Normalisation of the Gaussian wave packet

I have been trying to solve the question asking for the normalisation of the Gaussian wave packet's probability density given as $$\rho(x)=Ae^{-\lambda(x-a)^2}$$ The $$\rho(x)$$ is just the probability density not the actual Gaussian wave function. Now, proceeding as the normalisation condition that $$\int_{-\infty}^\infty\rho(x)dx = 1$$, I got the value $$A=\sqrt{\frac{\lambda}{\pi}}$$. Now when I try to find the expectation value I got confused due to the following reason:

1. The expectation value of $$x$$ is given as \begin{align}\langle x\rangle&\:=\int_{-\infty}^{\infty}x\rho(x)dx\\\implies\langle x\rangle&\:=\int_{-\infty}^{\infty}xAe^{-\lambda(x-a)^2}\end{align} Now will the expectation value be zero because the integrand is an odd function being $$A$$ a constant because the exponential function is considered as even function always?
2. But unlike 1. if I consider $$(x-a)^2=t$$ (let), and then $$(x-a)=\sqrt t$$ and $$dx = \frac{1}{2\sqrt t}dt$$ with $$x=\sqrt t+a$$ then what should I take the limits? What will be the value of $$t$$ for $$x=-\infty$$? But if I consider the limits, as usual, $$[-\infty,\infty]$$, then the expectation value comes out as $$\langle x\rangle\:=\frac{1}{\sqrt{\pi\lambda}}+a$$
3. Now when I consider $$(x-a)=t$$ only, then the expectation value comes out to be $$\langle x\rangle\:=\:a$$

Which one is correct and why? I know the value $$\langle x\rangle\:=a$$ or $$\langle x\rangle\:=0$$ seems to be correct somehow according to the function given but talking about the mathematical approach why can't I approach with the other methods and why the other approaches are not correct? Please provide your answers. And please also tell me the physical meaning of getting the expectation value of $$x$$ to be zero.

• You statement 1) is false $\exp\{-\lambda(x-a)^2\}$ is not an even function of $x$. Jun 29 at 20:44
• ... but it suggests how to compute it. Substitute $u=x-a$. Jun 29 at 20:46
• @march but what if I consider it's square as $u$, as in point 2), why it doesn't yield me the correct answer? I am little bit confused about this Jun 29 at 20:48
• @mikestone but can you please tell why? How can I check that? Jun 29 at 20:51
• Your method 2 doesn't work for a reason hinted at by your question of what the limits should be. To make that substitution work, you'd have to break the integral into two pieces, one from $-\infty$ to $a$ and one from $a$ to $\infty$, and be more careful about taking the square root of a square (because $\sqrt{b^2} = |b|$)! Jun 29 at 20:52

I am tempted to ignore the fact that you're asking multiple questions in one stack exchange question, because they all kind of revolve around this idea of calculating one thing in a bunch of different ways. But that's borderline for me, you might consider in the future splitting this up. Some observations:

(0) Pretty sure you are missing a factor of $\sqrt{2} from your first expression. (1) The exponential function is almost never considered an even function. The even part of the exponential function is $$\cosh x=\frac{e^{x}+e^{-x}}2$$ and the odd part is $$\sinh x=\frac{e^{x}-e^{-x}}2$$ , which is not zero. In this particular case you have a function of a square, $$f\big((x-a)^2\big)$$. This is “even about $$x=a$$” in the sense that it has a reflection symmetry, if we define the reflection about $$a$$ as $$\xi = a-(x-a)$$ then this is $$f\big((a-\xi)^2\big)=f\big((\xi-a)^2\big).$$ But, the other term you are multiplying by, plain $$x$$, is not even or odd about $$a$$. To fix this, you can just decompose it into an even and an odd part (about $$a$$) the same way that I did the exponential; you will find that the odd part is $$x-a$$ which indeed integrates to zero as you desire, but that leaves the even part $$a$$ which integrates to $$a$$. (2) Just go back to the start, when $$(x-a)^2 = t$$ and $$x\to-\infty$$ then $$t\to\infty.$$ “So the integration bounds are plus infinity to plus infinity?” Yes! And thus you can see that you need to be careful when you use these transforms that are not invertible! The proper thing to do is to identify where they go non-invertible and break them into chunks, so in this case start from $$\int_{-\infty}^\infty\mathrm dx~f(x)= \int_{-\infty}^a\mathrm dx~f(x) + \int_{a}^\infty\mathrm dx~f(x)$$ and then the $$u$$-substitution is invertible on those two branches separately. • As you said we have to see where the transform is going non-invertible, how can we check that? Jun 29 at 21:54 • Any monotonically increasing or decreasing function on some domain is invertible on that domain. So we want to look for places where this breaks down, and it turns out that this is the same set of places that you would look for an extremum—the critical points—because you can't have an extremum inside a domain where the function is monotonically increasing or decreasing. Seeing that$(x-a)^2$is continuous and its derivative is also continuous, that leaves only the places where the derivative is zero,$2(x-a)=0, x=a.\$ In practice people just know parabolas are symmetric about 0 though. Jun 29 at 22:28
• Thank you very much sir Jun 30 at 11:45

(1) Since $$\rho(x-a)$$ is even, then:

$$\langle x-a \rangle = 0$$

Therefore: $$\langle x-a \rangle = \langle x \rangle - \langle a \rangle=0$$

So you should verify:

$$\langle x \rangle = \langle a \rangle=a\langle 1 \rangle=a$$

The physical meaning of $$\langle x \rangle = \bar x$$ for any value of $$\bar x$$ means that the particle's average position is $$\bar x$$.

Note: You got away with it here only because

$$[\psi(x),\hat x]=0$$

but in general:

$$\langle \hat A \rangle\ne \int \hat A\rho(x)dx$$

You need use:

$$\langle \hat A \rangle = \int \psi^*(x)\hat A\psi(x)dx$$