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## Main Question or Discussion Point

I've been studying quantum mechanics, and working problems to get a feel for expectation values and what causes them to be real.

I was working the problem of finite 1D wells, when I came across a situation I did not understand.

A stationary state solution is made up of a forward and reverse travelling wave of the same amplitude in regions where the potential is constant.

When I try to compute the expectation values in these areas, I get a different answer depending on whether or not I sum up the waves before or after I apply the momentum operator. I am not sure what I am doing wrong ... or why the method works one way, but not the other, as the books I've read aren't clear on this point:

Consider a forward travelling wave with defined momentum:

[itex] \psi_f(x) = e^{ i k x}[/itex]

I can compute the contribution of momentum of this wave to a 1D well, using the following generic formula:

[itex]< p > = \int_{x_1}^{x_2} \psi_f^{*} (- i \hbar ){ \partial \over { \partial x }} \psi_f[/itex]

For the forward wave, I compute a positive momentum:

[itex] < p > = \int_{x_1 }^{x_2 } A e^{ - i k x } (- i \hbar ){ \partial \over { \partial x }} A e^{ i k x } [/itex]

[itex] < p > = \int_{x_1 }^{x_2 } A e^{ - i k x } (- i \hbar ){ \partial \over { \partial x }} A e^{ i k x } [/itex]

[itex] < p > = \int_{x_1 }^{x_2 } A e^{ i k x } (- i \hbar )( i k )A e^{ i k x } [/itex]

[itex] < p > = A^2 \int_{x_1 }^{x_2 } e^{ 0 } (- i \hbar )( i k ) [/itex]

[itex] < p > = A^2 \hbar k ( x_2 - x_1 ) [/itex]

If I were to use a reverse travelling wave, instead [itex]\psi_r(x) = e^{- i k x }[/itex], the solution is the same but with reversed sign:

[itex] \int_{x_1 }^{x_2 } A e^{ i k x } (- i \hbar ){ \partial \over { \partial x }} e^{ -i k x } [/itex]

[itex] < p > = \int_{x_1 }^{x_2 } e^{ -i k x } (- i \hbar )( i k ) A e^{ i k x } [/itex]

[itex] < p > = A^2 \int_{x_1 }^{x_2 } e^{ 0 } (i \hbar )( i k ) [/itex]

[itex] < p > = - A^2 \hbar k ( x_2 - x_1 ) [/itex]

If A is the normalisation constant for an infinite well, over all space, then [itex]A={1 \over \sqrt{ x_2 - x_1 }}[/itex] and [itex]<p>=- \hbar k[/itex]

That's the obvious, and correct result.

Since these solutions are all linear and hermitian, I'm also supposed to be able to add together solutions to make another solution.... but, that's where I get into trouble!

The sum of the solved expectation of momentums is clearly 0. The forward and reverse waves cancel each other's momentum out. A stationary state neither moves (on average) to the right or to the left, but simply oscillates in place. So, the sum of expectation values gives the intuitive result.

However, when I attempt to solve for momentum using the sum of the forward and reverse wave-functions [itex] \psi = \psi_f + \psi_r [/itex] ... I don't get a real number.

I thought that momentum being a "linear" Hermitian operator, meant it would produce the same result whether working on individual waves, and summing the results ... or a superposition of waves with a single result. Did I misunderstand something, or did I make a mistake in the following computation; because I get a purely imaginary result:

[itex] < p > = \int_{x_1}^{x_2} A (e^{ -i k x } + e^{ i k x } ) \cdot { - i \hbar \partial \over { \partial x } } A (e^{ i k x } + e^{ -i k x } )[/itex]

[itex] < p > = \int_{x_1}^{x_2} A (e^{ -i k x } + e^{ i k x } ) \cdot { - i \hbar } A ( i k e^{ i k x } - i k e^{ -i k x } )[/itex]

[itex] < p > = A^2 \hbar k \int_{x_1}^{x_2} ( e^{ -i k x } + e^{ i k x } ) ( e^{ i k x } - e^{ -i k x } )[/itex]

[itex] < p > = A^2 \hbar k \int_{x_1}^{x_2} (e^{ 2 i k x } - e^{ -2 i k x } ) [/itex]

[itex] < p > = A^2 \hbar k \int_{x_1}^{x_2} (i 2 sin( 2 k x) ) [/itex]

Since the right hand side of the equation is is purely imaginary, then the integral must be purely imaginary ... and so must the result.

The issue seems to be that sum of products, is not the same as product of sums:

( a + 1/a ) (-1/a + a) = a*a - 1/(a*a)

(a*-1/a) + (1/a*a) = 0

But momentum operator notation sets the problem up differently depending on whether it's passed individual wave components or sums.

How am I supposed to correctly do operatior notation on a superposition of solutions to the wave equation?

I was working the problem of finite 1D wells, when I came across a situation I did not understand.

A stationary state solution is made up of a forward and reverse travelling wave of the same amplitude in regions where the potential is constant.

When I try to compute the expectation values in these areas, I get a different answer depending on whether or not I sum up the waves before or after I apply the momentum operator. I am not sure what I am doing wrong ... or why the method works one way, but not the other, as the books I've read aren't clear on this point:

Consider a forward travelling wave with defined momentum:

[itex] \psi_f(x) = e^{ i k x}[/itex]

I can compute the contribution of momentum of this wave to a 1D well, using the following generic formula:

[itex]< p > = \int_{x_1}^{x_2} \psi_f^{*} (- i \hbar ){ \partial \over { \partial x }} \psi_f[/itex]

For the forward wave, I compute a positive momentum:

[itex] < p > = \int_{x_1 }^{x_2 } A e^{ - i k x } (- i \hbar ){ \partial \over { \partial x }} A e^{ i k x } [/itex]

[itex] < p > = \int_{x_1 }^{x_2 } A e^{ - i k x } (- i \hbar ){ \partial \over { \partial x }} A e^{ i k x } [/itex]

[itex] < p > = \int_{x_1 }^{x_2 } A e^{ i k x } (- i \hbar )( i k )A e^{ i k x } [/itex]

[itex] < p > = A^2 \int_{x_1 }^{x_2 } e^{ 0 } (- i \hbar )( i k ) [/itex]

[itex] < p > = A^2 \hbar k ( x_2 - x_1 ) [/itex]

If I were to use a reverse travelling wave, instead [itex]\psi_r(x) = e^{- i k x }[/itex], the solution is the same but with reversed sign:

[itex] \int_{x_1 }^{x_2 } A e^{ i k x } (- i \hbar ){ \partial \over { \partial x }} e^{ -i k x } [/itex]

[itex] < p > = \int_{x_1 }^{x_2 } e^{ -i k x } (- i \hbar )( i k ) A e^{ i k x } [/itex]

[itex] < p > = A^2 \int_{x_1 }^{x_2 } e^{ 0 } (i \hbar )( i k ) [/itex]

[itex] < p > = - A^2 \hbar k ( x_2 - x_1 ) [/itex]

If A is the normalisation constant for an infinite well, over all space, then [itex]A={1 \over \sqrt{ x_2 - x_1 }}[/itex] and [itex]<p>=- \hbar k[/itex]

That's the obvious, and correct result.

Since these solutions are all linear and hermitian, I'm also supposed to be able to add together solutions to make another solution.... but, that's where I get into trouble!

The sum of the solved expectation of momentums is clearly 0. The forward and reverse waves cancel each other's momentum out. A stationary state neither moves (on average) to the right or to the left, but simply oscillates in place. So, the sum of expectation values gives the intuitive result.

However, when I attempt to solve for momentum using the sum of the forward and reverse wave-functions [itex] \psi = \psi_f + \psi_r [/itex] ... I don't get a real number.

I thought that momentum being a "linear" Hermitian operator, meant it would produce the same result whether working on individual waves, and summing the results ... or a superposition of waves with a single result. Did I misunderstand something, or did I make a mistake in the following computation; because I get a purely imaginary result:

[itex] < p > = \int_{x_1}^{x_2} A (e^{ -i k x } + e^{ i k x } ) \cdot { - i \hbar \partial \over { \partial x } } A (e^{ i k x } + e^{ -i k x } )[/itex]

[itex] < p > = \int_{x_1}^{x_2} A (e^{ -i k x } + e^{ i k x } ) \cdot { - i \hbar } A ( i k e^{ i k x } - i k e^{ -i k x } )[/itex]

[itex] < p > = A^2 \hbar k \int_{x_1}^{x_2} ( e^{ -i k x } + e^{ i k x } ) ( e^{ i k x } - e^{ -i k x } )[/itex]

[itex] < p > = A^2 \hbar k \int_{x_1}^{x_2} (e^{ 2 i k x } - e^{ -2 i k x } ) [/itex]

[itex] < p > = A^2 \hbar k \int_{x_1}^{x_2} (i 2 sin( 2 k x) ) [/itex]

Since the right hand side of the equation is is purely imaginary, then the integral must be purely imaginary ... and so must the result.

The issue seems to be that sum of products, is not the same as product of sums:

( a + 1/a ) (-1/a + a) = a*a - 1/(a*a)

(a*-1/a) + (1/a*a) = 0

But momentum operator notation sets the problem up differently depending on whether it's passed individual wave components or sums.

How am I supposed to correctly do operatior notation on a superposition of solutions to the wave equation?