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# Posts Tagged ‘unitary operator’

## PHY356F: Quantum Mechanics I. Dec 7 2010 ; Lecture 11 notes. Rotations and Angular momentum.

Posted by peeterjoot on December 7, 2010

## This time. Rotations (chapter 26).

Why are we doing the math? Because it applies to physical systems. Slides of IBM’s SEM quantum coral and others shown and discussed.

PICTURE: Standard right handed coordinate system with point $(x,y,z)$. We’d like to discuss how to represent this point in other coordinate systems, such as one with the $x,y$ axes rotated to $x',y'$ through an angle $\phi$.

Our problem is to find in the rotated coordinate system from $(x,y,z)$ to $(x', y', z')$.

There’s clearly a relationship between the representations. That relationship between $x', y', z'$ and $x,y,z$ for a counter-clockwise rotation about the $z$ axis is

\begin{aligned}x' &= x \cos \phi - y \sin\phi \\ y' &= x \sin \phi + y \cos\phi \\ z' &= z\end{aligned} \hspace{\stretch{1}}(13.214)

Treat $(x,y,z)$ and $(x',y',z')$ like vectors and write

\begin{aligned}\begin{bmatrix}x' \\ y' \\ z' \end{bmatrix}=\begin{bmatrix}\cos \phi &- \sin\phi & 0 \\ \sin \phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(13.217)

Or

\begin{aligned}\begin{bmatrix}x' \\ y' \\ z' \end{bmatrix}=R_z(\phi)\begin{bmatrix}x \\ y \\ z \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(13.218)

\paragraph{Q: Is $R_z(\phi)$ a unitary operator?}

Definition $U$ is unitary if $U^\dagger U = \mathbf{1}$, where $\mathbf{1}$ is the identity operator. We take Hermitian conjugates, which in this case is just the transpose since all elements of the matrix are real, and multiply

\begin{aligned}(R_z(\phi))^\dagger R_z(\phi) &=\begin{bmatrix}\cos \phi & \sin\phi & 0 \\ -\sin \phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\cos \phi &- \sin\phi & 0 \\ \sin \phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix} \\ &=\begin{bmatrix}\cos^2 \phi + \sin^2\phi & -\sin\phi \cos\phi + \sin\phi \cos\phi & 0 \\ -\cos\phi \sin\phi + \cos\phi \sin\phi & \cos^2\phi + \sin^2 \phi & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ &=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ &= \mathbf{1}\end{aligned}

Apply the above to a vector $\mathbf{v} = (v_x, v_y, v_z)$ and write $\mathbf{v}' = (v_x', v_y', v_z')$. These are related as

\begin{aligned}\mathbf{v} = R_z(\phi) \mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.219)

Now we want to consider the infinitesimal case where we allow the rotation angle to get arbitrarily small. Consider this specific $z$ axis rotation case, and assume that $\phi$ is very small. Let $\phi = \epsilon$ and write

\begin{aligned}\mathbf{v}' &=\begin{bmatrix}v_x' \\ v_y' \\ v_z' \end{bmatrix}=R_z(\phi)\begin{bmatrix}v_x \\ v_y \\ v_z \end{bmatrix}=\begin{bmatrix}\cos \epsilon &- \sin\epsilon & 0 \\ \sin \epsilon & \cos\epsilon & 0 \\ 0 & 0 & 1\end{bmatrix} \mathbf{v} \\ &\approx\begin{bmatrix}1 &- \epsilon & 0 \\ \epsilon & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \mathbf{v} =\left(\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} +\begin{bmatrix}0 &- \epsilon & 0 \\ \epsilon & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \right)\mathbf{v} \end{aligned} \hspace{\stretch{1}}(13.220)

Define

\begin{aligned}S_z = i \hbar\begin{bmatrix}0 &- 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(13.222)

which is the generator of infinitesimal rotations about the $z$ axis.

Our rotated coordinate vector becomes

\begin{aligned}\mathbf{v}' &= \left(\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} +\frac{i \hbar \epsilon}{i\hbar}\begin{bmatrix}0 &- 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \right)\mathbf{v} \\ &=\left(\mathbf{1} + \frac{\epsilon}{i \hbar} S_z\right)\mathbf{v}\end{aligned}

Or

\begin{aligned}\mathbf{v}'=\left(\mathbf{1} - \frac{i \epsilon}{\hbar} S_z\right)\mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.223)

Many infinitesimal rotations can be combined to create a finite rotation via

\begin{aligned}\lim_{N \rightarrow \infty} \left( 1 + \frac{\alpha}{N} \right)^N = e^\alpha\end{aligned} \hspace{\stretch{1}}(13.224)

\begin{aligned}\alpha = -i \phi S_z/\hbar\end{aligned} \hspace{\stretch{1}}(13.225)

For a finite rotation

\begin{aligned}\mathbf{v}'=e^{ -i \frac{\phi S_z}{\hbar} }\mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.226)

Now think about transforming $g(x,y,z)$, an arbitrary function. Take $\epsilon$ is very small so that

\begin{aligned}x' &= x \cos \phi - y \sin\phi = x \cos \epsilon - y \sin\epsilon \approx x - y \epsilon \\ y' &= x \sin \phi + y \cos\phi = x \sin \epsilon + y \cos\epsilon \approx x \epsilon + y \\ z' &= z\end{aligned} \hspace{\stretch{1}}(13.227)

\paragraph{Question: Why can we assume that $\epsilon$ is small.}
\paragraph{Answer: We declare it to be small because it is simpler, and eventually build up to the general case where it is larger. We want to master the easy task before moving on to the more difficult ones.}

Our function is now transformed

\begin{aligned}g(x', y', z') \approx g( x - y \epsilon, y + x \epsilon, z) \\ &= g( x , y , z) - \epsilon y \frac{\partial {g}}{\partial {x}} + \epsilon x \frac{\partial {g}}{\partial {y}} + \cdots \\ &=\left( \mathbf{1} - \epsilon y \frac{\partial {}}{\partial {x}}+ \epsilon x \frac{\partial {}}{\partial {y}}\right)g( x, y ,z )\end{aligned}

Recall that the coordinate definition of the angular momentum operator is

\begin{aligned}L_z = -i \hbar \left( x \frac{\partial {}}{\partial {y}} - y \frac{\partial {}}{\partial {x}} \right) = x p_y - y p_x\end{aligned} \hspace{\stretch{1}}(13.230)

We can now write

\begin{aligned}g(x', y', z') &=\left( \mathbf{1} +\frac{-i \hbar \epsilon}{-i\hbar} \left(x \frac{\partial {}}{\partial {y}}- y \frac{\partial {}}{\partial {x}}\right)\right)g( x, y ,z ) \\ &=\left( \mathbf{1} +\frac{i \epsilon}{\hbar} L_z\right)g( x, y ,z )\end{aligned}

For a finite rotation with angle $\phi$ we have

\begin{aligned}g(x', y', z') =e^{i \frac{\phi L_z}{\hbar}}g( x, y ,z )\end{aligned} \hspace{\stretch{1}}(13.231)

\paragraph{Question: somebody says that the rotation is clockwise not counterclockwise.}

I didn’t follow the reasoning briefly mentioned on the board since it looks right to me. Perhaps this is the age old mixup between rotating the coordinates and the basis vectors. Review what’s in the text carefully. Can also check by

If you rotate a ket, and examine how the state representation of that ket changes under rotation, we have

\begin{aligned}{\lvert {x', y', z'} \rangle} = {\lvert {x - \epsilon y, y + \epsilon x, z} \rangle}\end{aligned} \hspace{\stretch{1}}(13.232)

Or

\begin{aligned}\left\langle{{\Psi}} \vert {{x', y', z'}}\right\rangle &= \Psi^{*}(x', y', z') \\ &=\Psi^{*}(x - \epsilon y, y + \epsilon x, z) \\ &=\Psi^{*}(x , y , z) - \epsilon \frac{\partial {\Psi^{*}}}{\partial {y}}+ \epsilon \frac{\partial {\Psi^{*}}}{\partial {x}} \\ &=\left( \mathbf{1} + \frac{i \epsilon} {\hbar} L_z \right) \Psi^{*}(x , y , z) \end{aligned}

Taking the complex conjugate we have

\begin{aligned}\Psi(x', y', z') \left( \mathbf{1} - \frac{i \epsilon} {\hbar} L_z \right) \Psi(x , y , z) \end{aligned} \hspace{\stretch{1}}(13.233)

For infinitesimal rotations about the $z$ axis we have for functions

\begin{aligned}\Psi(x', y', z') =e^{ - \frac{i \epsilon} {\hbar} L_z } \Psi(x , y , z) \end{aligned} \hspace{\stretch{1}}(13.234)

For finite rotations of a vector about the $z$ axis we have

\begin{aligned}\mathbf{v}'=e^{ - \frac{i \phi S_z} {\hbar} } \Psi(x , y , z) \mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.235)

and for functions

\begin{aligned}\Psi(x', y', z') =e^{ - \frac{i \phi L_z} {\hbar} } \Psi(x , y , z) \end{aligned} \hspace{\stretch{1}}(13.236)

Vatche has mentioned some devices being researched right now where there is an attempt to isolate the spin orientation so that, say, only spin up or spin down electrons are allowed to flow. There are some possible interesting applications here to Quantum computation. Can we actually make a quantum computing device that is actually usable? We can make NAND devices as mentioned in the article above. Can this be scaled? We don’t know how to do this yet.

Recall that one description of a “particle” that has both a position and spin representation is

\begin{aligned}{\lvert {\Psi} \rangle} = {\lvert {u} \rangle} \otimes {\lvert {s m} \rangle}\end{aligned} \hspace{\stretch{1}}(13.237)

where we have a tensor product of kets. One usually just writes the simpler

\begin{aligned}{\lvert {u} \rangle} \otimes {\lvert {s m} \rangle} \equiv {\lvert {u} \rangle} {\lvert {s m} \rangle} \end{aligned} \hspace{\stretch{1}}(13.238)

An example of the above is

\begin{aligned}\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \\ \end{bmatrix}= \Bigl( {\langle {\mathbf{r}} \rvert} {\langle { s m} \rvert} \Bigr) {\lvert {\Psi} \rangle}\end{aligned} \hspace{\stretch{1}}(13.239)

where $u_1$ is spin component one. For $s=1$ this would be $m=-1, 0, 1$.

Here we have also used

\begin{aligned}{\lvert {\mathbf{r}} \rangle}= {\lvert {x} \rangle}\otimes{\lvert {y} \rangle}\otimes{\lvert {z} \rangle} \\ &={\lvert {x} \rangle}{\lvert {y} \rangle}{\lvert {z} \rangle} \\ &={\lvert {x y z} \rangle}\end{aligned}

We can now ask the question of how this thing transforms. We transform each component of this as a vector. The transformation of

\begin{aligned}\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \end{bmatrix}\end{aligned}

results in

\begin{aligned}{\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \end{bmatrix}}'=e^{ -i \phi (S_z + L_z)/\hbar }\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(13.240)

Or with $J_z = S_z + L_z$

\begin{aligned}{\lvert {\Psi'} \rangle} = e^{-i \phi J_z/\hbar } {\lvert {\Psi} \rangle} \end{aligned} \hspace{\stretch{1}}(13.241)

Observe that this separates out nicely with the $S_z$ operation acting on the vector parts, and the $L_z$ operator acting on the functional dependence.

## Rotations using matrix exponentials

Posted by peeterjoot on July 27, 2010

# Motivation.

In [1] it is noted in problem 1.3 that any Unitary operator can be expressed in exponential form

\begin{aligned}U = e^{iC},\end{aligned} \hspace{\stretch{1}}(1.1)

where $C$ is Hermitian. This is a powerful result hiding away in this problem. I haven’t actually managed to prove this yet to my satisfaction, but working through some examples is highly worthwhile. In particular it is interesting to compute the matrix $C$ for a rotation matrix. One finds that the matrix for such a rotation operator is in fact one of the Pauli spin matrices, and I found it interesting that this falls out so naturally. Additionally, it is rather slick that one is able to so concisely express the rotation in exponential form, something that is natural and powerful in complex variable algebra, and also possible using Geometric Algebra using exponentials of bivectors. Here we can do it after all with nothing more than the plain old matrix algebra that everybody is already comfortable with.

# The logarithm of the Unitary matrix.

By inspection we can invert 1.1 for $C$, by taking the logarithm

\begin{aligned}C = -i \ln U.\end{aligned} \hspace{\stretch{1}}(2.2)

The problem becomes one of evaluating the logarithm, or even giving meaning to it. I’ll assume that the functions of matrices that we are interested in are all polynomial in powers of the matrix, as in

\begin{aligned}f(U) = \sum_k \alpha_k U^k,\end{aligned} \hspace{\stretch{1}}(2.3)

and that such series are convergent. Then using a spectral decomposition, possible since Unitary matrices are normal, we can write for diagonal $\Sigma = {\begin{bmatrix} \lambda_i \end{bmatrix}}_i$

\begin{aligned}U = V \Sigma V^\dagger,\end{aligned} \hspace{\stretch{1}}(2.4)

and

\begin{aligned}f(U) = V \left( \sum_k \alpha_k \Sigma^k \right) V^\dagger = V {\begin{bmatrix} f(\lambda_i) \end{bmatrix}}_i V^\dagger.\end{aligned} \hspace{\stretch{1}}(2.5)

Provided the logarithm has a convergent power series representation for $U$, we then have for our Hermitian matrix $C$

\begin{aligned}C = -i V (\ln \Sigma) V^\dagger\end{aligned} \hspace{\stretch{1}}(2.6)

## Evaluate this logarithm for an $x,y$ plane rotation.

Given the rotation matrix

\begin{aligned}U =\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.7)

We find that the eigenvalues are $e^{\pm i\theta}$, with eigenvectors proportional to $(1, \pm i)$ respectively. Our decomposition for $U$ is then given by
2.4, and

\begin{aligned}V &= \frac{1}{{\sqrt{2}}}\begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix} \\ \Sigma &=\begin{bmatrix}e^{i\theta} & 0 \\ 0 & e^{-i\theta}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.8)

Taking logs we have

\begin{aligned}C&=\frac{1}{2}\begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix}\begin{bmatrix}\theta & 0 \\ 0 & -\theta\end{bmatrix} \begin{bmatrix}1 & -i \\ 1 & i\end{bmatrix} \\ &=\frac{1}{2}\begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix}\begin{bmatrix}\theta & -i\theta \\ -\theta & -i\theta\end{bmatrix} \\ &=\begin{bmatrix}0 & -i\theta \\ i\theta & 0\end{bmatrix}.\end{aligned}

With the Pauli matrix

\begin{aligned}\sigma_2 =\begin{bmatrix}0 & -i \\ i & 0\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.10)

we then have for an $x,y$ plane rotation matrix just:

\begin{aligned}C = \theta \sigma_2\end{aligned} \hspace{\stretch{1}}(2.11)

and

\begin{aligned}U = e^{i \theta \sigma_2}.\end{aligned} \hspace{\stretch{1}}(2.12)

Immediately, since $\sigma_2^2 = I$, this also provides us with a trigonometric expansion

\begin{aligned}U = I \cos\theta + i \sigma_2 \sin\theta.\end{aligned} \hspace{\stretch{1}}(2.13)

By inspection one can see that this takes us full circle back to the original matrix form 2.7 of the rotation. The exponential form of
2.12 has a beauty that is however far superior to the plain old trigonometric matrix that we are comfortable with. All without any geometric algebra or bivector exponentials.

## Three dimensional exponential rotation matrices.

By inspection, we can augment our matrix $C$ for a three dimensional rotation in the $x,y$ plane, or a $y,z$ rotation, or a $x,z$ rotation. Those are, respectively

\begin{aligned}U_{x,y}&=\exp\begin{bmatrix}0 & \theta & 0 \\ -\theta & 0 & 0 \\ 0 & 0 & i\end{bmatrix} \\ U_{y,z}&=\exp\begin{bmatrix}i & 0 & 0 \\ 0 & 0 & \theta \\ 0 & -\theta & 0 \\ \end{bmatrix} \\ U_{x,z}&=\exp\begin{bmatrix}0 & 0 & \theta \\ 0 & i & 0 \\ -\theta & 0 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.14)

Each of these matrices can be related to each other by similarity transformation using the permutation matrices

\begin{aligned}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix},\end{aligned}

and

\begin{aligned}\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{bmatrix}.\end{aligned}

## Exponential matrix form for a Lorentz boost.

The next obvious thing to try with this matrix representation is a Lorentz boost.

\begin{aligned}L =\begin{bmatrix}\cosh\alpha & -\sinh\alpha \\ -\sinh\alpha & \cosh\alpha\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.17)

where $\cosh\alpha = \gamma$, and $\tanh\alpha = \beta$.

This matrix has a spectral decomposition given by

\begin{aligned}V &= \frac{1}{{\sqrt{2}}}\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix} \\ \Sigma &=\begin{bmatrix}e^\alpha & 0 \\ 0 & e^{-\alpha}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.18)

Taking logs and computing $C$ we have

\begin{aligned}C&=-\frac{i}{2}\begin{bmatrix}1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix}\alpha & 0 \\ 0 & -\alpha\end{bmatrix} \begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix} \\ &=-\frac{i}{2}\begin{bmatrix}1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix}\alpha & -\alpha \\ -\alpha & -\alpha\end{bmatrix} \\ &=i \alpha\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}.\end{aligned}

Again we have one of the Pauli spin matrices. This time it is

\begin{aligned}\sigma_1 =\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.20)

So we can write our Lorentz boost 2.17 as just

\begin{aligned}L = e^{-\alpha \sigma_1} = I \cosh\alpha - \sigma_1 \sinh\alpha.\end{aligned} \hspace{\stretch{1}}(2.21)

By inspection again, we can come full circle by inspection from this last hyperbolic representation back to the original explicit matrix representation. Quite nifty!

It occurred to me after the fact that the Lorentz boost is not Unitary. The fact that the eigenvalues are not a purely complex phase term, like those of the rotation is actually a good hint that looking at how to characterize the eigenvalues of a unitary matrix can be used to show that the matrix $C = -i V \ln \Sigma V^\dagger$ is Hermitian.

# References

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