# Application of the operator $\exp\left( \alpha \frac{\partial^2}{\partial q^2}\right)$

I need to apply the operator

$$\exp\left( \alpha \frac{\partial^2}{\partial q^2}\right) \tag{1} \label{1}$$

To the function

$$M(x) N(y +C_{1}p)\mathcal{F}[f(q)](p) \tag{2} \label{2}$$

where $$M(x)$$ and $$N(y + C_{1}p)$$ are arbitrary functions, $$C_{1}$$ is a real constant and $$\mathcal{F}[f(q)](p)=\frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty}f(q)e^{-iqp}dq \tag{3} \label{D.1}$$

is the Fourier transform of the arbitrary function $$f(q)$$; is important to note that $$q$$ and $$p$$ is a conjugate pair (analogous to the position and moment variables of quantum mechanics), such that, in a quantum mechanical sense, the $$y$$ and $$p$$ variables are entangled. Specifically I need to know if I am correctly applying the operator $$(\ref{1})$$ to the fuction $$(\ref{2})$$, my process is as follows:

First, I write the function $$(\ref{2})$$ with the explicit Fourier transform and I expand the operator ($$[\ref{1}]$$) as an infinite sum of derivatives

$$\sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n \frac{\partial^{2n}}{\partial q^{2n}} M(x) N(y +C_{1}p)\mathcal{F}[f(q)](p)$$ $$=\frac{1}{\sqrt{2\pi}} M(x)N(y+C_{1}p) \quad \sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n \frac{\partial^{2n}}{\partial q^{2n}} \int_{-\infty}^{\infty}f(q)e^{-iqp}dq \tag{4}\label{4}$$

Second, I introduce the derivative inside the integral $$=\frac{1}{\sqrt{2\pi}} M(x)N(y+C_{1}p) \quad \sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n \int_{-\infty}^{\infty} \frac{\partial^{2n}}{\partial q^{2n}} f(q)e^{-iqp}dq \tag{5}\label{5}$$

and then I use the derivative theorem of the Fourier transform:

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{\partial^{2n}}{\partial q^{2n}} f(q)e^{-iqp}dq =\frac{1}{\sqrt{2\pi}} (ip)^{2n} \int_{-\infty}^{\infty} f(q)e^{-iqp}dq \tag{D.2}$$

such that the line (\ref{5}) is

$$\frac{1}{\sqrt{2\pi}} M(x)N(y+C_{1}p) \quad \sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n (ip)^{2n} \int_{-\infty}^{\infty} f(q)e^{-iqp}dq$$

$$= M(x)N(y+C_{1}p)e^{-\alpha p^2} \quad \mathcal{F}[f(q)](p) \tag{7}\label{7}$$

where I have reversed the infinite sum, then, the last line is my final result.

So, my question is: Am I correctly applying operator ($$\ref{1}$$) to the function ($$\ref{2}$$)?

• Looks good to me. I feel like there might be some convergence questions hiding in this problem, but, assuming those can be ignored (which is probably a safe assumption since this seems like a physics question), I don't see any problems. I also haven't seen the derivative identity (D.2) before, but it seems like a straightforward application of integration by parts, assuming that $f$ and all its derivatives die at $\infty$. – Charles Hudgins Apr 25 '19 at 5:25
• This looks wrong. In particular, when you take the $\partial^{2n}/\partial q^{2n}$ of the Fourier transform, which is a function of $p$, you should get zero. – Jacky Chong Apr 25 '19 at 5:39
• @CharlesHudgins I think that your convergence questions are related to the integral limits in the definition of the Fourier transform (3), in order to avoid problems or confusions assume that $f(q)$ is a function such that the integral converges. The definition (D.2) is the simple derivative theorem of the Fourier transform (this question was ask here: math.stackexchange.com/questions/430858/…) – Julio Abraham Mendoza Fierro Apr 25 '19 at 5:57
• @JackyChong keep in mind that $q$ and $p$ are a pair of conjugate variables, that is, the function in $q$-space is related to the funtion in $p$-space through a Fourier transform – Julio Abraham Mendoza Fierro Apr 25 '19 at 6:03
• But you already integrated out the $q$ variable. What is left is a function of $p$ not $q$. – Jacky Chong Apr 25 '19 at 6:06

You need to convert the operator to its representation in p-space before applying it. This can be done by setting $$\frac{1}{i} \partial/\partial q = p$$.

The steps you take from line $$(4)$$ to line $$(5)$$ are not correct. The integral,

$$\int_{-\infty}^{\infty} f(q) e^{-iqp} \ dq,$$

is a function of $$p$$ not $$q$$. Any partial derivative of this with respect to $$q$$ would result in $$0$$.

$$\frac{\partial}{\partial q }\int_{-\infty}^{\infty} f(q) e^{-iqp} \ dq =0 .$$

To convince yourself of this consider a more elementary expression.

$$\int_0^1 x \ dx = \frac12$$

Does it make sense to take a partial derivative of both sides with respect to $$x$$? Following your procedure I would get,

$$\frac{\partial}{\partial x} \int_0^1 x \ dx = \frac{\partial}{\partial x} \frac12$$ $$\int_0^1 \frac{\partial}{\partial x} x \ dx = 0$$ $$\int_0^1 1 \ dx = 0$$ $$1 = 0$$

The correct procedure is to express the operator in terms of the basis that the state vector is represented in.

To derive the form of the operator in $$p$$-space we would do the following.

(1) Act on a generic $$q$$-state with the operator.

(2) Apply the Fourier transform to the result of (1)

(3) Determine what how the result of (2) is related to the Fourier transform of the initial state we acted on.

Step 1:

Let $$g(q)$$ be a function in our Hilbert space. Then we apply our operator $$\exp(\alpha \frac{\partial^2}{\partial q^2} )$$ to $$g(q)$$.

$$e^{\alpha \frac{\partial^2}{\partial q^2} } g(q) \quad \textbf{(a)}$$

Step 2:

We take the Fourier transform of the result.

$$\int e^{-ipq} e^{\alpha \frac{\partial^2}{\partial q^2} } g(q) \ dq \quad \textbf{(b)}$$

Step 3:

Using integration by parts it is possible to show that,

$$\int e^{-ipq} e^{\alpha \frac{\partial^2}{\partial q^2} } g(q) \ dq = \int e^{-ipq} e^{-\alpha p^2} g(q) \ dq$$

$$= e^{-\alpha p^2} \int e^{-ipq} g(q) \ dq$$

$$= e^{-\alpha p^2} \hat{g}(p) \quad \textbf{(c)}$$

We now see that our operator acts on functions in $$p$$-space as multipication by $$e^{-\alpha p^2}$$. We can now act on any $$p$$-space state by simply multiplying by $$e^{-\alpha p^2}$$.

Notice that at no point did I pull a derivative operation inside an integral $$\frac{\partial }{\partial q} \int \ dq \rightarrow \int \frac{\partial}{\partial q} \ dq$$; which would be an invalid step.

One last thing I would like to point out is that the integration variable in a definite integral is always a dummy variable. It doesn't take on a particular value, and never "survives" the integration process. This is what user Jacky Chong was referring to when he said that you "[...] integrated out the $$q$$ variable." One feature of dummy variables is that you can change their labels without changing the value of the mathematical expression.

For example,

$$\int f(q) e^{-iqp} \ dq = \int f(y) e^{-iyp} \ dy = \int f(k) e^{-ikp} \ dk ,$$

in the same way that,

$$\int_{0}^1 x^2 \ dx = \int_0^1 y^2 \ dy = \int_0^1 k^2 \ dk = \frac13.$$

Dummy variables also occur in sums, e.g. $$\sum_{i=1\dots 10} i^2 = \sum_{j=1\dots 10} j^2$$.

It is a definite red flag to be differentiating an integral with respect to its integration variable because that variable is a dummy variable.

• following your recommendation, the result I get in line 7 is recovered, but muy question is not how to apply the operator, rather it is to know if my application procedure (lines 4-7) is mathematically correct – Julio Abraham Mendoza Fierro Apr 25 '19 at 13:44
• Specifically my question is whether it is mathematically valid to introduce the derivative within the integral of the Fourier transform, and then apply the theorem (D.2) – Julio Abraham Mendoza Fierro Apr 25 '19 at 14:00
• Has my edit answered your question sufficiently? – Spencer Apr 26 '19 at 1:10
• is right, my procedure is wrong, however it gives the right result right? – Julio Abraham Mendoza Fierro Apr 26 '19 at 16:09
• If I have answered your question, it would now be appropriate to accept the answer. – Spencer Apr 26 '19 at 17:30