# Quantum operator derivation

1. Apr 14, 2013

### BruceW

Hi all! I was reviewing some basic quantum mechanics, and I was trying to 'derive' the equation
$$-i \hbar \frac{\partial}{\partial x} \psi_{(t,x)} = <x| \hat{P} | \psi >$$
using the commutator relation, and the form of the identity operator. OK, I know that the proper, mathematical way to prove the above equation, is by the Stone-von Neumann theorem. But anyway, I tried doing a sloppy, physicist's 'derivation', so that I could get a more intuitive sense for why it works. I am posting this partly so that it helps it stick in my brain, partly in case anyone has heard of any other interesting derivations and partly because I am not certain that all my steps are correct, so hopefully if I went wrong somewhere, maybe someone can kindly point it out. I do get to the right answer, but that doesn't mean all the steps were correct :)

Anyway, so assuming the commutator relation
$$\hat{X} \hat{P} - \hat{P} \hat{X} = i \hbar$$
And pre-multiply by an eigenvector of the X operator, and post-multiply by another (not necessarily the same) eigenvector of the X operator, we get:
$$<x| \hat{X} \hat{P} |x'> - <x| \hat{P} \hat{X} |x'> = i \hbar <x|x'>$$
For this derivation, I need to use this form of the identity operator:
$$\int dx \ |x><x|$$
(In this post, I am just considering 1d, and always assume an integral is over all space, unless explicitly stated otherwise). I'm also going to assume that:
$$<x|x'> = \delta (x-x')$$
Is the Dirac-delta function. And so, using this with the above equation, and using the fact that the X operator acts on its eigenvectors to give eigenvalues, we get:
$$(x-x') \ <x| \hat{P} |x'> = i \hbar \delta(x-x')$$
And from now, I will use the notation:
$$<x| \hat{P} |x'> = P_{(x,x')}$$
Which means our equation looks like:
$$(x-x') \ P_{(x,x')} = i \hbar \delta(x-x')$$
(I will refer to this as equation 1). And now, introducing the thing we are interested in:
$$<x| \hat{P} | \psi >$$
The first thing we can do, is to insert identity operators around the P operator, and use the notation:
$$<x'| \psi> = \psi_{(x')}$$
So that we get:
$$<x| \hat{P} | \psi > = \int dx' \ P_{(x,x')} \psi_{(x')}$$
Now, I'm going to define a new function, to put this equation into a useful form. The new function is defined by this equation:
$$\psi_{(x')} = (x-x') U_{(x,x')}$$
(Where the subscripts are variables which the function is dependent on). So now, using this substitution, we get:
$$<x| \hat{P} | \psi > = \int dx' \ P_{(x,x')} (x-x') U_{(x,x')}$$
Now, we can use equation 1, to get:
$$<x| \hat{P} | \psi > = \int dx' \ i \hbar \delta(x-x') U_{(x,x')}$$
And now, hooray, the Dirac-delta function does its thing, so then:
$$<x| \hat{P} | \psi > = i \hbar U_{(x,x)}$$
Notice that the function 'U' now depends on the two same variables, because of the integration with the Dirac-delta function. So now, we need to find what U(x,x) is. Right, so starting with our original definition, but I'm going to swap around the primes, just for convenience (it is still the same equation):
$$\psi_{(x)} = (x'-x) U_{(x',x)}$$
And now, again for convenience, I am going to stop writing the subscripts in the next few equations. Remember, psi and U still have the subscripts in the way I have just written them, and I should really write them out in full, but I think that would take up a lot of space. Now, doing a partial differentiation on both sides with respect to x', while keeping x constant, and keep in mind that psi does not vary when x is held constant, we get:
$$0 = U + (x'-x) \frac{\partial U}{\partial x'}$$
Also, doing a full derivative with respect to x on both sides gives:
$$\frac{d \psi}{dx} = ( \frac{dx'}{dx} - 1)U + (x'-x)( \frac{\partial U}{\partial x'} \frac{dx'}{dx} + \frac{\partial U}{\partial x} )$$
And now combining the last two equations gives:
$$\frac{d \psi}{dx} = -U + (x'-x) \frac{\partial U}{\partial x}$$
So now, writing it out properly with the subscripts:
$$U_{(x',x)} = (x'-x) \frac{\partial U_{(x',x)} }{\partial x} - \frac{d \psi_{(x)} }{dx}$$
So now, we see that:
$$U_{(x,x)} = - \frac{d \psi_{(x)} }{dx}$$
Which is awesome, because using this substitution for U(x,x) in the equation way back earlier, we have:
$$<x| \hat{P} | \psi > = -i \hbar \frac{d \psi}{dx}$$
Which is the result I was hoping for. Now, you'll notice that this is different to the standard equation (which is a partial differentiation, while keeping t constant). This is because I haven't introduced a time variable yet. It is not necessary for this derivation. And now, if I want to introduce a time variable, then the equation simply changes to:
$$<x| \hat{P} | \psi > = -i \hbar \frac{\partial \psi_{(t,x)}}{\partial x}$$
So x is being differentiated, while t is held constant. (which is the standard equation). Whew. So, what do you guys think? It is longer than I had hoped. Sorry about that. Hope you enjoyed reading it though.

Last edited: Apr 14, 2013
2. Apr 14, 2013

### Jano L.

Bruce, your procedure shows that you know the Dirac notation quite well. Unfortunately, this notation oversimplifies things and betrays students all the time. In the step

"Now, we can use equation 1, to get"

you substitute

$$P_{xx'} (x-x')$$

by

$$i\hbar \delta(x-x').$$

That would be OK if the function $U$ of $x'$ next to it was continuous in some neighborhood of the point $x$. But you define the function $U$ to have singularity around the point $x$ ; it behaves as $\frac{1}{x'-x}$. For such functions, the above substitution is not valid, because

$$\int dx' \delta(x-x') \frac{1}{x'-x}$$

is an invalid expression.

Before you try to circumvent this problem and find another derivation, consider how everything becomes much more clear and simple, if you turn the question around: given that the operator of momentum is $\hat p_x = i\hbar \frac{\partial}{\partial x}$, derive the action of $x\hat p_x - \hat p_x x$ on any function $\psi(x)$. No distributions, no kets, no problems :-)

3. Apr 14, 2013

### BruceW

You are right for sure that:
$$\delta(x-x') \frac{1}{x'-x}$$
Is a pretty messed up kind of expression. But I don't think it is necessarily invalid. I just did a google search, and found it being referred to as a 'derivative of the Dirac-delta function' whatever that means... I agree that I haven't shown that it works in a rigorous way. But rigor is definitely not something I was aiming for, hehe. It is probably the most mathematically questionable derivation ever! Well, maybe that's a bit of an exaggeration.

Also, yes, I totally agree that it is much easier to show that the commutation relation holds when the operator of momentum takes that form. Maybe I am a masochist? ;) was that too creepy? Is saying something like that against forum rules??!

4. Apr 15, 2013

### Jano L.

5. Apr 15, 2013

### vanhees71

The usual derivation uses the fact that
$$[\hat{x},\hat{p}]=\mathrm{i}$$
($\hbar=1$, one-dimensional motion) implies that the momentum operator is the generator of translations.

To prove this consider the generalized eigenvectors of the position operator $|x \rangle$. For the transformed vector we get
$$\hat{x} \exp(-\mathrm{i} \delta x \hat{p}) |x \rangle = \exp(-\mathrm{i} \delta x \hat{p}) \exp(\mathrm{i} \delta x \hat{p}) \hat{x} \exp(-\mathrm{i} \delta x \hat{p}) |x \rangle = \exp(-\mathrm{i} \delta x \hat{p}) [\hat{x}-\mathrm{i} \delta x [\hat{x},\hat{p}] + \mathcal{O}(\delta x^2)] |x \rangle.$$
Now due to the above assumed commutation relation we get
$$\hat{x} \exp(-\mathrm{i} \delta x \hat{p}) |x \rangle = (x+\delta x) \exp(-\mathrm{i} \delta x \hat{p}) |x \rangle.$$
This means that, up to a phase factor which we can choose conveniently to be 1, we have
$$\exp(-\mathrm{i} \delta x \hat{p} ) |x \rangle = |x+\delta x \rangle = |x \rangle -\mathrm{i} \delta x \hat{p} x \rangle + \mathcal{O}(\delta x^2).$$
From this you derive
$$\frac{\mathrm{d}}{\mathrm{d} x} |x \rangle =\lim_{\delta x \rightarrow 0} \frac{1}{\delta x} (|x + \delta x \rangle - |x \rangle)=-\mathrm{i} \hat{p} |x \rangle.$$
Taking the adjoint of this equation, gives
$$\frac{\mathrm{d}}{\mathrm{d} x} \langle x | = +\mathrm{i} \langle x | \hat{p}.$$
Multiplying from the right with an appropriate true Hilbert-space vector gives the position representation of the momentum operator to be
$$\langle x | \hat{p}| \psi \rangle:=\hat{p} \psi(x)=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} x} \psi(x),$$
and that's the formula you wanted to prove.

6. Apr 15, 2013

### BruceW

ah, very nice. And much quicker derivation :) So we define momentum as the generator of translations, and use the commutation relation, and this gives the position representation of the momentum operator. Generators turn up a lot in more advanced quantum theory, right? I've only used them a few times, but I feel like they are a useful 'tool'.

@ Jano L. Wikipedia seem to define the derivative of the Dirac-delta in this way:
$$x \delta'_{(x)} = - \delta_{(x)}$$
http://en.wikipedia.org/wiki/Dirac_delta_function So using their definition,
$$P_{(x,x')} = -i \hbar \delta'_{(x-x')}$$
And since
$$<x| \hat{P} | \psi > = \int dx' P_{(x,x')} \psi_{(x')}$$
we can sub in the derivative of the Dirac-delta:
$$<x| \hat{P} | \psi > = \int dx' (-)i \hbar \delta'_{(x-x')} \psi_{(x')}$$
Right, so now we have a convolution of psi with the derivative of the Dirac-delta, and wikipedia say that:
$$\delta' \ast f = f'$$
(i.e. the convolution of the derivative of the Dirac-delta with a general function gives the derivative of that function). Therefore, we have:
$$<x| \hat{P} | \psi > = -i \hbar \delta' \ast \psi = -i \hbar \psi'$$
Which is the formula I wanted to prove, and done a bit faster this time, by making use of the properties of the derivative of the Dirac-delta, instead of doing it the longer way like I did the first time. (But still it is not as fast as vanhees' method).

7. Apr 16, 2013

### Jano L.

Yes, the above procedure makes sense, but I would not call it a proof. It is rather an exercise in correct handling of the Dirac notation.. You derived differentiation as integration over $\delta'$, irregular disitrbution. The difficulty is that if you wanted to give the above manipulations solid mathematical meaning, you would spend many pages on it and I am afraid that you would need to use much less intuitive knowledge than $\hat p_x = i\hbar \frac{\partial}{\partial x}$ in the process, namely theory of distributions, questions of function spaces etc. So, as an exercise in the Dirac notation, its OK, but as a proof which is supposed to explain the complicated in terms of the simple, it is useless.

8. Apr 16, 2013

### Jano L.

Another thing came to my mind. Back to your procedure, the distribution

$$P_{xx'} = C\delta(x-x') - i\hbar \delta'(x-x')$$

satisfies

$$(x-x')P_{xx'} = i\hbar \delta (x-x')$$

too. So you have yet to find the value of $C$...

9. Apr 16, 2013

### BruceW

ah darn, you're right. My second derivation doesn't work. Also I am not happy with my first derivation. Even if we can assume that psi can be re-written as:
$$\psi_{(x)} = (x'-x) U_{(x',x)}$$
(Which I am now thinking is a mistake, because I don't think we can generally do this... Unless someone knows of a theorem which says that you can do this if psi is a differentiable function?) Anyway, even if we assume this is possible, then I don't think that this implies that:
$$U_{(x,x)} = - \frac{d \psi_{(x)} }{dx}$$
I think I was just tricking myself by re-writing things. d'oh. So anyway, I think the second derivation is looking better than the first. Even though there is that C, which is mysteriously unknown. Maybe if we make some assumptions about the wavefunction, then the C problem goes away? I'm having a hard time thinking what this assumption might be..