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I am confused about writing a Fock state in the momentum basis. Specifically, normalizing the wave function. Below summarizes my confusion:

Let's say we have a state $|\psi \rangle$ represented in the Fock basis as:

$$|\psi \rangle = |n\rangle$$

where $n$ is the particle number. Imagine this state describes a cluster of particles moving in space (for now assume all particles share the same momentum) and we wish to write it in the momentum basis. We can write:

$$|\psi\rangle = \sum_{\mathbf{k}}p_{\mathbf{k}}|n_{\mathbf{k}}\rangle$$

where $p_{\mathbf{k}}$ is the probability amplitude of finding the cluster of particles in momentum $\mathbf{k}$. This equation states that the particle cluster is in a superposition of momenta, but any one measurement of the particles will yield one particular momentum. Now suppose we wish to transision this to the continuous domain. Due to the integration measure $d^3k$ we must introduce a constant proportional to the quantization volume $V$ to be dimensionally correct (no dimensions). Namely:

$$|\psi\rangle = \alpha V \int p(\mathbf{k})|n_{\mathbf{k}}\rangle d^3 k$$

where $\alpha$ is some unitless constant and we have transitioned $p_{\mathbf{k}}$ (discrete) into $p(\mathbf{k})$ (continuous).

Let's calculate $\langle \psi | \psi \rangle$ to check if we get 1, and if we don't let's normalize.

$$\langle \psi | \psi \rangle = \alpha^2 V^2 \int\int p(\mathbf{k})p(\mathbf{k}')^*\langle n_{\mathbf{k}'}|n_{\mathbf{k}}\rangle d^3 k d^3 k' = \alpha^2 V^2 \int |p(\mathbf{k})|^2 d^3 k$$

So now, since $|p(\mathbf{k})|^2$ is a probability amplitude, the integral should be equal to 1 since the integral is over all $\mathbf{k}$. But if this occurs, then in order to obtain 1 for the entire inner product result, it must be the case that $\alpha = 1/V$, and if this is the case, then we get:

$$|\psi\rangle = \int p(\mathbf{k})|n_{\mathbf{k}}\rangle d^3 k$$

which is no longer dimensionless and goes against our earlier assertion that $\alpha$ is dimensionless. I feel as though I am missing something here...

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Firstly, the reason for the problem is a contradiction in the dimension of $p(\mathbf{k})$. In your definition

$$|\psi\rangle = \alpha V \int p(\mathbf{k})|n_{\mathbf{k}}\rangle \mathrm{d}^3 k$$

you implicitly assume that $p(\mathbf{k})$ is dimensionless (otherwise there would be no need for the $V$). But if $p(\mathbf{k})$ is dimensionless, then we cannot have

$$\int |p(\mathbf{k})|^2 \mathrm{d}^3 k = 1$$

because the LHS has the wrong dimensions!

So how should it be done? The way in which we transform to the continuous limit is by making the replacement

$$\sum_{\mathbf{k}} \rightarrow \frac{V}{(2\pi)^3}\int \mathrm{d}^3\mathbf{k}$$

where the pre-factor $V/(2\pi)^3$ is the density of points in $\mathbf{k}$-space (which can be found by e.g. considering periodic boundary conditions). If we do this, everything should work automatically, so we have

$$|\psi\rangle = \sum_{\mathbf{k}}p_{\mathbf{k}}|n_{\mathbf{k}}\rangle \rightarrow \frac{V}{(2\pi)^3}\int \mathrm{d}^3\mathbf{k} \, p({\mathbf{k}}) |n({\mathbf{k})}\rangle$$

and the normalization becomes

$$ 1 = \sum_{\mathbf{k}} \lvert p_{\mathbf{k}} \rvert^2 = \frac{V}{(2\pi)^3}\int \mathrm{d}^3\mathbf{k} \, \lvert p({\mathbf{k}})\rvert^2. $$

We must also change how the states are normalized in a similar way with the replacement

$$\langle n_{\mathbf{k'}} \vert n_{\mathbf{k}} \rangle = \delta_{\mathbf{k'},\mathbf{k}} \rightarrow \langle n({\mathbf{k'}}) \vert n({\mathbf{k}}) \rangle = \frac{(2\pi)^3}{V} \delta^{(3)}(\mathbf{k'}-\mathbf{k}) $$

in order to be consistent with the change from sum to integral. Finally, let's check if $\lvert \psi\rangle$ is normalised. We get

$$\langle\psi\vert\psi\rangle=\frac{V}{(2\pi)^3}\int \mathrm{d}^3\mathbf{k'} \frac{V}{(2\pi)^3}\int \mathrm{d}^3\mathbf{k} \,p(\mathbf{k'})^*p(\mathbf{k}) \underbrace{\langle n(\boldsymbol{k'})\vert n(\boldsymbol{k})\rangle}_{\frac{(2\pi)^3}{V} \delta^{(3)}(\mathbf{k'}-\mathbf{k})}=\frac{V}{(2\pi)^3}\int \mathrm{d}^3\mathbf{k} \, \lvert p({\mathbf{k}})\rvert^2=1.$$

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  • $\begingroup$ This is perfect, thanks so much. I completely understand now :) $\endgroup$
    – user41178
    Jan 9, 2023 at 2:29

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