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A real scalar free field of mass $m$ can be represented as:

$$\hat\phi(\mathbf{x}) = \int \frac{d^3\mathbf{k}}{(2\pi)^3\sqrt{2\omega_{\mathbf{k}}}}\hat a_{\mathbf{k}}e^{i(\omega_{\mathbf{k}}t - \mathbf{k} \cdot \mathbf{x})} + \hat a^\dagger_{\mathbf{k}} e^{-i(\omega_{\mathbf{k}} t + \mathbf{k} \cdot \mathbf{x})},$$

where $\omega_{\mathbf{k}} = \sqrt{|\mathbf{k}|^2 + m^2}$.

What wavefunction does the creation operator $\hat a^\dagger_{\mathbf{k}}$ create from the vacuum state? Is it the momentum eigenstate $e^{i\mathbf{k}\cdot\mathbf{x}}$? If so, how can I show this?

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  • $\begingroup$ Suggested reading: the beginning of Hatfield covers the needed relationships between QFT and QM. $\endgroup$ – Sean E. Lake Feb 16 '18 at 0:32
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The operator create a one-particle state that is the eigenstate of the momentum. This eigenstate is not represented by $e^{ikx}$ anymore since the states $|x\rangle$ are not orthogonal, so they don't provide a orthogonal basis to represent states. Even if you fix a space-like hypersurface, the $|x\rangle$'s will still be non-orthogonal among themselves.

You can check this by too ways: by assuming that $|x\rangle=\phi(x)|0\rangle$ and then calculating $\langle x|y\rangle$ or by try to construct $|x\rangle$ as a superposition of $|k\rangle$'s, and use the fact that $|k\rangle$'s provide an orthogonal basis. The last one is based on the fact that there is a unique superpostions of $|k\rangle$'s that give a set of states that behaves under translation as $U(b)|x\rangle=|x+b\rangle$.

To check if $a_k^{\dagger}|0\rangle$ is indeed the eigenstate of momentum is simple, just apply the operator of translation $U(b)$ in $\phi(x)|0\rangle$. You know that the field behaves as $U(b)\phi(x)U^{\dagger}(b)=\phi(x+b)$ and that the vacuum state is invariant under translation $U(b)|0\rangle=|0\rangle$, together with the fact that $U(b)$ is unitary $U(b)U^{\dagger}(b)=1$.

The main lesson is: there is no one-particle wave functions for particles in the space of $x$, there is just wave functions in the space of momenta $k$. The one particle states are all spanned by this basis $\mathcal{B}=\{|k\rangle\}$. The multi-particle states will be then tensor products.

Since in QFT is natural to consider multi-particle states, is better to represent all these multi-particle states into a single object, the wave-packetal: a functional of the field configuration $\Psi[\phi]$

You may ask why the Schrodinger wave function works so well for non-relativistic regime. Turns out that if the particle is massive, $\langle x|y\rangle$ decays as $e^{-mc(y-x)/h}$ at space-like separation, so at lengths $L\gg \hbar/mc$ we have $\langle x|y\rangle\approx 0$. Under this circumstances that the Schrodinger wave function works well. Note that the non-relativistic limit in quantum mechanics implies not only $v\ll c$ but also $\Delta x \gg h/mc$. This is so due the uncertainty principle $\Delta x \Delta p \sim \hbar$.

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    $\begingroup$ It's also worth adding that $\left< x | y \right>$ vanishes for all practical purposes (that is, still nonzero, but unmeasurably small) for spacelike separations $\Delta x$ much larger than the Compton wavelength of the particle (which in natural units is $\lambda \approx m^{-1}$). Thus, it makes perfect sense to talk about localized wavepackets of size greater than $\lambda$. $\endgroup$ – Prof. Legolasov Feb 16 '18 at 11:15
  • $\begingroup$ @SolenodonParadoxus thanks for remember me that. I gonna edit my answer $\endgroup$ – Nogueira Feb 17 '18 at 0:36
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I'm adding a draft calculation here, which shows that Nogueira@'s answer is indeed correct:

We are studying a free massive Klein-Gordon QFT. The field operator is given by

$$ \hat{\phi}({\bf x}, t) = \frac{1}{(2 \pi)^3} \int \frac{d^3 {\bf k}}{\sqrt{2 \omega_k}} \left( \hat{a}_k e^{i(\omega_k t - {\bf k}{\bf x})} + \hat{a}^{\dagger}_k e^{-i(\omega_k t - {\bf k}{\bf x})} \right). $$

The bosonic Fock space is generated by the continuum of copies of the oscillator algebra: $$ \left[ \hat{a}_k, \hat{a}^{\dagger}_l \right] = (2\pi)^3 \delta^{(3)}({\bf k} - {\bf l}), $$

and the vacuum state satisfying

$$ \hat{a}_k \left| 0 \right> = 0, \quad \forall {\bf k}. $$

One-particle states:

$$ \left| {\bf k} \right> = \hat{a}^{\dagger}_k \left| 0 \right>, $$ $$ \quad \left| {\bf x} \right> = \hat{\phi}(t_0, {\bf x}) \left| 0 \right> = \frac{1}{(2 \pi)^3} \int \frac{d^3 {\bf k}}{\sqrt{2 \omega_k}} e^{i {\bf k} {\bf x}} \left| {\bf k} \right>. $$

The inner product:

$$ \left< {\bf k} | {\bf l} \right> = \left< 0 \right| \hat{a}_l \hat{a}^{\dagger}_k \left| 0 \right> = \left< 0 \right| \left( \hat{a}^{\dagger}_k \hat{a}_l + (2\pi)^3 \delta^{(3)} ({\bf k} - {\bf l}) \right) \left| 0 \right> = (2\pi)^3 \delta^{(3)} ({\bf k} - {\bf l}). $$

$$ \left< {\bf x} | {\bf y} \right> = \frac{1}{(2 \pi)^6} \int \frac{d^3 {\bf k} \, d^3 {\bf l}}{\sqrt{4 \omega_k \omega_l}} e^{i (-{\bf k} {\bf x} + {\bf l} {\bf y})} \left< \bf{k} | \bf{l} \right> = \frac{1}{(2 \pi)^3} \int \frac{d^3 {\bf k} \, d^3 {\bf l}}{\sqrt{4 \omega_k \omega_l}} e^{i (-{\bf k} {\bf x} + {\bf l} {\bf y})} \delta^{(3)}({\bf k} - {\bf l}) = $$ $$ \frac{1}{(2 \pi)^3} \int \frac{d^3 {\bf k}}{2 \omega_k} e^{i {\bf k}({\bf y} - {\bf x})} = \frac{1}{2 (2\pi)^3} \intop_0^{\infty} d\rho \intop_0^{\pi} d \theta \intop_0^{2\pi} d\varphi \, \frac{\rho^2 \sin{\theta}e^{i \rho \Delta x \cos \theta}}{\rho^2 + m^2} = $$ $$ \frac{1}{8 \pi^2} \intop_0^{\infty} d\rho \intop_0^{\pi} d \theta \, \frac{\rho^2 \sin{\theta}e^{i \rho \Delta x \cos \theta}}{\rho^2 + m^2} = \frac{1}{4 \pi^2} \intop_0^{\infty} d\rho \, \frac{\rho \sin (\rho \Delta x)}{\Delta x (\rho^2 + m^2)} = \frac{e^{- m \Delta x}}{8 \pi \Delta x}. $$

$$ \left< {\bf x} | {\bf y} \right> = \frac{e^{-m \left| {\bf x} - {\bf y} \right|}}{8 \pi \left| {\bf x} - {\bf y} \right|} \neq \text{const} \cdot \delta^{(3)}({\bf x} - {\bf y}) !!! $$

The final formula agrees with what Nogueira@ wrote in his answer, and his interpretation follows from this.

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