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Prove that Operators are Hermitianby chris_avfc
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#1
Nov2612, 08:16 AM

P: 82

1. The problem statement, all variables and given/known data
Prove that i d/dx and d^2/dx^2 are Hermitian operators 2. Relevant equations I have been using page three of this document http://www.phys.spbu.ru/content/File...pe/qm0703.pdf and the formula there. 3. The attempt at a solution I have taken a photo and attached it as I'm not good with Latex and it will just look messy, hope that isn't a problem. I get stuck as I have done the integration for both sides of the equation, I get answers very similar, apart from the complex conjugates. I'm confused as to what do with them or if I've even done it right. Thank You 


#2
Nov2612, 09:32 PM

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Since this is a two part problem, I'll start with the first part, proving that i d/dx is Hermitian. First the goal. Your goal is to start with: [tex] \int \left( i \frac{d}{dx} \psi \right)^* \psi \ dx, [/tex] and finagle that until you can show that it equals [tex] \int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx [/tex] And you are correct that you want to use integration by parts to do this. Example (ii) on page 3 of your attachment pretty much does it for you. However, this example takes a few shortcuts. I'll try to give you some hints about how to do the math without taking the shortcuts. Hint 1: If A and B are complex variables, then (AB)* = A*B*. Hint 2: [tex] \left( \frac{d}{dx} U \right) V \ dx = \left( d U \right) V = V \ dU [/tex] The important part of that hint is to note that the "dx"s cancel. The example (ii) in your attachment doesn't follow through with this cancellation, but it could have. The reason it doesn't follow through with the cancellation is so that it doesn't have to add the "dx"s back in at the end of the calculations (or concern itself with the limits of integration, although those wouldn't change anyway for this particular problem). That's one of the shortcuts the example is doing. Hint 3: You don't really have to make the substitutions using U's, V's dU's and dV's if you don't want to. The equation you start with is so close to the integration by parts definition to begin with, it might just be easier to stick with [itex] \psi [/itex]'s, [itex] \psi^* [/itex]'s, [itex] d \psi [/itex]'s and [itex] d \psi^* [/itex]'s. (However if you'd like to make the substitutions to U's, V's dU's and dV's, you are certainly free to do so.) Hint 4: The example in your attachment leaves out the limits of integration so as not to clutter up the equations. But they are still there. The limits of integration are x = ∞ to ∞. You can't forget about those for this problem. Restating the general formula for integration by parts, without forgetting limits, we have [tex] \int_a^b U \ dV = UV _a^b  \int_a^b V \ dU [/tex]. I think you missed part of that in your attempted solution. You need to evaluate the UV term using the limits (the actual limits in this case are at x = ∞ to ∞). Evaluating that term is the crux of solving this part of the problem. After all that, refer to example (iii) for the d^{2}/dx^{2} operator. That example uses some shortcuts (and substitutions) too, but they are no less straightforward than the previous example. 


#3
Nov2712, 06:50 AM

P: 82

I still can't get it right though, I'm taking the constant [itex] i [/itex] or [itex] i^* [/itex] out of the differential, cancelling the [itex] dx [/itex], then doing integration by parts with what is left which is: [tex] i \int \psi^* d\psi [/tex] on the right hand side, and [tex] i^* \int \psi d\psi^* [/tex] on the left hand side. I'm ending up with: [tex]i(\psi \psi^*  \infty^2)[/tex] on the right hand side. and [tex]i^*(\psi \psi^*  \infty^2)[/tex] on the left hand side. The whole infinite thing makes me think I have done it wrong again. 


#4
Nov2712, 01:00 PM

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Prove that Operators are Hermitian
[tex] \int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx. [/tex] Once it looks like that you're finished. Of course there's nothing wrong with simplifying it [tex] \int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx = i \int \psi^* d \psi [/tex] and then "meeting in the middle." I'm just sayin', I think your instructor probably wants you to start with [tex] \int \left(i \frac{d}{dx} \psi \right)^* \psi \ dx [/tex] and end with [tex] \int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx. [/tex] (Note, I left the limits of integration off too, to avoid clutter. You can add them in if you wish.) But what you did is fine, so let's just continue. Hint 5: The complex conjugate of [itex] i [/itex] is [itex] i [/itex]. [tex] i^* = i [/tex] Let me help out a little more, setting up the integration by parts. [tex] i \left[ \int \psi d \psi^* \right] = i \left[ \psi \psi^* _{x=\infty}^{\infty}  \int \psi^* d \psi \right] [/tex] I mentioned before, you can't evaluate [itex] \int \psi^* d \psi [/itex]. But you can evaluate [itex] \psi \psi^* _{x=\infty}^{\infty} [/itex]. In order for [itex] \psi (x) [/itex] to be a valid wavefunction, it must be square integrable. And in order for it to be square integrable, [itex]  \psi (x) ^2 = \psi \psi^*(x) [/itex] must taper off to zero as x approaches infinity or negative infinity. I think you were assuming that [itex] \psi [/itex] approaches infinity as x approaches infinity. That's not true. (If it was, [itex] \psi [/itex] would not be a valid wavefunction.) 


#5
Nov2712, 01:18 PM

P: 82

But how do you get the [itex] dx [/itex] back in at the end, do you just insert them back? God knows how I am going to manage the rest of this sheet now, thank you though. By the way the link in your sig is such a great help, I've bookmarked it. 


#6
Nov2712, 01:26 PM

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#7
Nov2712, 01:54 PM

P: 82

So looking at example iii on the sheet I linked, how do they manage to move the [itex] p [/itex] out of the conjugate like that? 


#8
Nov2712, 03:31 PM

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[tex] \hat p = i \hbar \frac{d}{dx} [/tex] The example didn't explicitly show that. Instead example (iii) builds off of the previous example (which is almost the same as what we just did above). Building off of the previous example, you can use [tex] \left[ \left( \frac{d}{dx} \right) g(x) \right]^* =g^*(x) \left(\frac{d}{dx} \right) [/tex] twice. If you don't want to build off of the previous example, you can do example (iii) by noting that [tex] \hat T = \frac{\hat p ^2}{2m} = \frac{\hbar^2}{2m} \frac{d^2}{dx^2} [/tex] and do integration by parts twice, the long way. 


#9
Nov2712, 04:19 PM

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'Just to avoid any possible misinterpretation, when I claimed,



#10
Nov2812, 07:47 AM

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#11
Nov2812, 12:49 PM

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I'm sorry, but I made a big mistake. The last thing I intended to do was confuse you, but I think I might have done so.
In my two most recent posts, I mistakenly wrote, [tex] \left[ \left( \frac{d}{dx} \right) g(x) \right]^* =g^*(x) \left(\frac{d}{dx} \right) \mathrm{\ \ \ \ \ [sic]}[/tex] That's wrong! I forgot an [itex] i [/itex] in a couple places. What I should have wrote was [tex] \left[ \left(i \frac{d}{dx} \right) g(x) \right]^* =g^*(x) \left(i \frac{d}{dx} \right)[/tex] It turns out that [itex] i [/itex] is important. Why? because [itex] \frac{d}{dx} [/itex] by itself is not a Hermitian operator. That [itex] i [/itex] in front of it is needed to make it Hermitian. The reason it is allowed to pull the [itex] \left( i \frac{d}{dx} \right) [/itex] operator out from under the conjugate, and to the right (instead of to the left), is because it is Hermitian. If the operator was not Hermitian, we wouldn't be allowed to do that. Since I confused you once (by forgetting the [itex] i [/itex]), let me try to make up for it by helping out a little more. As I mentioned before, there are a couple ways to approach the second part of the problem statement. You could prove it by integrating by parts, twice. Or alternately, you could prove it by building off the first part of the problem statement. Following the path taken by page 3, (iii) if your attachment, I'll help you with the latter approach. Although you're not allowed to pull out a [itex] \frac{d^2}{dx^2} [/itex] from underneath the conjugate, and to the right, in this particular problem (until you prove it is Hermitian), you can pull out a [itex] \left( i \frac{d}{dx} \right) [/itex], because you've already proven that that operator is Hermitian. Let's start with [tex] \int \left( \frac{d^2}{dx^2} \psi \right)^* \psi \ dx. [/tex] The first thing we can do is split up the [itex] \frac{d^2}{dx^2} [/itex] into two [itex] \frac{d}{dx} [/itex] operations. [tex] \int \left[ \frac{d}{dx} \left( \frac{d}{dx} \psi \right) \right]^* \psi \ dx. [/tex] Next we note that [itex] i^2 = 1 [/itex], and multiply everything in the square brackets by [itex] i^2 [/itex]. It should be obvious now that that doesn't change anything, because we're just multiplying everything within the brackets by 1. [tex] \int \left[ i^2 \frac{d}{dx} \left( \frac{d}{dx} \psi \right) \right]^* \psi \ dx. [/tex] Let's redistribute the [itex] i [/itex]'s. [tex] \int  \left[ i \frac{d}{dx} \left(i \frac{d}{dx} \psi \right) \right]^* \psi \ dx. [/tex] Now we can start pulling out [itex] \left( i \frac{d}{dx} \right) [/itex] terms, and to the right. I can't do the whole problem for you but I'll show you how to do the first one (you'll have to do it twice, total). [tex] \int  \left[ \color{red}{i \frac{d}{dx}} \left(i \frac{d}{dx} \psi \right) \right]^* \psi \ dx = \int  \left[ \left( i \frac{d}{dx} \psi \right)^* \color{red}{\left(i \frac{d}{dx} \right)} \right] \psi \ dx.[/tex] Now let me just get rid of a set of brackets. [tex] \int  \left( i \frac{d}{dx} \psi \right)^* \left(i \frac{d}{dx} \right) \psi \ dx.[/tex] Now you take it from here. Start by pulling out a [itex] \left( i \frac{d}{dx} \right) [/itex] term from under the conjugate, and to the right, which is now in red: [tex] \int  \left( \color{red}{i \frac{d}{dx}} \psi \right)^* \left(i \frac{d}{dx} \right) \psi \ dx.[/tex] Simplify after that. Once you get the answer, I suggest doing the problem again from the beginning, without using this thread as a guide, just so you know you can do it on your own. Good luck! 


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