7
$\begingroup$

In deriving the Klein Gordon equation one starts out with the relativistic energy relation $E^2 = p^2 + m^2$ and substitutes the quantum momentum operator that corresponds to non-relativistic QM, $\hat{p}= -i \frac {∂}{∂x}$ . I was wondering what justifies the use of this when this quantum operator was derived non-relativistically, but the $p$ in the $E^2$ equation is relativistic. I can't imagine how one would go about deriving a quantum operator corresponding to $\gamma mv$.

$\endgroup$
7
  • $\begingroup$ The momentum operator is not derived (neither in non-relativistic QM nor in relativistic QM). The correspondance $\boldsymbol p\to-i\nabla$ is more or less a what if. KG is not derived, but postulated (the same applies to Schrodinger). Always remember that you can derive Classical Mechanics from QM, and not the other way around. $\endgroup$ Dec 14, 2015 at 19:24
  • $\begingroup$ NB: Relativistic quantum mechanics is not quantum field theory (since you tagged it that way). $\endgroup$
    – ACuriousMind
    Dec 14, 2015 at 19:26
  • $\begingroup$ @AccidentalFourierTransform I see what you are saying but why would we accept the postulate of the momentum operator having the same form whether relativistic or not, when it is clear that relativity has a different form of momentum from classical. $\endgroup$ Dec 14, 2015 at 19:33
  • 4
    $\begingroup$ @MarcusQuinnRodriguezTenes we dont accept the postulate just because. We do because it leads certain predictions, which are afterwards validated by experiment. $\endgroup$ Dec 14, 2015 at 19:35
  • 1
    $\begingroup$ @MarcusQuinnRodriguezTenes The form of the 3-momentum does not change when going from nonrelativistic to relativistic physics. What changes is how momentum transforms between frames. $\endgroup$
    – Praan
    Dec 15, 2015 at 17:07

5 Answers 5

3
$\begingroup$

@knzhou is correct for the translation aspect, but it is also possible to derive this equality from canonical quantization.

It is easiest to see by considering the Polyakov action for a relativistic particle

$$S = \frac 12 \int d\tau\ \frac{1}{e} \dot x^\mu \dot x_\mu - e m^2$$

Then picking a time direction, we can define the momenta

$$p_\mu = \frac{\dot x_\mu}{e}$$ $$p_e = 0$$

With the Hamiltonian

\begin{eqnarray} H &=& \frac 12 [p_\mu x^\mu(p) - \frac{1}{e} \dot x^\mu \dot x_\mu + e m^2]\\ &=& \frac e2( p^\mu p_\mu + m^2)\\ &=& \frac e2 (E - p^2 + m^2) \end{eqnarray}

After some gauge fixing where we set $e = 1$, and some use of constraints, this is just equivalent to $E = p^2 + m^2$. Furthermore, we need to compute the Poisson brackets for the basic phase space quantities. This will be

\begin{eqnarray} \left\{x^\mu, p_\nu \right\} = \delta^\mu_{\nu} \end{eqnarray}

since $x$ and $p$ have the same definition as in non-relativistic mechanics. From there, we simply apply the Dirac quantization

$$\left\{x^\mu, p_\nu \right\} \to \frac{1}{i\hbar} [\hat x^\mu, \hat p_\nu]$$

or in other words,

$$[\hat x^\mu, \hat p_\nu] = \delta^\mu_\nu i\hbar$$

via the Stone-von Neumann theorem, we know that the following are representations of such operators :

\begin{eqnarray} \hat x^\mu \psi &=& x^\mu \psi\\ \hat p_\nu \psi &=& -i\hbar \partial_\nu \psi \end{eqnarray}

$\endgroup$
1
$\begingroup$

The momentum operator is defined as the generator of translations, so we have $$\hat{p} = - i \partial_x$$ in all cases, relativistic or not, in the same way that the Hamiltonian always generates time translations, $H |\psi \rangle = i \partial_t |\psi \rangle$.

The only thing that relativity changes is the relationship between $\hat{p}$ and the velocity, i.e. we have $$\hat{p} = \gamma m \hat{v}$$ where the velocity operator is defined as $$\hat{v} = \frac{\partial H(p, q)}{\partial p}\bigg|_{p = \hat{p}, q = \hat{q}}$$ and gives the local velocity of the particle. So if we had instead tried to define $\hat{p} = m \hat{v}$ in all cases it would indeed be incompatible with relativity, but that's not what we're doing.

$\endgroup$
1
$\begingroup$

The operator $ \hat{p}=-i \hbar \frac {∂}{∂x} \tag{1}$ is valid in both cases.

We end up with this operator my cascading from the math of what we consider to be a “well behaved” wave.

From De Broglie’s wave mechanics we could associate to a particle with momentum $p$ a wave with wave number $k$ linked to the momentum by $p=\gamma mv=h/ \lambda=\hbar k \tag{2}$

Equation (2) is relativistic.

To this point no one knows (“knew?”) what exactly that wave or its actual form are. However, for its form we could postulate it. Starting from a plane wave $$ψ=e^{i(kx+ω t)} \tag{3}$$ The derivative of this equation with respect to $x$ is $$\frac {∂ }{∂x} ψ =ikψ \tag{4}$$

Defining the operator $$\hat{k}=-i\frac {∂ }{∂x} \tag{5}$$

one gets the Schrodinger equation for the operator $\hat{k}$, or $$\hat{k}ψ =kψ \tag{6}$$ which is always valid, as long as we have the plane wave in mind when “deriving” our wave mechanics. Now, replacing $k$ by $\hat{k}$ in (2) and considering that that equation is valid in both situations, the operator (1) can be used in both cases for the same reasons.

PS: the operator (1) or (5) are not actually derived. They are postulated and checking their consequences with the experimental results will improve/reduce their merits.

$\endgroup$
0
$\begingroup$

You are absolutely right. It is necessary to replace the usual pulse with the momentum operator $p_{Nonrelativistic} \mapsto -i\hbar \triangledown$ in the expression for the coupling of the energy and momentum of the SRT, then we obtain a completely different equation M2. Which has no flaws in the Klein-Gordon equation.
$$ \Delta \Psi -\frac{1}{\hbar^{2}}\left [ \frac{m^{4}c^{6}}{\left ( E-U\left ( \overrightarrow{r}\right ) \right )^{2}}-m^{2}c^{2} \right ]\Psi =0 $$

$\endgroup$
0
$\begingroup$

In the first place $p=\gamma m v$ is only valid when one ignores interactions. For a particle in an electromagnetic field, the momentum is $p=\gamma m v + eA$

In the second place one wouldn't confound an operator $\hat{p}$ with an eigenvalue $p$. Both are related by

$$\hat{p} \Psi = p \Psi$$

For free particles, the value of the momentum $p$ will be $p=\gamma m v$ or $p=m v$ depending of the state of the particle, but the operator is the same

$$\hat{p} =-i \hbar \frac {∂}{∂x}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.