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Let $H$ be the operator $ -\frac{d^{2}}{dx^{2}} $ and let its domain be $$\{f\in L^{2}(\mathbb{R},d\lambda)\text{ }:\int_{-\infty}^{\infty}|xF[f(x)]|^{2}dx<\infty\} $$ where $F$ is the Fourier Transform.

Let $h(x)$ be the state vector $h(x)=\begin{cases} \frac{1}{\sqrt{2}} & x\in[0,2]\\ 0 & \text{else }\end{cases}$.

Find the probability that a measurement will return values in $[\frac{1}{2},1]$.

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Here is my own response, which I am very skeptical of.

The General Idea

  1. Find the eigenspace of of the opeartor $H$ spanned by its eigenvectors (eigfunctions) $\psi_{k}(x)$.

  2. Express $h(x)$ as a linear combination of the eigenfunctions: $h(x)=\sum C_{k}\psi_{k}(x)$.

  3. Based on this document, I believe that the if $\{C_{k}\}$ is the set of all possible eigenvalues, then the probability a measurement of the observable on $H$ on the state returns value in the set $$\Omega\subseteq\{C_{k}\}$$ is

$$\frac{\sum_{k\in\Omega}|C_{k}|^{2}}{\sum_{k\in\{C_{k}\}}|C_{k}|^{2}}$$

The Calculation

1.Eigenspace of $H$

Let $\lambda$ be the eigenvalue of $H$, then $$H[f]=\lambda f(x)\implies f''(x)=-\lambda f(x)$$

$$f(x)=c_{1}e^{i\sqrt{\lambda}x}+c_{2}e^{-i\sqrt{\lambda}x}$$

Thus the eigenvalues of $H$ are $\lambda\in(-\infty,+\infty)$

while correspoinding eigenfunctions for each eigenvalue $\lambda$ are $e^{i\sqrt{\lambda}x}$ and $e^{-i\sqrt{\lambda}x}$.

2.Decomposition of the state function

For our state $$h(x)=\begin{cases} \frac{1}{\sqrt{2}} & x\in[0,2]\\ 0 & \text{else}\end{cases}$$

we can decompose into the the space spanned by eigenfunctions of $H$ using Fourier Transform, which states that in general:

$$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f(x)}e^{ikx}dx$$and $$\hat{f(x)}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

For our state, Fourier Transformation produces

$$h(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{h(x)}e^{ikx}dk$$ where $$\hat{h(x)}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}h(x)e^{-ikx}dx=\frac{1}{\sqrt{2\pi}}\int_{0}^{2}\frac{1}{\sqrt{2}}e^{-ikx}dx=\frac{1}{2\sqrt{\pi}}\frac{1}{-ik}e^{-ikx}|_{0}^{2}$$ $$=\frac{i}{2k\sqrt{\pi}}(e^{-2ik}-1)$$

Thus

$$h(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{i}{2k\sqrt{\pi}}(e^{-2ik}-1)e^{ikx}dk$$

Expressing $h(x)$ as a linear combination of the basis of the eigenspace of $H$, which is $\{e^{ikx}\}$, we have $$h(x)=\sum_{k\in\mathbb{R}}C_{k}e^{ikx}$$

where $$C_{k}=\frac{i(e^{-2ik}-1)}{2\sqrt{2}\pi k}$$

3.The probability

To have a measure result in $[\frac{1}{2},1]$, $\lambda=[\frac{1}{2},1]$ means that the eigenfuctions are $$e^{\pm i\sqrt{\lambda}x}=e^{\pm ikx}$$ where $$k\in[-1,-\frac{1}{\sqrt{2}}]\cup[\frac{1}{\sqrt{2}},1]=\Omega$$.

The probability that this event occurs is

$$\sum_{k\in\Omega}|C_{k}|^{2}/\sum_{k\in\mathbb{R}}|C_{k}|^{2}$$

Since the summation is over a continuous range of k, (this is where the the descrete idea from this document is extended in an continuous case, which I am not sure if it is correct), we can compute it using an integral:

$$\sum_{k\in\text{all}k}|C_{k}|^{2}=\int_{-\infty}^{\infty}\left|\frac{i(e^{-2ik}-1)}{2\sqrt{2}\pi k}\right|^{2}dk$$

$$=\int_{-\infty}^{\infty}\frac{e^{-2ik}-1}{2\sqrt{2}\pi k}\cdot\frac{e^{2ik}-1}{2\sqrt{2}\pi k}dk$$ $$=\int_{-\infty}^{\infty}\frac{1-e^{-2ik}-e^{2ik}+1}{8\pi^{2}k^{2}}dk$$ $$=\frac{1}{8\pi^{2}}\int_{-\infty}^{\infty}\frac{2-2\cos2k}{k^{2}}dk=$$ $$\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty}\frac{1-\cos2k}{k^{2}}dk=0.159154 $$

and similarly $$\sum_{k\in\Omega}|C_{k}|^{2}=\int_{-1}^{-1/\sqrt{2}}\left|\frac{e^{-2ik}-1}{2\sqrt{2}\pi k}\right|^{2}dk+\int_{\frac{1}{\sqrt{2}}}^{1}\left|\frac{e^{-2ik}-1}{2\sqrt{2}\pi k}\right|^{2}dk$$ $$\frac{1}{2\pi^{2}}\int_{1/\sqrt{2}}^{1}\frac{1-\cos2k}{k^{2}}dk=0.00230997$$

Thus the probability is $$=0.00230996/0.159154=0.145$$ or 14.5%

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I am posting up my supposed answers. I am only asking because I am very skeptical of my own answer, which is almost entirely based on theory in this page: vergil.chemistry.gatech.edu/notes/quantrev/node20.html, except it is extended into a continuous case. –  Xiaowen Li Oct 28 '11 at 8:24
    
There are a few too many typos in this to warrant checking it in detail (e.g. $\lambda^{i\sqrt\lambda x}$), but the general approach seems OK, except for the double integral in the last equation -- presumably that's also a typo? Also it's rather unusual to define the Laplacian as minus the second derivative. If you put some more of your own time and effort into making this less error-ridden, I'll be happy to put some more of my time and effort into checking it. By the way, I'd suggest using more display-style equations; they're a lot easier to read. –  joriki Oct 28 '11 at 10:42
    
+1 for the much improved edited post! :-) –  joriki Oct 28 '11 at 18:02
    
Thanks for the advice Joriki. You were right, it required a lot of editing. –  Xiaowen Li Oct 28 '11 at 18:07
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1 Answer

up vote 2 down vote accepted

The calculations and result seem to be correct (except that you've got the integration bounds for $[-1,-1/\sqrt2]$ the wrong way around, and the end result is $0.145$, not $0.146$; the fraction leading to that value is OK, though).

But you're quite right to be skeptical about the way you generalized the discrete result to the continuous case. The continuous case is somewhat tricky in that the individual eigenfunctions in the continuous spectrum are not normalizable (try normalizing $\mathrm e^{\mathrm ikx}$). The formulas involving sums that you found for the discrete case assume that you have an orthonormal basis of discrete eigenstates; then the probability can be calculated simply as a weighted average of the eigenvalues with the squared amplitudes as weights. In the continuous case one would first need to define what it means for a continuous spectrum of eigenfunctions to form an orthonormal basis. There are various levels of rigour at which this can be handled. A common approach in physics is not to worry about this too much and to write completeness and orthonormality relations using delta "functions":

$$\int|k\rangle\langle k|\mathrm dk=1\;,$$ $$\int\langle k|k'\rangle\mathrm dk=\delta(k-k')\;.$$

This can be made more rigorous using the theory of distributions, where the delta "function" becomes a well-defined distribution. Another approach is to associate projection operators with subsets of the continuous spectrum. You didn't have to deal with any of this because things work out nicely for the Fourier transform, but I'd maintain that healthy skepticism with respect to replacing discrete sums by "sums over $\mathbb R$" and the like.

By the way, you can put the amplitude

$$C_{k}=\frac{\mathrm i(\mathrm e^{-2\mathrm ik}-1)}{2\sqrt{2}\pi k}$$

into a more recoginzable form by noting that the given wave function $h(x)$ is just a rectangular pulse shifted away from the origin, and that shift corresponds to a phase factor in Fourier space, and if you take out that phase factor you should get the Fourier transform of a rectangular pulse, which is the sinc function. And indeed:

$$ \begin{eqnarray} C_{k} &=& \frac{\mathrm i(\mathrm e^{-2\mathrm ik}-1)}{2\sqrt{2}\pi k}\\ &=& \mathrm e^{-\mathrm ik}\frac{\mathrm i(\mathrm e^{-\mathrm ik}-e^{\mathrm ik})}{2\sqrt{2}\pi k}\\ &=& \mathrm e^{-\mathrm ik} \frac1{\sqrt2\pi k}\frac{(\mathrm e^{\mathrm ik}-e^{-\mathrm ik})}{2\mathrm i}\\ &=& \mathrm e^{-\mathrm ik} \frac1{\sqrt2\pi}\frac{\sin k}k\\ &=& \mathrm e^{-\mathrm ik} \frac1{\sqrt2\pi}\operatorname{sinc}k\;. \end{eqnarray} $$

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Its beyond words how much I appreciate your help Joriki. And thanks for the insight you provided. –  Xiaowen Li Oct 28 '11 at 19:39
    
@XiaowenLi: You're welcome! –  joriki Oct 28 '11 at 19:45
    
Its beyond words how much I appreciate your help Joriki. And thanks for the insight you provided. As for the orthogonality concerns, I was worried about that initially, then I saw this wierd proof that $$\int_{-\infty}^{\infty}e^{ikx}dx=2\pi\delta(k)$$, (from page 289 of this math.umn.edu/~olver/pd_/ft.pdf) which means that the basis for this eigenspace is orthogonal. (It actually makes much more sense if we interprets it as a distribution like you said). Actually I just realized that the normalization constant is just $\frac{1}{2\pi}$. –  Xiaowen Li Oct 28 '11 at 19:49
    
By the way do know why the domain of this operator is defined that way? It seems a little counter-intuitive. –  Xiaowen Li Oct 28 '11 at 19:51
    
@XiaowenLi: You've got the absolute value sign and the integral mixed up in that definition. I'm not sure exactly why it's defined exactly that way, I'm not an expert on this sort of thing, but I'm pretty sure it's related to the fact that differentiating in real space corresponds to multiplying by the coordinate in Fourier space; in the expectation value of the Laplacian you can move one differentiation onto each factor by integration by parts, so then you integrate over $|x\hat f(x)|^2$ in Fourier space, so that condition states that the expectation value of the Laplacian exists. –  joriki Oct 28 '11 at 19:57
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