# How is the initial condition term $\frac{1}{\sqrt{2\pi}}\phi(k)$ derived for free particles?

I'm reading Griffiths' introduction to quantum mechanics. It is explained that solutions to the Schrodinger equation are expressed as linear combinations of separable solutions $$\{\psi_n(x)\}$$. $$\Psi(x,0)=\sum_{n=1}^{\infty}c_n\psi_n(x)e^{-i\frac{E_n}{\hbar}t}$$ While I understand that $$c_n$$ represents the wavefunction's initial conditions at $$t=0$$, its analogue in the case of a free particle $$\phi(k)$$ baffles me.

Correct me if I'm wrong. According to Griffiths, the transition from $$\Psi_k(x,t)=A\exp\bigg[i\bigg(kx-\frac{\hbar k^2}{2m}t\bigg)\bigg]\implies\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\exp\bigg[i\bigg(kx-\frac{\hbar k^2}{2m}t\bigg)\bigg]dk$$ is written in a way so that the $$\Psi(x,t)$$ is now normalizable. However, the free particle cannot exist in a stationary state and, hence, no longer has a definite energy, so each term does not represent anything physical since $$k$$ is continuous.

My question is: can someone can clarify why this integral is evaluated with respect to $$k$$ instead of $$x$$ unless Plancherel's theorem is applied for $$\phi(k)$$? $$\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Psi(x,0)e^{-ikx}dx$$

My other question is can someone explain where the constant $$1/\sqrt{2\pi}$$ comes from in the following analogue (I'm not that familiar with Fourier analysis)? $$c_n\rightarrow\frac{1}{\sqrt{2\pi}}\phi(k)$$

Interpreting the Fourier transform as an expansion in a basis

Ignore the time-dependence for the moment. Some pure maths stuffs from Fourier analysis allows us to write any square-integrable function $$f(x)$$ in the form $$f(x) = \int_{-\infty}^{\infty}dk\,\tilde{f}(k)\frac{e^{ikx}}{\sqrt{2\pi}} = \int_{-\infty}^{\infty}dk\,\tilde{f}(k)\psi_k(x)\,,$$ where $$\tilde{f}(k) = \int_{-\infty}^{\infty}dx \,f(x)\frac{e^{-ikx}}{\sqrt{2\pi}} = \int_{-\infty}^{\infty}dx\,\psi_k^*(x){f}(x)\,.$$ This is essentially a statement of Plancherel's theorem. Now, we can come to a linear-algebraic interpretation of these two statements by comparing what we've done to the case for for discrete (or even finite-dimensional) spaces. The idea is that we interpret the first integral as an expansion in a basis (the basis being the free-particle eigenstates, which are plane waves) and the second integral as the formula for computing these coefficients. (There really is a dual relationship here, though, in the sense that we can just as easily interpret them in the opposite way, with the second being an expansion in a basis and the first being the coefficients, but leave that be for now.)

Here is the comparison. For the case of a discrete basis, we have a set of normalized basis functions $$\psi_n(x)$$, and we can expand an arbitrary square-integrable function $$f$$ as a linear combination $$f(x) = \sum_n c_n\psi_n(x)\,,$$ where $$c_n = \int_{-\infty}^{\infty}dx\, \psi_n^*(x)f(x)\,.$$ We compare directly the forms as follows: \begin{align*} \begin{array}{ccccc} f(x) & =& \displaystyle\int_{\infty}^{\infty} dk & \tilde{f}(k) & \psi_k(x)\\[1em] f(x) & =& \displaystyle\sum_n & c_n & \psi_n(x) \end{array} \end{align*} and \begin{align*} \begin{array}{ccccc} \tilde{f}(k)& =& \displaystyle\int_{\infty}^{\infty} dx & \psi_k^*(x) & f(x)\\[1em] c_n & =&\displaystyle\int_{-\infty}^{\infty}dx & \psi_n^*(x) & {f}(x) \end{array} \end{align*} We can see that the sum over $$k$$ corresponds to the sum over $$n$$, the function $$\tilde{f}(k)$$ corresponds to $$c_n$$ (so really it's an expansion coefficient), and the basis function $$\psi_k(x)=\frac{e^{ikx}}{\sqrt{2\pi}}$$ corresponds to the basis element $$\psi_n(x)$$. This should make the correspondence clear. The first set of expressions represent expansions of $$f$$ in an "orthonormal basis", and the second set of expressions represent the computation of the expansion coefficients. (Note importantly that the factor of $$1/\sqrt{2\pi}$$ is not grouped with the expansion coefficient but instead goes along with the complex exponential! The reason why we group the $$\sqrt{2\pi}$$ together with the exponential will be explained below.)

The expansion of an initial state in the energy-eigenbasis of the free-particle

Now, suppose we are considering a free-particle system, in which case the eigenstates of the Hamiltonian are given by $$\psi_k(x) = \frac{e^{ikx}}{\sqrt{2\pi}}\,,$$ where the corresponding eigenvalue is indexed by $$k$$ and is given by $$E_k=\frac{\hbar^2k}{2m}\,.$$ These are states of well-defined and definite energy (modulo the problem of them being non-normalized, but deal with that later). According to the prescription for how we solve the time-dependent Schrodinger equation, given an initial state $$\Psi(x,0)$$, we expand it in the energy eigenstates as $$\Psi(x,0) = {"{\sum_k}"}c_k \psi_k(x)\,,$$ and tack on the time-dependence in the usual way to get $$\Psi(x,t) = {"{\sum_k}"}c_k \psi_k(x)e^{-iE_kt/\hbar}\,.$$ However, we don't have a discrete basis, so what really have is $$\Psi(x,0) = \int_{-\infty}^{\infty}dk\,\tilde{\Psi}(k,0)\psi_k(x) \Longrightarrow \Psi(x,0) = \int_{-\infty}^{\infty}dk\,\tilde{\Psi}(k,0)\psi_k(x)e^{-iE_kt/\hbar}\,$$ where $$E_k$$ is as above and $$\tilde{\Psi}(k,0)$$ is computed as explained above. (As a matter of notation, I have used $$\tilde{\Psi}(k,0)$$ for what the OP--and Griffiths--calls $$\phi(k)$$ in order to make it clear that this "function" is really the set of expansion coefficients of the state $$\Psi(x,0)$$ in the free-particle basis.) This reduces to the OP's--and Griffith's--expression if we plug in $$\frac{e^{ikx}}{\sqrt{2\pi}}$$ for $$\psi_k(x)$$.

The choice of normalization

The reason that the $$\sqrt{2\pi}$$ is there is to ensure that the state is "normalized" in the continuous-variable sense. That is, since $$\delta(k-k') = \int_{-\infty}^{\infty}dx\,\frac{e^{-i(k-k')x}}{2\pi}\,,$$ is a representation of the Dirac-delta function, then we can write that the inner product of two such free-particle eigenstates is $$\langle\psi_{k}|\psi_{k'}\rangle =\int_{-\infty}^{\infty}dx\,\psi_k^*(x)\psi_{k'}(x) =\int_{-\infty}^{\infty}dx\,\left(\frac{e^{ikx}}{\sqrt{2\pi}}\right)^*\frac{e^{ik'x}}{\sqrt{2\pi}} = \int_{-\infty}^{\infty}dx\,\frac{e^{-i(k-k')x}}{2\pi} =\delta(k-k')\,.$$ Similarly, $$\int_{-\infty}^{\infty}dk\,\frac{e^{i(x-x')k}}{2\pi} =\delta(x-x')\,.$$ Using this, we can write \begin{align*} f(x) &= \int_{-\infty}^{\infty}dx'\,\delta(x-x') f(x') = \int_{-\infty}^{\infty}dx'\,\int_{-\infty}^{\infty}dk\,\frac{e^{i(x-x')k}}{2\pi}f(x') \\ &= \int_{-\infty}^{\infty}dk\,\frac{e^{ikx}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx'\,\frac{e^{-ikx'}}{\sqrt{2\pi}}f(x') \\ &= \int_{-\infty}^{\infty}dk\,\frac{e^{ikx}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx'\,\psi_k^*(x')f(x') \\ &= \int_{-\infty}^{\infty}dk\,\frac{e^{ikx}}{\sqrt{2\pi}}\tilde{f}(k) \\ &= \int_{-\infty}^{\infty}dk\,\tilde{f}(k)\psi_k(x)\,. \end{align*} Thus, the "justification"/"derivation" of the Fourier transform and its inverse, and the choice for where the $$\sqrt{2\pi}$$ goes.

But, one more thing. That choice of where the $$\sqrt{2\pi}$$ goes has the following consequence. It guarantees that if $$f(x)$$ is normalized in $$x$$-space, $$\tilde{f}(k)$$ is normalized in $$k$$-space! This allows us to interpret $$\tilde{f}(k)$$ as the momentum-space representation of the wave function (once we've identified $$p=\hbar k$$), which is a representation of the wave function that reveals momentum information about the state in the same way that the position-space wave function reveals position information. To be explicit, $$|f(x)|^2dx$$ is the probability that when we measure the position of the particle, we get the result that it is in the interval $$[x,x+dx]$$, and so $$|\tilde{f}(k)|^2dk$$ is the probability that when we measure the momentum of the particle, we get the result that it is in the interval $$[\hbar k,\hbar k+\hbar dk]$$.