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Thread: Quantum 2

  1. #1
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    Quantum 2

    1.) Derive the expression of $\displaystyle \hat{p}$ in $\displaystyle x$-space. Prove $\displaystyle \hat{p}$ is linear. Then, prove it is hermitian.

    2.) Find the following commutators, showing two different ways for finding them:

    $\displaystyle [x,\hat{H}]$

    $\displaystyle [\hat{p}, \hat{H} + x]$

    (For first way use $\displaystyle [x,p] = i\bar{h}$, and second way using properties of commutators).

    3.) Show $\displaystyle Ae^{-ikx}\ \ \ A, k \in \mathbb{R}$ is an eigenfunction of $\displaystyle \hat{p}$. Determine the eigenvalue.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DiscreteW View Post
    1.) Derive the expression of $\displaystyle \hat{p}$ in $\displaystyle x$-space. Prove $\displaystyle \hat{p}$ is linear. Then, prove it is hermitian.

    2.) Find the following commutators, showing two different ways for finding them:

    $\displaystyle [x,\hat{H}]$

    $\displaystyle [\hat{p}, \hat{H} + x]$

    (For first way use $\displaystyle [x,p] = i\bar{h}$, and second way using properties of commutators).

    3.) Show $\displaystyle Ae^{-ikx}\ \ \ A, k \in \mathbb{R}$ is an eigenfunction of $\displaystyle \hat{p}$. Determine the eigenvalue.
    I'll get back to you on these.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DiscreteW View Post
    1.) Derive the expression of $\displaystyle \hat{p}$ in $\displaystyle x$-space. Prove $\displaystyle \hat{p}$ is linear. Then, prove it is hermitian.
    I can do this fairly easily in the "bra-ket" representation, but for some reason I'm getting a migraine from trying to rewrite it in terms of wavefunctions. I'll either get back to you on this, or I won't.

    Quote Originally Posted by DiscreteW View Post
    2.) Find the following commutators, showing two different ways for finding them:

    $\displaystyle [x,\hat{H}]$

    $\displaystyle [\hat{p}, \hat{H} + x]$

    (For first way use $\displaystyle [x,p] = i\bar{h}$, and second way using properties of commutators).
    I'm going to leave the "hats" off. We know these are operators. (And besides, the x needs one, too!)

    I'm sure one of the ways is
    $\displaystyle [ x, H ] = xH - Hx$

    $\displaystyle = x \left ( \frac{p^2}{2m} + V \right ) - \left ( \frac{p^2}{2m} + V \right )x$

    Now expand this out, do the expansion for $\displaystyle [ x, p^2 ] $, note that if $\displaystyle V = V(x)$ then $\displaystyle [ x, V(x) ] = 0$, etc.

    I'm not sure if this is the other way you are thinking of. It may be shown that
    $\displaystyle [ x, G(p) ] = i \hbar \frac{dG}{dp}$
    and
    $\displaystyle [ F(x), p ] = i \hbar \frac{dF}{dx}$

    Quote Originally Posted by DiscreteW View Post
    3.) Show $\displaystyle Ae^{-ikx}\ \ \ A, k \in \mathbb{R}$ is an eigenfunction of $\displaystyle \hat{p}$. Determine the eigenvalue.
    If $\displaystyle \psi$ is an eigenvalue of the momentum operator p, then
    $\displaystyle p \psi = \lambda \psi$
    where $\displaystyle \lambda$ is some constant.

    We know that
    $\displaystyle p = -i \hbar \frac{d}{dx}$

    So solve
    $\displaystyle -i \hbar \frac{d}{dx} \left ( Ae^{-ikx} \right ) = \lambda \left ( Ae^{-ikx} \right )$
    for $\displaystyle \lambda$

    -Dan
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  4. #4
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    I haven't been able to "derive" $\displaystyle \hat{p}$, but I proved $\displaystyle \hat{p}$ is hermitian! Took a while:

    We know it's hermitian if:

    $\displaystyle
    \int_{-\infty}^{\infty}\psi^*(\hat{p}\,\psi)~dx = \left[\int_{-\infty}^{\infty}\psi^*(\hat{p}\,\psi)\right]^*dx$

    $\displaystyle = \int_{-\infty}^{\infty}\psi(\hat{p}^*\psi^*)~dx$

    And my proof is:

    We know that $\displaystyle \hat{p} = -ih\frac{d}{dx}$. So, we have:

    $\displaystyle \int_{-\infty}^{\infty} \psi^*\left(-i\hbar\frac{d}{dx}\right)\psi~dx = -i\hbar\int_{-\infty}^{\infty} \psi^*\frac{d}{dx}\psi~dx$

    $\displaystyle = -i\hbar\int_{-\infty}^{\infty}\psi^*d\psi$


    Using integration by parts. Recall, it is $\displaystyle uv - \int{v\,du}$. Let $\displaystyle u = \psi^*$, and $\displaystyle v = \psi$. The rest is trivial. Hence, we have:

    $\displaystyle -i\hbar\int_{-\infty}^{\infty}\psi^*d\psi = -i\hbar\left(\psi^*\psi |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi d\,\psi^* dx\right)$

    $\displaystyle = -i\hbar\left(0-\int_{-\infty}^{\infty}\psi\,\frac{d}{dx}\psi^*dx\right)$

    $\displaystyle = i\hbar\int_{-\infty}^{\infty}\psi\,\frac{d}{dx}\psi^*dx$


    Bring $\displaystyle i\hbar$ back into the integral and group terms (now it's negative, as we'll be taking the conjugate:

    $\displaystyle \int_{-\infty}^{\infty}\psi\left(-i\hbar\frac{d}{dx}\psi\right)^*dx$

    And thus we've shown that:

    $\displaystyle \int_{-\infty}^{\infty}\psi^*\left(-i\hbar\frac{d}{dx}\right)\psi~dx = \int_{-\infty}^{\infty}\left(-i\hbar\frac{d}{dx}\psi\right)^*\psi~dx$

    which is the definition of herminicity. Thus, the proof is complete.

    But, my proof for showing it's linear needs some help:

    Proof:

    $\displaystyle \left(-i\hbar\frac{d}{dx}\right)\,[c_1\,\psi + c_2\psi] = \left(-c_1i\hbar\frac{d\psi}{dx}\right) - \left(c_2i\hbar\frac{d\psi}{dx}\right)$

    We can factor the constants out, since it'll just be another constant. But not sure how to show its linear from here.

    -------------------------


    Struggling on 2 still, but I'm working on it.


    -------------------------

    Found $\displaystyle \lambda$ with your guidance. Lots of Quantum this week . Thanks topsquark.
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  5. #5
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    Still need help with the linear part, but I worked on #2. It's quite messy, and took a long time, but I'm hoping you could help.

    For the first method, we use the identity $\displaystyle [\hat{A},\hat{B}\hat{C}] = [\hat{A},\hat{B}]\hat{C} + \hat{B}[\hat{A},\hat{C}]$. Also, we use the fact that $\displaystyle [x,p] = i\hbar$. Thus, we have:

    $\displaystyle [x,\hat{H}]\psi = [x,\frac{p^2}{2m}]\psi$
    $\displaystyle = [x,p\cdot\frac{p}{2m}]\psi$
    $\displaystyle = [x,p]\frac{p}{2m}\psi + p[x,\frac{p}{2m}]\psi$
    $\displaystyle = i\hbar\frac{p}{2m}\psi + p[x,p\cdot\frac{1}{2m}]\psi$
    $\displaystyle = i\hbar\frac{p}{2m}\psi + p[x,p]\frac{1}{2m}\psi + p[x,\frac{1}{2m}]\psi$
    $\displaystyle = i\hbar\frac{p}{2m}\psi + \frac{p}{2m}i\hbar + p\frac{x}{2m}\psi + \frac{x}{2m}\psi$
    $\displaystyle = \boxed{2i\hbar\frac{p}{2m}\psi + 2\frac{px}{2m}\psi}$

    Second method:

    $\displaystyle [x,\hat{H}]\psi = x\hat{H}\psi - \hat{H}x\psi$
    $\displaystyle = x \left ( \frac{p^2}{2m} + V(x) \right )\psi - \left ( \frac{p^2}{2m} + V(x) \right )x\psi$
    $\displaystyle = \frac{xp^2\psi}{2m} + x V(x)\psi - \frac{p^2x\psi}{2m} - V(x) x\psi$

    Obviously need help with it, but it looks somewhat right.
    -----------------------------------------------------

    Next one:

    For the second one, the first method I use uses the multiplication property above, in addition to $\displaystyle [\hat{A},\hat{B}\hat{C}] = [\hat{A},\hat{B}]\hat{C} + \hat{B}[\hat{A},\hat{C}]$. Hence, we have:

    $\displaystyle
    [\hat{p},\hat{H}]\psi = [\hat{p},\hat{H}]\psi + [\hat{p},x]\psi$
    $\displaystyle =[i\hbar\frac{\partial}{\partial p},\frac{p^2}{2m} + V]\psi + [i\hbar\frac{\partial}{\partial p},x]\psi$
    $\displaystyle =[i\hbar\frac{\partial}{\partial p}, \frac{p^2}{2m}]\psi + [i\hbar\frac{\partial}{\partial p}, V]\psi + [i\hbar\frac{\partial}{\partial p},x]\psi$
    $\displaystyle =[i\hbar\frac{\partial}{\partial p}, p\cdot \frac{p}{2m}]\psi + [i\hbar\frac{\partial}{\partial p}, V]\psi + [i\hbar\frac{\partial}{\partial p},x]\psi$
    $\displaystyle = [i\hbar\frac{\partial}{partial p}\frac{p}{2m}\psi + p[i\hbar\frac{\partial}{\partial p},\frac{p}{2m}]\psi + [i\hbar\frac{\partial}{\partial p}, V]\psi + [i\hbar\frac{\partial}{\partial p},x]\psi$
    $\displaystyle = \left(i\hbar\frac{\partial}{\partial p}p + pi\hbar\frac{\partial}{\partial p}\right)\frac{p}{2m}\psi + p\left(i\hbar\frac{\partial}{\partial p}\frac{p}{2m} + \frac{p}{2m}i\hbar\frac{\partial}{\partial p}\right)\psi + [i\hbar\frac{\partial}{\partial p}, V]\psi + [i\hbar\frac{\partial}{\partial p},x]\psi$
    $\displaystyle = i\hbar\frac{\partial}{\partial p}\frac{p^2}{2m}\psi + pi\hbar\frac{\partial}{\partial p}\frac{p}{2m}\psi + pi\hbar\frac{\partial}{\partial p}\frac{p}{2m}\psi + \frac{p}{2m}i\hbar\frac{\partial}{\partial p}\psi + [i\hbar\frac{\partial}{\partial p}, V]\psi + [i\hbar\frac{\partial}{\partial p},x]\psi$

    Then of course we need to expand the other 2 brackets that I didn't do above at the end. But look how insanely long and complicated this is getting.


    Second method:

    $\displaystyle [\hat{p}, \hat{H} + x]\psi = \hat{p}(\hat{H}+x)\psi + (\hat{H}+x)\hat{p}\psi$
    $\displaystyle = i\hbar\frac{\partial}{\partial p} \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)\psi + \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)i\hbar\frac{\partial}{\partial p}\psi$


    Ahhhh!
    Last edited by DiscreteW; Mar 8th 2008 at 12:17 PM.
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