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Using Peskin+Schroeder as a reference. Bear with me, there may be multiple mistakes in my discussion. But the underlying question should be clear - it's really just the title.

By analyzing the Lagrangian, we get that the fermion field operator in QFT has units of $\mathrm{Energy}^{3/2}$. Then the field operator eigenmode decomposition is (3.99): $$ \psi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}\sum_s\left( a_\mathbf{p}^s u^s(p)e^{-ip\cdot x}+ b_\mathbf{p}^{s\dagger} v^s(p)e^{ip\cdot x} \right).\tag{3.99} $$ Since $u$ and $v$ have units of $\mathrm{Energy}^{1/2}$, the particle creation operators must have units $\mathrm{Energy}^{-3/2}$. This is also confirmed in the anticommutation relation. But then there's this line: "The one-particle states (are:)" (3.106) $$ |\mathbf{p},s\rangle\equiv\sqrt{2 E_\mathbf{p}} a_\mathbf{p}^{s\dagger}|0\rangle.\tag{3.106} $$ If states have the same units all the time, then the left and right side have different units. So maybe states have different units if they have different numbers of particles? Later in Eq. 3.107, the inner product is given, which is another hint:

$$ \langle \mathbf{p},r|\mathbf{q},s\rangle = 2E_\mathbf{p}(2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q})\delta^{rs}.\tag{3.107} $$ Here we can at least see that the units of the right hand side are $\mathrm{Energy}^{-2}$, so the states and their adjoints don't have opposite units. Certainly not how it works in quantum mechanics, where inner products are unitless.

But I have another more wild suggestion. For this I need to switch to real-valued scalar fields. The state in QFT can be expressed as a functional, and inner products must be integrated along all possible field values for all points in space: $$ \langle a|b\rangle=\int d\phi(x_1)\int d\phi(x_2)\cdots a[\phi(x_1),\phi(x_2),\cdots]b^\dagger[\phi(x_1),\phi(x_2),\cdots]. $$ This would seem to indicate that the functionals $a$ and $b$ have units of $\mathrm{Energy}^{-\infty}$, since (I think?) this expression should evaluate to some finite power of energy, and the scalar field has units of energy. This leads me to the silliest resolution - Eq. 3.106 is consistent because both sides have units of $\mathrm{Energy}^{-\infty}$.

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This is one of those annoying things about relativistic quantum field theory. You pick up extra dimensions from the continuum normalization, and even more dimensions from the conventional factors of $\sqrt{2E}$ sprinkled all over the place (which are conceptually useful, as discussed here).

Here are some facts that should resolve the confusion:

  • The vacuum state is definitely normalized as usual, $\langle 0 | 0 \rangle = 1$, so $[|0 \rangle] = 0$.
  • The projection operator onto the one-particle Hilbert space is $$1_{\mathrm{one-particle}} = \int \frac{d^3 \mathbf{p}}{(2 \pi)^3 \, 2 E_p} \, |\mathbf{p} \rangle \langle \mathbf{p} |$$ from which we read off $[|\mathbf{p} \rangle] = -1$.
  • Repeating the same logic for the two-particle Hilbert space, we have $[|\mathbf{p}, \mathbf{q}\rangle] = -2$, and so on.
  • Mechanically, this happens because you add particles using $\sqrt{2 E_p} a_{\mathbf{p}}^\dagger$. We have $[a_{\mathbf{p}}] = -3/2$ from continuum normalization, so the overall dimension is $-1$.

I wouldn't recommend using the path integral to infer units. There's a lot of stuff hidden in the path integral measure, and the path integral only gives amplitudes; you'd have to do more work to infer units of states.

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  • $\begingroup$ Technically, the equation $\langle 0 | 0 \rangle=1$ doesn't imply that $ | 0 \rangle$ and $\langle 0 |$ both have units $E^0$ right? Because in (3-D single particle) quantum mechanics $ | \psi \rangle$ has units of distance$^{-3/2}$ (well at least the wave function does) and the adjoint has units of distance$^{+3/2}$. That is to say, the state and its adjoint can have different units. $\endgroup$ Commented Oct 16 at 18:18
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    $\begingroup$ @AXensen No, in the usual NRQM setup a generic physical state $|\psi \rangle$ is dimensionless. The wavefunction is dimensionful because it's $\psi(x) = \langle x | \psi \rangle$ and the unphysical basis states $|x \rangle$ are dimensionful because of continuum normalization, $\langle x | y \rangle = \delta(x-y)$. I guess you could have a different convention where kets and bras have different dimensions, but I haven't heard of that used before. $\endgroup$ Commented Oct 16 at 18:23
  • $\begingroup$ Yeah... you're right about that. Thanks for the clarification. $\endgroup$ Commented Oct 16 at 18:32
  • $\begingroup$ @AXensen It's a fair confusion, e.g. in differential geometry the one-form $dr$ (in spherical coordinates) has dimension $1$ while the basis vector $\partial / \partial r$ has dimension $-1$. I just haven't seen analogous things done in QM. $\endgroup$ Commented Oct 16 at 18:45
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+1 to knzhou's answer but I would like to expand a little about it.

Absolutely also go and check knzhou's own link to his earlier answer, where he discusses both the relativistic $_\text{rel}\!\left<x\right|$ and Newton-Wigner $_\text{NW}\!\left<x\right|$ "position eigenstates" because the latter is the thing that connects smoothly with quantum mechanics and will also be the one that brings in the units to the standard representation of wavefunctions, as you likely have already seen in the comments.

However, now I would like to invoke David Tong's notes where, right after Equation (2.141), he points out that this has something to do with conventions, because some other people define the creation and annihilation operators differently.

Following his notes religiously, though, it is seen that $\left<\text{vac}|\text{vac}\right>=1$ is taken as a given, for both NR QM and SR QFT. In Equations (2.85) and (2.87), it can be seen that in the QM QHO situation, the linear combination of $a$ and $a^\dagger$ implies that they have the same units, and $[H]=[\omega],\quad[q]=\left[\dfrac1{\sqrt\omega}\right],\quad[p]=[\sqrt\omega]$ and then those imply that $[a]=1=[a^\dagger]$.

Beware that the awkward units of $q$ and $p$ are because in Equation (2.83) the mass term is absorbed into them, causing the mayhem in the units that is really not there. In the later bits, we will see that this is not what David Tong himself uses, because it is extremely inconvenient in SR QFT to have the masses absorbed this way.

That was QM QHO. Now we move to SR QFT. Checking with Equations (2.93) and (2.94), we see yet again that $[a]=[a^\dagger]$ and then Equation (2.97) strongly suggests that $[H]=[\omega]$ and thus $[a]=[a^\dagger]=\dfrac1{\sqrt{[p]^3}}$. We can then check by substituting this guess into Equations (2.93) and (2.94) and comparing with Equation (2.95) and see that the units now all check out. To fix the units of momentum and position, we note that Equation (2.135), which is just the Einstein energy-momentum-mass relation, asserts that $[p]=[H]=[E]=[m]=[\omega]$ and then that would mean that in Equation (2.95) we must have $[q]=[x]=\dfrac1{[H]}$

This is then internally self-consistent, because when you substitute all of these into each other, you will realise that $[\phi(x)]=[H]$ and $[\pi(x)]=[H]^2$ and that means that in the horribly big expression of $$H=\frac12\int\mathrm d^3x\ \pi^2+(\nabla\phi)^2+m^2\phi^2$$ everything is dimensionally consistent, and all is well.

This also means that the other convention that David Tong briefly mentioned, would have $[a]=1=[a^\dagger]$ and that is also nice in its own way.

N.B. David Tong's convention disagrees with knzhou at Equation (2.117), where David Tong has $[\left|\vec p\right>]=\dfrac1{\sqrt{[H]^3}}$ and the reason for this is that he is using Equation (2.133) instead of what knzhou used.

What I really hope to impress upon you, is that this thing is very much convention-dependent, and you might want to cross-check with the convention that you might be using, whether they agree or disagree.

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