\textbookShankar, chapter 11

We first need to address the question “what is a symmetry in quantum mechanics?”. If we work in the standard Schrödinger picture↓ with time-dependent wave functions, and time-independent operators [E]  [E] This allows for time-dependent external forces. --there are alternatives-- we must be able to link the symmetry directly to the wave function, which is of course the most fundamental quantity in QM. At the same time we can also link it to the behaviour of general matrix elements $$\langle\psi|\hat{O}|\phi\rangle,$$ if we change the coordinates in the both wave functions under, e.g., one of the transformations discussed above. The question we now need to ask ourselves, is “how would we see a symmetry if we transform the argument of both wave functions in a matrix element under a transformation in a symmetry group?”. The answer is simple: since the matrix element describes physics, the matrix element itself should be invariant (unchanged) for any pair of wave functions if our system (particles and operator) have a symmetry. In most cases the operator that we shall use is the Hamiltonian.

What we have shown above is again the active view of such transformations in quantum mechanics, where we transform the wave functions. We shall analyse below the detailed behaviour of such transformations, but we shall assume that we can write [F]  [F] In mathematical physics we often use the term “a representation” of the transformation for the calculation of the operator on the wave function giving the action of a symmetry operator on the coordinates. We are looking at a linear representation here. $$\phi(O\vec{r},t)=\hat{T}_{O}\phi(\vec{r},t),$$ where $\hat{T}_{O}$ is a linear operator giving the effect of the change of coordinates, but by acting on the wave function, rather than acting on its arguments. This is a new intermediate stage--and it may not even be clear that such an operator actually exists; we shall construct these explicitly. We now derive the passive view from Eq. (↓): $$\begin{align*} \int d\vec{r}\,\psi(O\vec{r},t)^{*}\hat{H}\phi(O\vec{r},t) & =\int d\vec{r}\,\left[\hat{T}_{O}\psi(\vec{r},t)\right]^{*}\hat{H}\left[\hat{T}_{O}\phi(\vec{r},t)\right]\\ & =\int d\vec{r}\,\left(\psi(\vec{r},t)^{*}\hat{T}_{O}^{\dagger}\right)\hat{H}\left(\hat{T}_{O}\phi(\vec{r},t)\right)\\ & =\int d\vec{r}\,\psi(\vec{r},t)^{*}\left(\hat{T}_{O}^{\dagger}\hat{H}\hat{T}_{O}\right)\phi(\vec{r},t).\end{align*}$$ In the quantum passive view we transform operators (rather than the wave function), and we find that the operators transform according to $$\hat{H}\rightarrow\hat{T}_{O}^{\dagger}\hat{H}\hat{T}_{O}.$$ Using the passive view, if we have a symmetry, we must require that the transformed operator is invariant↓, i.e., is the same as the original one, $$\hat{H}=\hat{T}_{O}^{\dagger}\hat{H}\hat{T}_{O}.$$

In the case of a continuous symmetry, one with a set of parameters that vary continuously (which are all transformations but the discrete ones above), we can take the continuous parameter arbitrarily small, where the transformation has almost no effect, and analyse its behaviour when acting on the coordinates inside a wave function using a Taylor expansion. Note that since we transform coordinates, all our transformations carry the minus sign common to active transformations.

Consider a translation over a very small distance $\vec{\epsilon}$ in passive form, [G]  [G] We ignore the effect of a potential phase acquired by the wave function when we make such a transformation. See [1] and [3]. $$\begin{align*} \psi(T_{\vec{\epsilon}}\vec{r}) & =\psi(\vec{r}+\vec{\epsilon})\\ & \approx\psi(\vec{r})+\vec{\epsilon}\cdot\vec{\nabla}\psi(\vec{r})\\ & =\psi(\vec{r})+\frac{i}{\hbar}\vec{\epsilon}\cdot(\hat{\vec{p}}\psi(\vec{r})).\end{align*}$$ We use this expression and say that the operator $\hat{\vec{p}}$ is the generator of translations.

If we expand the expectation value (↓) above (suppressing time dependence) to first order in $\vec{\epsilon}$ , we find that $$\begin{align*} \int d\vec{r}\psi(T_{\vec{\epsilon}}\vec{r})^{*}\hat{H}(r)\phi(T_{\vec{\epsilon}}\vec{r})=\int d\vec{r}\psi(\vec{r})^{*}\hat{H}(r)\phi(\vec{r}),\end{align*}$$ or, expanding the left-hand side to first order in $\vec{\epsilon}$ and equating the coefficient of $\vec{\epsilon}$ to zero, $$\begin{align*} \int d\vec{r}\,-\frac{i}{\hbar}(\vec{p}\psi(\vec{r}))^{*}\hat{H}(r)\phi(\vec{r})+\int d\vec{r}\,\psi(\vec{r})^{*}\hat{H}(r)\left(\frac{i}{\hbar}\vec{p}\right)\phi(\vec{r}) & =0,\\ -\frac{i}{\hbar}\int d\vec{r}\,\psi(\vec{r})^{*}[\vec{p},\hat{H}(r)]\phi(\vec{r}) & =0.\end{align*}$$ Since $\psi$ and $\phi$ are arbitrary functions, we must have $$[\vec{p},\hat{H}(r)]=0,$$ and thus the generator of the symmetry↓ commutes with the Hamiltonian if it is compatible with the symmetry.

This one is easy! Look at $t\rightarrow t+\delta t$ ,

This one is a little harder; we use the following three rotation matrices $$\begin{align*} R_{x}(\phi_{1}) & =\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & \cos(\phi_{1}) & \sin(\phi_{1})\\ 0 & -\sin(\phi_{1}) & \cos(\phi_{1})\end{array}\right),\\ R_{y}(\phi_{2}) & =\left(\begin{array}{ccc} \cos(\phi_{2}) & 0 & -\sin(\phi_{2})\\ 0 & 1 & 0\\ \sin(\phi_{2}) & 0 & \cos(\phi_{2})\end{array}\right),\\ R_{z}(\phi_{3}) & =\left(\begin{array}{ccc} \cos(\phi_{3}) & \sin(\phi_{3}) & 0\\ -\sin(\phi_{3}) & \cos(\phi_{3}) & 0\\ 0 & 0 & 1\end{array}\right).\end{align*}$$ If we make the angles small, we find that $$\begin{align*} R_{x}(\delta\phi_{1}) & =I+\delta\phi_{1}\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{array}\right)=I+\delta\phi_{1}\mathrm{S}_{1},\\ R_{y}(\delta\phi_{2}) & =I+\delta\phi_{2}\left(\begin{array}{ccc} 0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{array}\right)=I+\delta\phi_{2}\mathrm{S}_{2},\\ R_{z}(\delta\phi_{3}) & =I+\delta\phi_{3}\left(\begin{array}{ccc} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{array}\right)=I+\delta\phi_{2}\mathrm{S}_{3}.\end{align*}$$ We now calculate the effect on $\psi$ of a general transformation  [H]  [H] The dot product of $\delta\vec{\phi}$ and $\vec{S}$ denotes the matrix $\delta\vec{\phi}\cdot\vec{\mathrm{S}}=\delta\phi_{1}\mathrm{S}_{1}+\delta\phi_{12}\mathrm{S}_{2}+\delta\phi_{3}\mathrm{S}_{3}$ . $$\psi\left(\vec{r}+(\delta\vec{\phi}\cdot\vec{\mathrm{S})}\vec{r}\right)\approx\psi(\vec{r})+\sum_{i}\left((\delta\vec{\phi}\cdot\vec{\mathrm{S}})\vec{r}\right)_{i}\frac{\partial}{\partial r_{i}}\psi(\vec{r}).$$ If we analyse each of the three $\delta\phi_{i}$ ’s separately, we find for instance for a rotation about $x$ that $$\begin{align*} \psi(R_{x}(\delta\phi_{1})\vec{r}) & \approx\psi(\vec{r})+\delta\phi_{1}\sum_{ij}\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{array}\right)_{ij}r_{j}\nabla_{i}\psi(\vec{r})\\ & =\psi(\vec{r})+\delta\phi_{1}\left[z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}\right]\psi(\vec{r})\\ & =\psi(\vec{r})+\frac{i}{\hbar}\delta\phi_{1}\left[z\hat{p}_{y}-y\hat{p}_{z}\right]\psi(\vec{r})\\ & =\psi(\vec{r})-\frac{i}{\hbar}\delta\phi_{1}\hat{L}_{x}\psi(\vec{r}).\end{align*}$$

We find that the three generators of rotational symmetry are $\hat{L}_{x},\hat{L}_{y},\hat{L}_{z}$ . Each commutes with the Hamiltonian, but they do not commute with each other,  [I]  [I] See also example sheet $$[\hat{L}_{i},\hat{L}_{j}]=i\hbar\epsilon_{ijk}L_{k}$$ ↓

Discrete symmetries don’t have generators, since there is no small parameter to expand in. If they are realised by an action on the arguments of the wave function (again, excluding an external phase that could occur in general) they are rather simple. We then define the linear symmetry operator by its action on the wave function.

For space inversion we define the “parity operator↓” $\hat{P}$ (sometimes denoted as $\hat{\Pi}$ ) that inverts the coordinates $$\begin{align} \hat{P}\psi(\vec{r},t) & =\psi(-\vec{r},t),\\ \hat{P^{2}}\psi(\vec{r},t) & =\hat{P}\psi(-\vec{r},t)=\psi(\vec{r},t)\label{eq:Sym:paritysq}\end{align}$$ The fact that $\hat{P^{2}}=I$ , as we see from (↓) shows that we have eigenvalues $1$ and $-1$ : $$\begin{align*} \hat{P}\psi(\vec{r},t) & =p\psi(\vec{r},t)=\psi(-\vec{r},t)\implies\\ \hat{P^{2}}\psi(\vec{r},t) & =p^{2}\psi(\vec{r},t)=\psi(\vec{r},t)\implies\\ p^{2} & =1.\end{align*}$$ Thus in a problem with parity invariance, the wave function is either even or odd under space inversion. We say it has positive or negative parity. We leave the fact that momentum changes sign as an exercise.

For time reversal we define the time reversal operator↓ $\hat{T}$ $$\begin{align} \hat{T}\psi(\vec{r},t) & =\psi(\vec{r},-t),\\ {\hat{T}}^{2}\psi(\vec{r},t) & =\psi(\vec{r},t)\label{eq:Sym:Tsq}\end{align}$$ The fact that $\hat{T}^{2}=I$ , as we see from (↓) shows again that we have eigenvalues $1$ and $-1$ .

You should be aware from your previous work in quantum mechanics that two operators that commute can be simultaneously diagonalised; i.e., a state can be an energy eigenstate and a symmetry eigenstate at the same time.

Even though we will not linger on Lorentz symmetry for too long before having discussed relativistic wave equations, it is easy to show that given the wave function $\psi(\rvec{r})$ and a small velocity boost along the $x$ axis, $$\begin{equation} \Lambda(-\delta\beta,0,0)=\left(\begin{array}{cccc} 1 & -\delta\beta & 0 & 0\\ -\delta\beta & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{array}\right),\label{eq:Sym:boost}\end{equation}$$ we find $$\begin{align*} \psi(\Lambda\rvec{r}) & =\psi(\rvec r)-\delta\beta\left(x_{0}\partial_{x_{1}}+x_{1}\partial_{x_{0}}\right)\psi(\rvec r)\\ & =\psi(\rvec r)-\frac{i}{\hbar}\delta\beta\left(x_{0}\hat{p}_{1}+x_{1}\hat{p}_{0}\right)\psi(\rvec r).\end{align*}$$ We shall leave this expression at this stage: we have carelessly defined the operator $\hat{p}_{0}=\partial_{x_{0}}$ ; since we know there is a link between time derivatives and the action of the Hamiltonian, this needs to be done much more carefully, as will be done later on.