept of a wave function was introduced with the Schroedinger equation, the wave function is today understood in a broader sense as the function that is produced from a state when projected onto a coordinate basis. In other words, the wave function is given by $$ \psi(x) = \langle x|\psi\rangle . $$ The nature of the coordinate basis is such that when they are used with the individual spinor states, the result is the original Dirac spinor solutions $$ \langle x|p_s\rangle = u_s \exp(-ipx) . $$ In that sense, the Dirac spinor solution is the (unnormalizable) wave function of the individual plane wave states. Because it is not normalizable, it does not represent a physical wave function. However, all physical wave functions can be represented as superpositions of these Dirac spinor solutions: $$ \psi(x) = \sum_s \int u_s \exp(-ipx) F_s(p)\ \text{d}p . $$

What about a measurement? A measurement is represented by $$ \langle \hat{A}\rangle = \langle\psi| \hat{A}|\psi\rangle , $$ where $\hat{A}$ is the observable represented by a Hermitian operator. It can be the number operator defined in terms of the fermion field operators $$\hat{n}(x) = \Psi^{\dagger}(x)\Psi(x) . $$ When you substitute this operator into the expression for the measurement and apply it to the single particle fermion state you'll obtain an expression for the probability distribution of observing a single electron at a given location $x$.


The Born rule is the probability postulate of quantum mechanics.

Postulates are extra axioms used by physics theories in order to pick those mathematical solutions that will describe data and predict future data.

It does not apply on the wavefunctions, it defines the way wavefunctions predict the behavior of data by giving the probability distribution of the data by $Ψ^*Ψ$.

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    $\begingroup$ OP is asking whether or not this rule can be applied to spinors. $\endgroup$
    – joseph h
    Jun 28, 2021 at 5:30
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    $\begingroup$ @josephh Since spinors are a part of solutions of quantum mechanical equations and all quantum mechanical solutions have to follow the quantum mechanical postulates it obviously applies for $Ψ^*Ψ$ . . The "object" does not follow the Born rule, the $Ψ^*Ψ$ has to obey it axiomatically is what I am saying. $\endgroup$
    – anna v
    Jun 28, 2021 at 6:24
  • $\begingroup$ Are your $\Psi$'s here spinors or wave functions? $\endgroup$
    – joseph h
    Jun 28, 2021 at 6:27
  • $\begingroup$ @josephh The dirac wave equeation solutions afaik are spinors $\endgroup$
    – anna v
    Jun 28, 2021 at 8:27
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    $\begingroup$ @josephh This is your personal view. "Soon after in 1928, Dirac found an equation from the first successful unification of special relativity and quantum mechanics applied to the electron, now called the Dirac equation. In this, the wave function is a spinor " en.wikipedia.org/wiki/Wave_function#Historical_background $\endgroup$
    – anna v
    Jun 28, 2021 at 9:58

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