commutator

2D SHO xy perturbation

December 7, 2015 phy1520 , , , , ,

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Q: [1] pr. 5.4

Given a 2D SHO with Hamiltonian

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:20}
H_0 = \inv{2m} \lr{ p_x^2 + p_y^2 } + \frac{m \omega^2}{2} \lr{ x^2 + y^2 },
\end{equation}

  • (a)
    What are the energies and degeneracies of the three lowest states?

  • (b)
    With perturbation

    \begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:40}
    V = m \omega^2 x y,
    \end{equation}

    calculate the first order energy perturbations and the zeroth order perturbed states.

  • (c)
    Solve the \( H_0 + \delta V \) problem exactly, and compare.

A: part (a)

Recall that we have

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:60}
H \ket{n_1, n_2} =
\Hbar\omega
\lr{
n_1 + n_2 + 1
}
\ket{n_1, n_2},
\end{equation}

So the three lowest energy states are \( \ket{0,0}, \ket{1,0}, \ket{0,1} \) with energies \( \Hbar \omega, 2 \Hbar \omega, 2 \Hbar \omega \) respectively (with a two fold degeneracy for the second two energy eigenkets).

A: part (b)

Consider the action of \( x y \) on the \( \beta = \setlr{ \ket{0,0}, \ket{1,0}, \ket{0,1} } \) subspace. Those are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:200}
\begin{aligned}
x y \ket{0,0}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{0,0} \\
&=
\frac{x_0^2}{2} \lr{ b + b^\dagger } \ket{1,0} \\
&=
\frac{x_0^2}{2} \ket{1,1}.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:220}
\begin{aligned}
x y \ket{1, 0}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{1,0} \\
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \ket{1,1} \\
&=
\frac{x_0^2}{2} \lr{ \ket{0,1} + \sqrt{2} \ket{2,1} } .
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:240}
\begin{aligned}
x y \ket{0, 1}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{0,1} \\
&=
\frac{x_0^2}{2} \lr{ b + b^\dagger } \ket{1,1} \\
&=
\frac{x_0^2}{2} \lr{ \ket{1,0} + \sqrt{2} \ket{1,2} }.
\end{aligned}
\end{equation}

The matrix representation of \( m \omega^2 x y \) with respect to the subspace spanned by basis \( \beta \) above is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:260}
x y
\sim
\inv{2} \Hbar \omega
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{bmatrix}.
\end{equation}

This diagonalizes with

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:300}
U
=
\begin{bmatrix}
1 & 0 \\
0 & \tilde{U}
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:320}
\tilde{U}
=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1 \\
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:340}
D =
\inv{2} \Hbar \omega
\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \\
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:360}
x y = U D U^\dagger = U D U.
\end{equation}

The unperturbed Hamiltonian in the original basis is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:380}
H_0
=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & 2 I
\end{bmatrix},
\end{equation}

So the transformation to the diagonal \( x y \) basis leaves the initial Hamiltonian unaltered

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:400}
\begin{aligned}
H_0′
&= U^\dagger H_0 U \\
&=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & \tilde{U} 2 I \tilde{U}
\end{bmatrix} \\
&=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & 2 I
\end{bmatrix}.
\end{aligned}
\end{equation}

Now we can compute the first order energy shifts almost by inspection. Writing the new basis as \( \beta’ = \setlr{ \ket{0}, \ket{1}, \ket{2} } \) those energy shifts are just the diagonal elements from the \( x y \) operators matrix representation

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:420}
\begin{aligned}
E^{{(1)}}_0 &= \bra{0} V \ket{0} = 0 \\
E^{{(1)}}_1 &= \bra{1} V \ket{1} = \inv{2} \Hbar \omega \\
E^{{(1)}}_2 &= \bra{2} V \ket{2} = -\inv{2} \Hbar \omega.
\end{aligned}
\end{equation}

The new energies are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:440}
\begin{aligned}
E_0 &\rightarrow \Hbar \omega \\
E_1 &\rightarrow \Hbar \omega \lr{ 2 + \delta/2 } \\
E_2 &\rightarrow \Hbar \omega \lr{ 2 – \delta/2 }.
\end{aligned}
\end{equation}

A: part (c)

For the exact solution, it’s possible to rotate the coordinate system in a way that kills the explicit \( x y \) term of the perturbation. That we could do this for \( x, y \) operators wasn’t obvious to me, but after doing so (and rotating the momentum operators the same way) the new operators still have the required commutators. Let

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:80}
\begin{aligned}
\begin{bmatrix}
u \\
v
\end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} \\
&=
\begin{bmatrix}
x \cos\theta + y \sin\theta \\
-x \sin\theta + y \cos\theta
\end{bmatrix}.
\end{aligned}
\end{equation}

Similarly, for the momentum operators, let
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:100}
\begin{aligned}
\begin{bmatrix}
p_u \\
p_v
\end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
p_x \\
p_y
\end{bmatrix} \\
&=
\begin{bmatrix}
p_x \cos\theta + p_y \sin\theta \\
-p_x \sin\theta + p_y \cos\theta
\end{bmatrix}.
\end{aligned}
\end{equation}

For the commutators of the new operators we have

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:120}
\begin{aligned}
\antisymmetric{u}{p_u}
&=
\antisymmetric{x \cos\theta + y \sin\theta}{p_x \cos\theta + p_y \sin\theta} \\
&=
\antisymmetric{x}{p_x} \cos^2\theta + \antisymmetric{y}{p_y} \sin^2\theta \\
&=
i \Hbar \lr{ \cos^2\theta + \sin^2\theta } \\
&=
i\Hbar.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:140}
\begin{aligned}
\antisymmetric{v}{p_v}
&=
\antisymmetric{-x \sin\theta + y \cos\theta}{-p_x \sin\theta + p_y \cos\theta} \\
&=
\antisymmetric{x}{p_x} \sin^2\theta + \antisymmetric{y}{p_y} \cos^2\theta \\
&=
i \Hbar.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:160}
\begin{aligned}
\antisymmetric{u}{p_v}
&=
\antisymmetric{x \cos\theta + y \sin\theta}{-p_x \sin\theta + p_y \cos\theta} \\
&= \cos\theta \sin\theta \lr{ -\antisymmetric{x}{p_x} + \antisymmetric{y}{p_p} } \\
&=
0.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:180}
\begin{aligned}
\antisymmetric{v}{p_u}
&=
\antisymmetric{-x \sin\theta + y \cos\theta}{p_x \cos\theta + p_y \sin\theta} \\
&= \cos\theta \sin\theta \lr{ -\antisymmetric{x}{p_x} + \antisymmetric{y}{p_p} } \\
&=
0.
\end{aligned}
\end{equation}

We see that the new operators are canonical conjugate as required. For this problem, we just want a 45 degree rotation, with

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:460}
\begin{aligned}
x &= \inv{\sqrt{2}} \lr{ u + v } \\
y &= \inv{\sqrt{2}} \lr{ u – v }.
\end{aligned}
\end{equation}

We have
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:480}
\begin{aligned}
x^2 + y^2
&=
\inv{2} \lr{ (u+v)^2 + (u-v)^2 } \\
&=
\inv{2} \lr{ 2 u^2 + 2 v^2 + 2 u v – 2 u v } \\
&=
u^2 + v^2,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:500}
\begin{aligned}
p_x^2 + p_y^2
&=
\inv{2} \lr{ (p_u+p_v)^2 + (p_u-p_v)^2 } \\
&=
\inv{2} \lr{ 2 p_u^2 + 2 p_v^2 + 2 p_u p_v – 2 p_u p_v } \\
&=
p_u^2 + p_v^2,
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:520}
\begin{aligned}
x y
&=
\inv{2} \lr{ (u+v)(u-v) } \\
&=
\inv{2} \lr{ u^2 – v^2 }.
\end{aligned}
\end{equation}

The perturbed Hamiltonian is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:540}
\begin{aligned}
H_0 + \delta V
&=
\inv{2m} \lr{ p_u^2 + p_v^2 }
+ \inv{2} m \omega^2 \lr{ u^2 + v^2 + \delta u^2 – \delta v^2 } \\
&=
\inv{2m} \lr{ p_u^2 + p_v^2 }
+ \inv{2} m \omega^2 \lr{ u^2(1 + \delta) + v^2 (1 – \delta) }.
\end{aligned}
\end{equation}

In this coordinate system, the corresponding eigensystem is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:560}
H \ket{n_1, n_2}
= \Hbar \omega \lr{ 1 + n_1 \sqrt{1 + \delta} + n_2 \sqrt{ 1 – \delta } } \ket{n_1, n_2}.
\end{equation}

For small \( \delta \)

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:580}
n_1 \sqrt{1 + \delta} + n_2 \sqrt{ 1 – \delta }
\approx
n_1 + n_2
+ \inv{2} n_1 \delta
– \inv{2} n_2 \delta,
\end{equation}

so
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:600}
H \ket{n_1, n_2}
\approx \Hbar \omega \lr{ 1 + n_1 + n_2 + \inv{2} n_1 \delta – \inv{2} n_2 \delta
} \ket{n_1, n_2}.
\end{equation}

The lowest order perturbed energy levels are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:620}
\ket{0,0} \rightarrow \Hbar \omega
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:640}
\ket{1,0} \rightarrow \Hbar \omega \lr{ 2 + \inv{2} \delta }
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:660}
\ket{0,1} \rightarrow \Hbar \omega \lr{ 2 – \inv{2} \delta }
\end{equation}

The degeneracy of the \( \ket{0,1}, \ket{1,0} \) states has been split, and to first order match the zeroth order perturbation result.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

PHY1520H Graduate Quantum Mechanics. Lecture 16: Addition of angular momenta. Taught by Prof. Arun Paramekanti

November 17, 2015 phy1520 , , ,

[Click here for a PDF of this post with nicer formatting]

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 3 content.

Addition of angular momenta

  • For orbital angular momentum

    \begin{equation}\label{eqn:qmLecture16:20}
    \begin{aligned}
    \hat{\BL}_1 &= \hat{\Br}_1 \cross \hat{\Bp}_1 \\
    \hat{\BL}_1 &= \hat{\Br}_1 \cross \hat{\Bp}_1,
    \end{aligned}
    \end{equation}

    We can show that it is true that

    \begin{equation}\label{eqn:qmLecture16:40}
    \antisymmetric{L_{1i} + L_{2i}}{L_{1j} + L_{2j}} =
    i \Hbar \epsilon_{i j k} \lr{ L_{1k} + L_{2k} },
    \end{equation}

    because the angular momentum of the independent particles commute. Given this is it fair to consider that the sum

    \begin{equation}\label{eqn:qmLecture16:60}
    \hat{\BL}_1 + \hat{\BL}_2
    \end{equation}

    is also angular momentum.

  • Given \( \ket{l_1, m_1} \) and \( \ket{l_2, m_2} \), if a measurement is made of \( \hat{\BL}_1 + \hat{\BL}_2 \), what do we get?

    Specifically, what do we get for

    \begin{equation}\label{eqn:qmLecture16:80}
    \lr{\hat{\BL}_1 + \hat{\BL}_2}^2,
    \end{equation}

    and for
    \begin{equation}\label{eqn:qmLecture16:100}
    \lr{\hat{L}_{1z} + \hat{L}_{2z}}.
    \end{equation}

    For the latter, we get

    \begin{equation}\label{eqn:qmLecture16:120}
    \lr{\hat{L}_{1z} + \hat{L}_{2z}}\ket{ l_1, m_1 ; l_2, m_2 }
    =
    \lr{ \Hbar m_1 + \Hbar m_2 } \ket{ l_1, m_1 ; l_2, m_2 }
    \end{equation}

    Given
    \begin{equation}\label{eqn:qmLecture16:140}
    \hat{L}_{1z} + \hat{L}_{2z} = \hat{L}_z^{\textrm{tot}},
    \end{equation}

    we find
    \begin{equation}\label{eqn:qmLecture16:160}
    \begin{aligned}
    \antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_1^2} &= 0 \\
    \antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_2^2} &= 0 \\
    \antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_{1z}} &= 0 \\
    \antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_{1z}} &= 0.
    \end{aligned}
    \end{equation}

    We also find

    \begin{equation}\label{eqn:qmLecture16:180}
    \begin{aligned}
    \antisymmetric{(\hat{\BL}_1 + \hat{\BL}_2)^2}{\BL_1^2}
    &=
    \antisymmetric{\hat{\BL}_1^2 + \hat{\BL}_2^2 + 2 \hat{\BL}_1 \cdot
    \hat{\BL}_2}{\BL_1^2} \\
    &=
    0,
    \end{aligned}
    \end{equation}

    but for
    \begin{equation}\label{eqn:qmLecture16:200}
    \begin{aligned}
    \antisymmetric{(\hat{\BL}_1 + \hat{\BL}_2)^2}{\hat{L}_{1z}}
    &=
    \antisymmetric{\hat{\BL}_1^2 + \hat{\BL}_2^2 + 2 \hat{\BL}_1 \cdot
    \hat{\BL}_2}{\hat{L}_{1z}} \\
    &=
    2 \antisymmetric{\hat{\BL}_1 \cdot \hat{\BL}_2}{\hat{L}_{1z}} \\
    &\ne 0.
    \end{aligned}
    \end{equation}

Classically if we have measured \( \hat{\BL}_{1} \) and \( \hat{\BL}_{2} \) then we know the total angular momentum as sketched in fig. 1.

fig. 1. Classical addition of angular momenta.

fig. 1. Classical addition of angular momenta.

In QM where we don’t know all the components of the angular momenum simultaneously, things get fuzzier. For example, if the \( \hat{L}_{1z} \) and \( \hat{L}_{2z} \) components have been measured, we have the angular momentum defined within a conical region as sketched in fig. 2.

fig. 2. Addition of angular momenta given measured L_z

fig. 2. Addition of angular momenta given measured L_z

Suppose we know \( \hat{L}_z^{\textrm{tot}} \) precisely, but have impricise information about \( \lr{\hat{\BL}^{\textrm{tot}}}^2 \). Can we determine bounds for this? Let \( \ket{\psi} = \ket{ l_1, m_2 ; l_2, m_2 } \), so

\begin{equation}\label{eqn:qmLecture16:220}
\begin{aligned}
\bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi}
&=
\bra{\psi} \hat{\BL}_1^2 \ket{\psi}
+ \bra{\psi} \hat{\BL}_2^2 \ket{\psi}
+ 2 \bra{\psi} \hat{\BL}_1 \cdot \hat{\BL}_2 \ket{\psi} \\
&=
l_1 \lr{ l_1 + 1} \Hbar^2
+ l_2 \lr{ l_2 + 1} \Hbar^2
+ 2
\bra{\psi} \hat{\BL}_1 \cdot \hat{\BL}_2 \ket{\psi}.
\end{aligned}
\end{equation}

Using the Cauchy-Schwartz inequality

\begin{equation}\label{eqn:qmLecture16:240}
\Abs{\braket{\phi}{\psi}}^2 \le
\Abs{\braket{\phi}{\phi}}
\Abs{\braket{\psi}{\psi}},
\end{equation}

which is the equivalent of the classical relationship
\begin{equation}\label{eqn:qmLecture16:260}
\lr{ \BA \cdot \BB }^2 \le \BA^2 \BB^2.
\end{equation}

Applying this to the last term, we have

\begin{equation}\label{eqn:qmLecture16:280}
\begin{aligned}
\lr{ \bra{\psi} \hat{\BL}_1 \cdot \hat{\BL}_2 \ket{\psi} }^2
&\le
\bra{ \psi} \hat{\BL}_1 \cdot \hat{\BL}_1 \ket{\psi}
\bra{ \psi} \hat{\BL}_2 \cdot \hat{\BL}_2 \ket{\psi} \\
&=
\Hbar^4
l_1 \lr{ l_1 + 1 }
l_2 \lr{ l_2 + 2 }.
\end{aligned}
\end{equation}

Thus for the max we have

\begin{equation}\label{eqn:qmLecture16:300}
\bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi}
\le
\Hbar^2 l_1 \lr{ l_1 + 1 }
+\Hbar^2 l_2 \lr{ l_2 + 1 }
+ 2 \Hbar^2 \sqrt{ l_1 \lr{ l_1 + 1 } l_2 \lr{ l_2 + 2 } }
\end{equation}

and for the min
\begin{equation}\label{eqn:qmLecture16:360}
\bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi}
\ge
\Hbar^2 l_1 \lr{ l_1 + 1 }
+\Hbar^2 l_2 \lr{ l_2 + 1 }
– 2 \Hbar^2 \sqrt{ l_1 \lr{ l_1 + 1 } l_2 \lr{ l_2 + 2 } }.
\end{equation}

To try to pretty up these estimate, starting with the max, note that if we replace a portion of the RHS with something bigger, we are left with a strict less than relationship.

That is

\begin{equation}\label{eqn:qmLecture16:320}
\begin{aligned}
l_1 \lr{ l_1 + 1 } &< \lr{ l_1 + \inv{2} }^2 \\ l_2 \lr{ l_2 + 1 } &< \lr{ l_2 + \inv{2} }^2 \end{aligned} \end{equation} That is \begin{equation}\label{eqn:qmLecture16:340} \begin{aligned} \bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi} &< \Hbar^2 \lr{ l_1 \lr{ l_1 + 1 } + l_2 \lr{ l_2 + 1 } + 2 \lr{ l_1 + \inv{2} } \lr{ l_2 + \inv{2} } } \\ &= \Hbar^2 \lr{ l_1^2 + l_2^2 + l_1 + l_2 + 2 l_1 l_2 + l_1 + l_2 + \inv{2} } \\ &= \Hbar^2 \lr{ \lr{ l_1 + l_2 + \inv{2} } \lr{ l_1 + l_2 + \frac{3}{2} } - \inv{4} } \end{aligned} \end{equation} or \begin{equation}\label{eqn:qmLecture16:380} l_{\textrm{tot}} \lr{ l_{\textrm{tot}} + 1 } < \lr{ l_1 + l_2 + \inv{2} } \lr{ l_1 + l_2 + \frac{3}{2} } , \end{equation} which, gives \begin{equation}\label{eqn:qmLecture16:400} l_{\textrm{tot}} < l_1 + l_2 + \inv{2}. \end{equation} Finally, given a quantization requirement, that is \begin{equation}\label{eqn:t:1} \boxed{ l_{\textrm{tot}} \le l_1 + l_2. } \end{equation} Similarly, for the min, we find \begin{equation}\label{eqn:qmLecture16:440} \begin{aligned} \bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi} &>
\Hbar^2
\lr{
l_1 \lr{ l_1 + 1 }
+ l_2 \lr{ l_2 + 1 }
– 2 \lr{ l_1 + \inv{2} } \lr{ l_2 + \inv{2} }
} \\
&=
\Hbar^2
\lr{
l_1^2 + l_2^2 %+ \cancel{l_1 + l_2}
– 2 l_1 l_2
– \inv{2}
}
\end{aligned}
\end{equation}

This will be finished Thursday, but we should get

\begin{equation}\label{eqn:t:2}
\boxed{
\Abs{l_1 – l_2} \le l_{\textrm{tot}} \le l_1 + l_2.
}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Third update of aggregate notes for phy1520, Graduate Quantum Mechanics.

November 9, 2015 phy1520 , , , , , , , , , , , , , , , ,

I’ve posted a third update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 13, my solutions for the third problem set, and some additional worked practice problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

Position operator in momentum space representation

November 8, 2015 phy1520 , ,

[Click here for a PDF of this post with nicer formatting]

A derivation of the position space representation of the momentum operator \( -i \Hbar \partial_x \) is made in [1], starting with the position-momentum commutator. Here I’ll repeat that argument for the momentum space representation of the position operator.

What we want to do is expand the matrix element of the commutator. First using the definition of the commutator

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:20}
\bra{p’} X P – P X \ket{p”}
=
i \Hbar \braket{p’}{p”}
=
i \Hbar \delta{p’ – p”},
\end{equation}

and then by inserting an identity operation in a momentum space basis

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:40}
\begin{aligned}
\bra{p’} X P – P X \ket{p”}
&=
\int dp
\bra{p’} X \ket{p}\bra{p} P \ket{p”}
-\int dp
\bra{p’} P \ket{p}\bra{p} X \ket{p”} \\
&=
\int dp
\bra{p’} X \ket{p} p \delta(p – p”)
-\int dp
p \delta(p’ – p)
\bra{p} X \ket{p”} \\
&=
\bra{p’} X \ket{p”} p”
–
p’ \bra{p’} X \ket{p”}.
\end{aligned}
\end{equation}

So we have

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:60}
\bra{p’} X \ket{p”} p”
–
p’ \bra{p’} X \ket{p”}
=
i \Hbar \delta{p’ – p”}.
\end{equation}

Because the RHS is zero whenever \( p’ \ne p” \), the matrix element \( \bra{p’} X \ket{p”} \) must also include a delta function. Let

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:80}
\bra{p’} X \ket{p”} = \delta(p’ – p”) X(p”).
\end{equation}

Because \ref{eqn:positionOperatorInMomentumSpace:60} is an operator equation that really only takes on meaning when applied to a wave function and integrated, we do that

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:100}
\int dp” \delta(p’ – p”) X(p”) p” \psi(p”)
–
\int dp” p’ \delta(p’ – p”) X(p”) \psi(p”)
=
\int dp” i \Hbar \delta{p’ – p”} \psi(p”),
\end{equation}

or
\begin{equation}\label{eqn:positionOperatorInMomentumSpace:120}
i \Hbar \psi(p’)
=
X(p’) p’ \psi(p’)
–
p’
X(p’) \psi(p’).
\end{equation}

Provided \( X(p’) \) operates on everything to its right, this equation is solved by setting

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:140}
\boxed{
X(p’) = i \Hbar \PD{p’}{}.
}
\end{equation}

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

PHY1520H Graduate Quantum Mechanics. Lecture 12: Symmetry (cont.). Taught by Prof. Arun Paramekanti

November 5, 2015 phy1520 , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering chap. 4 content from [1].

Parity (review)

\begin{equation}\label{eqn:qmLecture12:20}
\hat{\Pi} \hat{x} \hat{\Pi} = – \hat{x}
\end{equation}
\begin{equation}\label{eqn:qmLecture12:40}
\hat{\Pi} \hat{p} \hat{\Pi} = – \hat{p}
\end{equation}

These are polar vectors, in contrast to an axial vector such as \( \BL = \Br \cross \Bp \).

\begin{equation}\label{eqn:qmLecture12:60}
\hat{\Pi}^2 = 1
\end{equation}

\begin{equation}\label{eqn:qmLecture12:80}
\Psi(x) \rightarrow \Psi(-x)
\end{equation}

If \( \antisymmetric{\hat{\Pi}}{\hat{H}} = 0 \) then all the eigenstates are either

  • even: \( \hat{\Pi} \) eigenvalue is \( + 1 \).
  • odd: \( \hat{\Pi} \) eigenvalue is \( – 1 \).

We are done with discrete symmetry operators for now.

Translations

Define a (continuous) translation operator

\begin{equation}\label{eqn:qmLecture12:100}
\hat{T}_\epsilon \ket{x} = \ket{x + \epsilon}
\end{equation}

The action of this operator is sketched in fig. 1.

lecture12Fig1

fig. 1. Translation operation.

 

This is a unitary operator

\begin{equation}\label{eqn:qmLecture12:120}
\hat{T}_{-\epsilon} = \hat{T}_{\epsilon}^\dagger = \hat{T}_{\epsilon}^{-1}
\end{equation}

In a position basis, the action of this operator is

\begin{equation}\label{eqn:qmLecture12:140}
\bra{x} \hat{T}_{\epsilon} \ket{\psi} = \braket{x-\epsilon}{\psi} = \psi(x – \epsilon)
\end{equation}

\begin{equation}\label{eqn:qmLecture12:160}
\Psi(x – \epsilon) \approx \Psi(x) – \epsilon \PD{x}{\Psi}
\end{equation}

\begin{equation}\label{eqn:qmLecture12:180}
\bra{x} \hat{T}_{\epsilon} \ket{\Psi}
= \braket{x}{\Psi} – \frac{\epsilon}{\Hbar} \bra{ x} i \hat{p} \ket{\Psi}
\end{equation}

\begin{equation}\label{eqn:qmLecture12:200}
\hat{T}_{\epsilon} \approx \lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{p} }
\end{equation}

A non-infinitesimal translation can be composed of many small translations, as sketched in fig. 2.

fig. 2. Composition of small translations

fig. 2. Composition of small translations

For \( \epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a \), the total translation operator is

\begin{equation}\label{eqn:qmLecture12:220}
\begin{aligned}
\hat{T}_{a}
&= \hat{T}_{\epsilon}^N \\
&= \lim_{\epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a }
\lr{ 1 – \frac{\epsilon}{\Hbar} \hat{p} }^N \\
&= e^{-i a \hat{p}/\Hbar}
\end{aligned}
\end{equation}

The momentum \( \hat{p} \) is called a “Generator” generator of translations. If a Hamiltonian \( H \) is translationally invariant, then

\begin{equation}\label{eqn:qmLecture12:240}
\antisymmetric{\hat{T}_{a}}{H} = 0, \qquad \forall a.
\end{equation}

This means that momentum will be a good quantum number

\begin{equation}\label{eqn:qmLecture12:260}
\antisymmetric{\hat{p}}{H} = 0.
\end{equation}

Rotations

Rotations form a non-Abelian group, since the order of rotations \( \hatR_1 \hatR_2 \ne \hatR_2 \hatR_1 \).

Given a rotation acting on a ket

\begin{equation}\label{eqn:qmLecture12:280}
\hatR \ket{\Br} = \ket{R \Br},
\end{equation}

observe that the action of the rotation operator on a wave function is inverted

\begin{equation}\label{eqn:qmLecture12:300}
\bra{\Br} \hatR \ket{\Psi}
=
\bra{R^{-1} \Br} \ket{\Psi}
= \Psi(R^{-1} \Br).
\end{equation}

Example: Z axis normal rotation

Consider an infinitesimal rotation about the z-axis as sketched in fig. 3(a),(b)

lecture12Fig3

fig 3(a). Rotation about z-axis.

fig 3(b). Rotation about z-axis.

fig 3(b). Rotation about z-axis.

\begin{equation}\label{eqn:qmLecture12:320}
\begin{aligned}
x’ &= x – \epsilon y \\
y’ &= y + \epsilon y \\
z’ &= z
\end{aligned}
\end{equation}

The rotated wave function is

\begin{equation}\label{eqn:qmLecture12:340}
\tilde{\Psi}(x,y,z)
= \Psi( x + \epsilon y, y – \epsilon x, z )
=
\Psi( x, y, z )
+
\epsilon y \underbrace{\PD{x}{\Psi}}_{i \hat{p}_x/\Hbar}
–
\epsilon x \underbrace{\PD{y}{\Psi}}_{i \hat{p}_y/\Hbar}.
\end{equation}

The state must then transform as

\begin{equation}\label{eqn:qmLecture12:360}
\ket{\tilde{\Psi}}
=
\lr{
1
+ i \frac{\epsilon}{\Hbar} \hat{y} \hat{p}_x
– i \frac{\epsilon}{\Hbar} \hat{x} \hat{p}_y
}
\ket{\Psi}.
\end{equation}

Observe that the combination \( \hat{x} \hat{p}_y – \hat{y} \hat{p}_x \) is the \( \hat{L}_z \) component of angular momentum \( \hat{\BL} = \hat{\Br} \cross \hat{\Bp} \), so the infinitesimal rotation can be written

\begin{equation}\label{eqn:qmLecture12:380}
\boxed{
\hatR_z(\epsilon) \ket{\Psi}
=
\lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{L}_z } \ket{\Psi}.
}
\end{equation}

For a finite rotation \( \epsilon \rightarrow 0, N \rightarrow \infty, \phi = \epsilon N \), the total rotation is

\begin{equation}\label{eqn:qmLecture12:420}
\hatR_z(\phi)
=
\lr{ 1 – \frac{i \epsilon}{\Hbar} \hat{L}_z }^N,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture12:440}
\boxed{
\hatR_z(\phi)
=
e^{-i \frac{\phi}{\Hbar} \hat{L}_z}.
}
\end{equation}

Note that \( \antisymmetric{\hat{L}_x}{\hat{L}_y} \ne 0 \).

By construction using Euler angles or any other method, a general rotation will include contributions from components of all the angular momentum operator, and will have the structure

\begin{equation}\label{eqn:qmLecture12:480}
\boxed{
\hatR_\ncap(\phi)
=
e^{-i \frac{\phi}{\Hbar} \lr{ \hat{\BL} \cdot \ncap }}.
}
\end{equation}

Rotationally invariant \( \hat{H} \).

Given a rotationally invariant Hamiltonian

\begin{equation}\label{eqn:qmLecture12:520}
\antisymmetric{\hat{R}_\ncap(\phi)}{\hat{H}} = 0 \qquad \forall \ncap, \phi,
\end{equation}

then every

\begin{equation}\label{eqn:qmLecture12:540}
\antisymmetric{\BL \cdot \ncap}{\hat{H}} = 0,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture12:560}
\antisymmetric{L_i}{\hat{H}} = 0,
\end{equation}

Non-Abelian implies degeneracies in the spectrum.

Time-reversal

Imagine that we have something moving along a curve at time \( t = 0 \), and ending up at the final position at time \( t = t_f \).

fig. 4. Time reversal trajectory.

fig. 4. Time reversal trajectory.

Imagine that we flip the direction of motion (i.e. flipping the velocity) and run time backwards so the final-time state becomes the initial state.

If the time reversal operator is designated \( \hat{\Theta} \), with operation

\begin{equation}\label{eqn:qmLecture12:580}
\hat{\Theta} \ket{\Psi} = \ket{\tilde{\Psi}},
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture12:600}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(t)} = \ket{\Psi(0)},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture12:620}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(0)} = \ket{\Psi(-t)}.
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.