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Thanks if you take the time to read this. Here is the problem statement: problem statement

The problem I'm getting is that I'm not getting the kinetic energy diagonal when I convert to the coordinates that diagonalize the potential energy. If you scroll all the way down you should see there's a cross derivative term involving the partial derivatives w.r.t $q_1$ and $q_2$. I haven't been able to spot any significant mistakes. So is there some other way to simultaneously diagonalize potential and the conjugate momenta in the new coordinates? Is there something I'm missing? How does the $\frac{1}{2(2M + m_2)}$ factor come into play?

First we make a transformation in the x coordinates so that the $a$'s go away. Namely $$\begin{bmatrix} x_1^\prime \\x_2^\prime \\ x_3^\prime\end{bmatrix} = \begin{bmatrix} x_1 \\x_2 \\ x_3 \end{bmatrix} - \begin{bmatrix} a/2 \\-a/2 \\ -3a/2\end{bmatrix} $$ And I will refer to the $x_i^\prime$ coordinates without the prime hereafter. $$(x_2 - x_1)^2 + (x_3 - x_2)^2 = x_1^2 +x_2^2 - 2x_1x_2 + x_2^2+ x_3^2 - 2x_2x_3 $$ There are multiple choices for the matrix which below which would produce the above algebraic expression. We want to choose the one which Hermitian, so that the corresponding transformation is unitary. $$= \begin{bmatrix} x_1 &x_2 & x_3\end{bmatrix}\begin{bmatrix} 1 &-1 &0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\x_2 \\ x_3\end{bmatrix} $$ $$det\begin{bmatrix} 1 - \lambda &-1 &0 \\ -1 &2- \lambda &-1 \\ 0 & -1 & 1- \lambda \end{bmatrix} = \left(1 - \lambda \right)^2\left(2 - \lambda \right) - \left(1 - \lambda \right) - \left(1 - \lambda \right) $$ \begin{align*} = \left(1 - \lambda \right)\left[\left(1 - \lambda \right)\left(2 - \lambda \right) -2 \right] &= \left(1 - \lambda \right)\left[2 - 3\lambda + \lambda^2 -2 \right] \\ & =\lambda\left(1 - \lambda \right)\left(\lambda - 3 \right) \end{align*} So the set of eigenvalues is $\left\{0, 1, 3 \right\}$. $$U = \begin{bmatrix} 1/\sqrt{3} &1/\sqrt{3} &1/\sqrt{3} \\ 1/\sqrt{2} &0 &-1/\sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \end{bmatrix}$$ $$ U^{T}\begin{bmatrix} 0 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 3 \end{bmatrix}U = \begin{bmatrix} 1 &-1 &0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}$$ $$\begin{bmatrix} q_1 \\q_2 \\ q_3\end{bmatrix} = U \begin{bmatrix} x_1 \\x_2 \\ x_3\end{bmatrix}$$ Next begins the calculation of the partial derivatives of $x_i$ in the new coordinate frame, as the first step to finding the KE in terms of the momenta conjugate to the new coordinates $q_i$. $$\begin{bmatrix}\frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3}\end{bmatrix} = \begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}U$$ $$KE = -\hbar^2\begin{bmatrix}\frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3}\end{bmatrix} \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix} \begin{bmatrix}\frac{\partial }{\partial x_1} \\ \frac{\partial }{\partial x_2} \\ \frac{\partial }{\partial x_3}\end{bmatrix} $$ $$ = -\hbar^2\begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}U \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix}U^{T}\begin{bmatrix}\frac{\partial }{\partial q_1} \\ \frac{\partial }{\partial q_2} \\ \frac{\partial }{\partial q_3}\end{bmatrix} $$ $$ U \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix}U^{T}$$ $$= \begin{bmatrix} 1/\sqrt{3} &1/\sqrt{3} &1/\sqrt{3} \\ 1/\sqrt{2} &0 &-1/\sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \end{bmatrix}\begin{bmatrix} \frac{1}{2M\sqrt{3}} & \frac{1}{2M\sqrt{2}} &\frac{1}{2M\sqrt{6}} \\ \frac{1}{2m_2\sqrt{3}} & 0 &-\frac{2}{2m_2\sqrt{6}} \\ \frac{1}{2M\sqrt{3}} & -\frac{1}{2M\sqrt{2}} & \frac{1}{2M\sqrt{6}} \end{bmatrix}$$ $$ = \begin{bmatrix} \frac{1}{3M} + \frac{1}{6m_2} & 0 & \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right) \\ 0 & \frac{1}{2M} & 0 \\ \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right) & 0 & \frac{1}{6M} + \frac{1}{3m_2} \end{bmatrix} $$ $$\frac{KE}{-\hbar^2} = \begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix} \begin{bmatrix}\left(\frac{1}{3M} + \frac{1}{6m_2} \right)\frac{\partial }{\partial q_1} + \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial }{\partial q_3}\\ \frac{1}{2M}\frac{\partial }{\partial q_2} \\ \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial }{\partial q_1}+ \left(\frac{1}{6M} + \frac{1}{3m_2} \right)\frac{\partial }{\partial q_3}\end{bmatrix} $$ $$= \left(\frac{1}{3M} + \frac{1}{6m_2} \right)\frac{\partial^2 }{\partial q_1^2} + \frac{1}{2M}\frac{\partial^2 }{\partial q_2^2} + \left(\frac{1}{6M} + \frac{1}{3m_2} \right)\frac{\partial^2 }{\partial q_3^2}$$ $$+ \frac{\sqrt{2}}{3}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial^2 }{\partial q_1 \partial q_3} $$ So to me the fact that there is a cross term means that I won't able to express the Hamiltonian in terms of three independent oscillators.

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  • $\begingroup$ From reading Goldstein about the corresponding classical problem, I seem to have gathered that the choice of U was not unique, and that I can somehow add the further restriction that the potential energy is diagonal. $\endgroup$ – WhatIAm Oct 28 '13 at 3:27
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$\newcommand{\oh}[0]{\frac{1}{2}}$As far as I can tell, this question doesn't involve quantum at all. It is essentially a question about how to find normal modes. You try and fail to find a normal mode. This is because you restrict yourself to orthogonal coordinate transformations. We will see what I mean.

Let's start with the minimal example, two masses $m_1$ and $m_2$ connect by a spring with spring constant $k$. The lagrangian is $L = \oh m_1 \dot{x}_1^2 + \oh m_2 \dot{x}_2^2 - \oh k (x_1 - x_2)^2.$ The solution is hopefully elementary at this point: we know the center of mass moves with constant velocity and the relative coordinate's motion can be found using the idea of reduced mass $\mu$ defined by $\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$. If $r = x_1 - x_2$, then $\ddot{r} = - \omega^2 r$, where $\omega^2 = \frac{k}{\mu}$. Thus our normal mode coordinates are $q_1 =\frac{1}{m_1 + m_2}(m_1 x_1 + m_2 x_2)$ and $q_2 = x_1 - x_2$. Notice the transformation between coordinates is not orthogonal. This is necessary to simultaneously diagonalize both the mass term and potential term in the lagrangian. Now that you have seen how this works with a familiar example, let's move on to the general case.

I will adapt the discussion in landau mechanics 3rd edition starting at the bottom of page 65. In general the hamiltonian is $L = \oh \dot{x}_i m_{ij} \dot{x}_j - \oh x_i V_{ij} x_j$. The equation of motion (EOM) is $m_{ij} \ddot{x}_j = - V_{ij} x_j$. We seek a solution where each coordinate oscillates with the same frequency: $x_j(t) = A_j \sin(\omega t + \phi_j)$. Plugging this into the EOM we get $\omega^2 m_{ij} A_j - V_{ij} A_j = 0$. i.e. $A_j$ is a null vector of $\omega^2 m_{ij} - V_{ij}$. This is only possible if $\omega$ takes on such a value where the determinant of $\omega^2 m_{ij} - V_{ij}$ is zero. The number of $\omega^2$'s which give zero determinant will be equal to the dimensionality of our system. We will index this set of $\omega$'s by the greek letter $\alpha$.

Now say we have found a $\omega_\alpha$ where the determinant of $\omega_\alpha^2 m_{ij} - V_{ij}$ is zero. Then we must find the $A_{k\alpha}$ which is a null vector of $\omega_\alpha^2 m_{ij} - V_{ij}$. Notice that the $A_{k\alpha}$ will not be orthogonal because if $v_{k\alpha}$ are the solution to the eigenvalue problem for $ V_{ij} m^{-1}_{ij}$, then $A_{k\alpha} = m_{kj} v_{j \alpha}$, and the $A_{k\alpha}$ will not in general be orthogonal, because the $v_{k \alpha}$ are.

Having done this for each $\alpha$, we find that the most general solution is $x_k(t) = A_{k \alpha} q_\alpha \sin(\omega_\alpha t - \phi_\alpha)$. The $q_\alpha$ are seen to be the normal mode coordinates.

I will leave it to you to apply this strategy to your problem. Work out the two mass problem as a sanity check first if you want.

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  • $\begingroup$ I do not understand your argument for why $A_{k\alpha}$ is not orthogonal. Also, there is some difference with the quantum problem because it seems to me partial derivatives do not transform like velocities. However I think due to the symmetry of $m_{ij}$ this doesn't matter. I had already found the $\omega_{\alpha}^2$ classically so I can proceed from there. $\endgroup$ – WhatIAm Oct 28 '13 at 17:38
  • $\begingroup$ I said that the $A_{k\alpha}$ are the null vectors of $\omega^2_\alpha m_{ij} - V_{ij}$. You can check that if $v_{j\alpha}$ are the eigenvectors of $V_{ij} m^{-1}_{ij}$, then $A_{i \alpha} = m_{ij} v_{j\alpha}$. We know the $v_{j\alpha}$ are orthogonal, i.e., $v_{j\alpha}v_{j\beta} = \delta_{\alpha \beta}$. This means that the $A_{i \alpha}$ will not in general be orthogonal: $A_{i \alpha} A_{i \beta} = v_{i \alpha} m_{ji} m_{jk} v_{k \beta}$, which is the inner product of the $v$'s using a quadratic form $m_{ji} m_{jk}$. The $v$'s cannot diagonalize both these quadratic forms at once. $\endgroup$ – Brian Moths Oct 28 '13 at 22:57
  • $\begingroup$ partial derivatives do not transform like velocities. This would in fact be important because we are using a non-orthogonal change of coordinates even though the matrices themselves are symmetric. Fortunately, a derivate is not supposed to be like a velocity but a momentum, and derivatives and momenta transform the same way. Thus the classical description does correspond to the quantum description. $\endgroup$ – Brian Moths Oct 28 '13 at 23:01
  • $\begingroup$ By the way, I was under the impression that the plan of attack for this problem was to find the canonical coordinates classically, express the classical hamiltonian in these coordinates, quantize the system, the find the energies. Note that once the hamiltonian has been expressed in terms of canonical coordinates, quantizing and finding energy levels is trivial because it is just independent harmonic oscillators. $\endgroup$ – Brian Moths Oct 28 '13 at 23:03

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