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Let me start from path integral formulation in quantum mechanics and quantum field theory. In QM, we have $$ U(x_b,x_a;T) = \langle x_b | U(T) |x_a \rangle= \int \mathcal{D}q e^{iS} \tag{1} $$ $|x_a \rangle$ is an eigenstate of position operator $\hat{x}$.

In QFT we have $$ U(\phi_b,\phi_a;T) = \langle \phi_b | U(T) |\phi_a \rangle= \int \mathcal{D}\phi e^{iS} \tag{2} $$ $| \phi_a \rangle $ is an eigenstate of field operator $\hat{\phi}(x)$.

By analogy with QM, it is tempting to relate $$|\phi \rangle \leftrightarrow |x \rangle \tag{3} $$

However, in Peskin and Schroeder's QFT, p24, by computing it is said

$$ \langle 0 | \phi(\mathbf{x}) | \mathbf{p} \rangle = e^{i \mathbf{p} \cdot \mathbf{x}} \tag{2.42} $$ We can interpret this as the position-space representation of the single-particle wavefunction of the state $| \mathbf{p} \rangle$, just as in nonrelativistic quantum mechanics $\langle \mathbf{x} | \mathbf{p} \rangle \propto e^{i \mathbf{p} \cdot \mathbf{x}} $ is the wavefunction of the state $|\mathbf{p}\rangle$.

Based on the quoted statement, seems $$\hat{\phi}(x) | 0 \rangle \leftrightarrow | x \rangle \tag{4} $$

If relations (3) and (4) are both correct, I should have $$\hat{\phi} ( \hat{\phi} | 0 \rangle ) = \phi(x) ( \hat{\phi}| 0 \rangle ) \tag{5}$$ seems Eq. (5) is not correct. At least I cannot derive Eq. (5).

How to reconcile analogies (3) and (4)?

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It's not clear what you're trying to write with eq. 5, how are you defining $\phi$ without a hat? –  bechira Jul 23 at 13:22
    
I mean $\hat{\phi}(x_1) | \phi_1 \rangle = \phi_1(x_1) | \phi_1 \rangle$, without hat is the eigenvalue of field operator. –  user26143 Jul 23 at 13:30
1  
Gotcha. No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2 |0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct. –  bechira Jul 23 at 14:36
    
Thanks. So $\langle 0 | \hat{\phi}(x) | p \rangle$ is just an analogy of $\langle x | p \rangle$. It does not imply $\hat{\phi}(x) | 0 \rangle$ is an eigenstate of $\hat{\phi}(x)$ anyway. –  user26143 Jul 23 at 14:43
    
Yes that's right. –  bechira Jul 23 at 15:04

2 Answers 2

up vote 2 down vote accepted
  1. No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct

  2. To obtain some analogy of $| x\rangle$, you can just take a fourier transform of $a^\dagger(p)$ to get $a^\dagger(x)$, and $a^\dagger(x)|0 \rangle \equiv |x \rangle$ is the best analogy of $|x \rangle$ that I can think of

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You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230.

Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$).

With slightly more precision: the Fock space is isomorphic to an $L^2$ space where $\hat{\phi}$ acts as the multiplication by the function $x$ (is a "variable" of the $L^2$ space), and $\hat{\pi}$ as the (functional) derivative $-i\frac{\delta}{\delta x}$; and in this context you can define the "eigenfunctions" (they do not belong to the $L^2$ obviously) $\lvert\phi\rangle$ and $\lvert\pi\rangle$ with the usual meaning as (infinite dimensional) position and momentum eigenfunctions. The precise construction is detailed in the reference above.

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