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BruceW

Homework Helper

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Hi all! I was reviewing some basic quantum mechanics, and I was trying to 'derive' the equation

[tex]-i \hbar \frac{\partial}{\partial x} \psi_{(t,x)} = <x| \hat{P} | \psi >[/tex]

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

[tex]\hat{X} \hat{P} - \hat{P} \hat{X} = i \hbar [/tex]

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:

[tex]<x| \hat{X} \hat{P} |x'> - <x| \hat{P} \hat{X} |x'> = i \hbar <x|x'> [/tex]

For this derivation, I need to use this form of the identity operator:

[tex]\int dx \ |x><x| [/tex]

(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:

[tex]<x|x'> = \delta (x-x') [/tex]

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:

[tex](x-x') \ <x| \hat{P} |x'> = i \hbar \delta(x-x') [/tex]

And from now, I will use the notation:

[tex]<x| \hat{P} |x'> = P_{(x,x')} [/tex]

Which means our equation looks like:

[tex](x-x') \ P_{(x,x')} = i \hbar \delta(x-x') [/tex]

(I will refer to this as equation 1). And now, introducing the thing we are interested in:

[tex]<x| \hat{P} | \psi >[/tex]

The first thing we can do, is to insert identity operators around the P operator, and use the notation:

[tex]<x'| \psi> = \psi_{(x')} [/tex]

So that we get:

[tex]<x| \hat{P} | \psi > = \int dx' \ P_{(x,x')} \psi_{(x')}[/tex]

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:

[tex]\psi_{(x')} = (x-x') U_{(x,x')} [/tex]

(Where the subscripts are variables which the function is dependent on). So now, using this substitution, we get:

[tex]<x| \hat{P} | \psi > = \int dx' \ P_{(x,x')} (x-x') U_{(x,x')} [/tex]

Now, we can use equation 1, to get:

[tex]<x| \hat{P} | \psi > = \int dx' \ i \hbar \delta(x-x') U_{(x,x')} [/tex]

And now, hooray, the Dirac-delta function does its thing, so then:

[tex]<x| \hat{P} | \psi > = i \hbar U_{(x,x)} [/tex]

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):

[tex]\psi_{(x)} = (x'-x) U_{(x',x)} [/tex]

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:

[tex]0 = U + (x'-x) \frac{\partial U}{\partial x'} [/tex]

Also, doing a full derivative with respect to x on both sides gives:

[tex]\frac{d \psi}{dx} = ( \frac{dx'}{dx} - 1)U + (x'-x)( \frac{\partial U}{\partial x'} \frac{dx'}{dx} + \frac{\partial U}{\partial x} ) [/tex]

And now combining the last two equations gives:

[tex]\frac{d \psi}{dx} = -U + (x'-x) \frac{\partial U}{\partial x} [/tex]

So now, writing it out properly with the subscripts:

[tex]U_{(x',x)} = (x'-x) \frac{\partial U_{(x',x)} }{\partial x} - \frac{d \psi_{(x)} }{dx} [/tex]

So now, we see that:

[tex]U_{(x,x)} = - \frac{d \psi_{(x)} }{dx} [/tex]

Which is awesome, because using this substitution for U(x,x) in the equation way back earlier, we have:

[tex]<x| \hat{P} | \psi > = -i \hbar \frac{d \psi}{dx} [/tex]

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:

[tex]<x| \hat{P} | \psi > = -i \hbar \frac{\partial \psi_{(t,x)}}{\partial x} [/tex]

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.

[tex]-i \hbar \frac{\partial}{\partial x} \psi_{(t,x)} = <x| \hat{P} | \psi >[/tex]

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

[tex]\hat{X} \hat{P} - \hat{P} \hat{X} = i \hbar [/tex]

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:

[tex]<x| \hat{X} \hat{P} |x'> - <x| \hat{P} \hat{X} |x'> = i \hbar <x|x'> [/tex]

For this derivation, I need to use this form of the identity operator:

[tex]\int dx \ |x><x| [/tex]

(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:

[tex]<x|x'> = \delta (x-x') [/tex]

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:

[tex](x-x') \ <x| \hat{P} |x'> = i \hbar \delta(x-x') [/tex]

And from now, I will use the notation:

[tex]<x| \hat{P} |x'> = P_{(x,x')} [/tex]

Which means our equation looks like:

[tex](x-x') \ P_{(x,x')} = i \hbar \delta(x-x') [/tex]

(I will refer to this as equation 1). And now, introducing the thing we are interested in:

[tex]<x| \hat{P} | \psi >[/tex]

The first thing we can do, is to insert identity operators around the P operator, and use the notation:

[tex]<x'| \psi> = \psi_{(x')} [/tex]

So that we get:

[tex]<x| \hat{P} | \psi > = \int dx' \ P_{(x,x')} \psi_{(x')}[/tex]

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:

[tex]\psi_{(x')} = (x-x') U_{(x,x')} [/tex]

(Where the subscripts are variables which the function is dependent on). So now, using this substitution, we get:

[tex]<x| \hat{P} | \psi > = \int dx' \ P_{(x,x')} (x-x') U_{(x,x')} [/tex]

Now, we can use equation 1, to get:

[tex]<x| \hat{P} | \psi > = \int dx' \ i \hbar \delta(x-x') U_{(x,x')} [/tex]

And now, hooray, the Dirac-delta function does its thing, so then:

[tex]<x| \hat{P} | \psi > = i \hbar U_{(x,x)} [/tex]

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):

[tex]\psi_{(x)} = (x'-x) U_{(x',x)} [/tex]

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:

[tex]0 = U + (x'-x) \frac{\partial U}{\partial x'} [/tex]

Also, doing a full derivative with respect to x on both sides gives:

[tex]\frac{d \psi}{dx} = ( \frac{dx'}{dx} - 1)U + (x'-x)( \frac{\partial U}{\partial x'} \frac{dx'}{dx} + \frac{\partial U}{\partial x} ) [/tex]

And now combining the last two equations gives:

[tex]\frac{d \psi}{dx} = -U + (x'-x) \frac{\partial U}{\partial x} [/tex]

So now, writing it out properly with the subscripts:

[tex]U_{(x',x)} = (x'-x) \frac{\partial U_{(x',x)} }{\partial x} - \frac{d \psi_{(x)} }{dx} [/tex]

So now, we see that:

[tex]U_{(x,x)} = - \frac{d \psi_{(x)} }{dx} [/tex]

Which is awesome, because using this substitution for U(x,x) in the equation way back earlier, we have:

[tex]<x| \hat{P} | \psi > = -i \hbar \frac{d \psi}{dx} [/tex]

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:

[tex]<x| \hat{P} | \psi > = -i \hbar \frac{\partial \psi_{(t,x)}}{\partial x} [/tex]

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.

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