# How the quantum Hamiltonian changes under a transformation?

Let's say that I have an Hamiltonian $$H(k)$$ in momentum space and I consider a transformation (to be concrete let's say time reversal $$\mathcal{T}$$). We say that this is a symmetry, if $$\mathcal{T} H(k) \mathcal{T^{-1}}=H(-k).$$

My question is (in general): Why does the Hamiltonian transforms under a generic transformation $$\mathcal{T}$$ as $$\mathcal{T} H(k) \mathcal{T^{-1}}$$? Where does this transformation law come from? What confuses me about this is that I thought that under a change of basis $$\psi \rightarrow U\psi$$ the hamiltonian (or any other operator) would change according to $$U^{\dagger}HU$$. So what is going on?

• Is the question about why one would use a generic transformation $\mathcal{T}$ instead of a unitary $U$, or about why the rule with $\mathcal{T}$ has the form transformation-opertor-[inverse transformation] versus the rule with $U$ having the form [inverse transformation]-opertor-transformation? Feb 28, 2022 at 18:20
• @QuantumMechanic the second one, so about why the rule with T has the form transformation-opertor-[inverse transformation] versus the rule with U having the form [inverse transformation]-opertor-transformation Feb 28, 2022 at 20:39

One can always define a second unitary operator $$V=U^\dagger$$ to write $$U^\dagger H U= V H V^\dagger.$$ This actually holds true for all of group theory, where by definition the inverse of a group element must also be a member of the group. As such, we can equally well inspect the set of transformations $$\mathfrak{g}^{-1}\mathfrak{h}\mathfrak{g}$$ or $$\mathfrak{g}\mathfrak{h}\mathfrak{g}^{-1}$$ for a set of group elements $$\mathfrak{g}$$ because they will lead to the same overall set of transformations.

The important thing mentioned by OP is to match up the transformations on operators with transformations on things like quantum states (like rotating vectors versus rotating coordinate systems). Indeed when a state undergoes the transformation $$|\psi\rangle\to U|\psi\rangle$$, this is equivalent to observables (operators) undergoing the transformation $$O\to U^\dagger O U$$. However, this is also equivalent to the density operator transformaing according to $$\rho\to U\rho U^\dagger$$, to ensure consistency for pure states $$\rho=|\psi\rangle\langle \psi|$$. So one has to keep track of what is being transformed in order to choose the correct group element $$U$$ or $$V$$.

Time reversal has the nice property that if you apply it twice you get back what you started with. I might assume that $$\mathcal{T}^2 H\mathcal{T}^{-2}=H$$ implies that $$\mathcal{T}^2$$ must be a phase times the identity matrix. It turns out that $$\mathcal{T}^2=\pm 1$$, so $$\mathcal{T}=\pm\mathcal{T}^{-1} \quad\Rightarrow\quad \mathcal{T} H\mathcal{T}^{-1}=\mathcal{T}^{-1} H\mathcal{T}.$$ This is similar to the parity operator in that it doesn't matter whether you apply the operator or the inverse of the operator, you get the same final result. [Intuitively, the inverse of time reversal is just time reversal (up to a phase, that part isn't intuitive), so it doesn't matter if you reverse the time or undo the time reversal.]

• @QuantumMechanics I'm not entirely convinced, and probably I don't completely understand your arguments. It now makes sense to me that if we transform $\psi$ with U (not necessarily unitary) as $U \psi$ then we would like $O\psi$ to transform in the same way (O generic operator), so $O^{'}=UOU^{-1}$. But if I look at the matrix element of O I have $(<\psi|U^{\dagger}OU|\psi>)$ so the two things seem contradictory. Feb 28, 2022 at 22:47
• @Mathew if you transform your state as $U|\psi\rangle$, it is as if you did nothing to your state and transformed your operators as $U^\dagger O U$; that ensures that the transformed expectation values are consistent. This is the equivalence between predictions of the Schrodinger and Heisenberg pictures. The idea of having states $|\psi\rangle$ and $O|\psi\rangle$ both transform by applying the same unitary is physically different and you'll have to convince me in a particular situation that it is really what is physically happening Mar 1, 2022 at 13:54
• @Mathew the statement $U|\psi\rangle$ and $U^\dagger OU$ is "transform what I started with according to $U$." This is what happens with unitary evolution in textbook quantum, where $U=\exp(-iHt/\hbar)$ for Hamiltonian $H$. The other statement is the statement $U|\psi\rangle$ and $U OU^\dagger$ is "transform whatever I have now (I don't know what I have now) according to $U$." This is the case when your current state might be $|\psi\rangle$ and might be some previously transformed state $O|\psi\rangle$. Mar 1, 2022 at 14:58
• Maybe final comment: the update rule when you just want to transform your current state is to do both $U|\psi\rangle$ and $UOU^\dagger$, just to be sure that you correctly transform any state $O|\psi\rangle$. The update rule for making predictions after unitary evolution is to either do $U|\psi\rangle$ or $U^\dagger O U$. Mar 1, 2022 at 16:33
• @QuantumMechanics So, if I understood correctly, if you update both $\psi$ as $U \psi$ and all operators O as $UOU^{-1}$, that means you are changing your basis but you are still looking at the $\textbf{same}$ state and operator (in the new basis). Whereas if you either do $U \psi$ or $U^{+}OU$ you are "evolving" to $\textbf{different}$ states (Shrodinger picture),or operators (Heisenber picture). Mar 1, 2022 at 17:22

Under the action of a unitary operator $$U$$, all operators $$O$$ transform as $$O \to U O U^{-1}$$ and states transform as $$|\psi \rangle \to U | \psi \rangle$$. The transformation law for $$O$$ takes that form so that the state $$O | \psi \rangle$$ transforms in exactly the same way as $$| \psi \rangle$$.

• but if $\psi \rightarrow U \psi$ then the matrix representation of the operator O becomes $<\psi |U^{\dagger}O U |{\psi} >$, so shouldn't the operator change as $O \rightarrow U^{\dagger}OU$ ? Feb 28, 2022 at 14:57
• It may be worth pointing out that for a unitary transformation $U^{-1} = U^\dagger$ so the two forms are equivalent, but the time reversal operator is not unitary, so we must write $\mathcal{T}^{-1}$ Feb 28, 2022 at 14:57
• @BySymmetry but to be equivalent shouldn't it be $U^{-1}OU$ rather then $UOU^{-1}$? Feb 28, 2022 at 15:01
• Seconding @Mathew, why do you want your state $O|\psi\rangle$ to always transform in the same way as $|\psi\rangle$? These are not the same physical process, because they each lead to different expectation values. Mar 1, 2022 at 14:54
• Continuing the previous comment, you might use $UOU^\dagger$ for all previously applied operators $O$ to ensure that your current state $O|\psi\rangle$ gets updated by left-multiplication with $U$, but you'd use $U^\dagger OU$ if you have some initial state $|\psi\rangle$ that then gets updated as $U|\psi\rangle$ and you want to instead put the update rule onto the operators Mar 1, 2022 at 15:03

I would like to add the following description based on Wigner's theorem, as described on page 62 in the book "quantum mechanics a modern development":

Theorem (Wigner). Any mapping of the vector space onto itself that preserves the value of $$|\langle\phi \mid \psi\rangle|$$ may be implemented by an operator $$U$$:

\begin{aligned} |\psi\rangle \rightarrow\left|\psi^{\prime}\right\rangle &=U|\psi\rangle \\ |\phi\rangle \rightarrow\left|\phi^{\prime}\right\rangle=U|\phi\rangle \end{aligned} (3.1)

with $$U$$ being either unitary (linear) or antiunitary (antilinear).

The transformation of state vectors, of the form (3.1), is accompanied by a transformation $$A \rightarrow A^ \prime$$ of the operators for observables. It must be such that the transformed observables bear the same relationship to the transformed states as did the original observables to the original states. In particular, if $$A\left|\phi_{n}\right\rangle=a_{n}\left|\phi_{n}\right\rangle$$, then $$A^{\prime}\left|\phi_{n}^{\prime}\right\rangle=a_{n}\left|\phi_{n}^{\prime}\right\rangle$$. Substitution of $$\left|\phi_{n}^{\prime}\right\rangle=U\left|\phi_{n}\right\rangle$$, using (3.1), yields $$A^{\prime} U\left|\phi_{n}\right\rangle=a_{n} U\left|\phi_{n}\right\rangle$$, and hence $$U^{-1} A^{\prime} U\left|\phi_{n}\right\rangle=a_{n}\left|\phi_{n}\right\rangle$$. Subtracting this from the original eigenvalue equation yields $$\left(A-U^{-1} A^{\prime} U\right)\left|\phi_{n}\right\rangle=0$$. Since this equation holds for each member of the complete set $$\left\{\left|\phi_{n}\right\rangle\right\}$$, it holds for an arbitrary vector, and therefore $$\left(A-U^{-1} A^{\prime} U\right)=0$$. Thus the desired transformation of operators that accompanies (3.1) is

$$A \rightarrow A^{\prime}=U A U^{-1}$$. (3.2)

Unitary transformations are motivated by the fact that a pure phase of a quantum state is not observable since $$|\psi|^2 = |\psi e^{i\varphi}|^2$$. This means that the 'physics" doesn't change if we apply a unitary transformation to the system. Here physics refers to three related things

1. the probability density should be unaffected $$\langle \psi |\psi\rangle = \langle \tilde\psi |\tilde\psi\rangle$$
2. the eigenvalues of some operator $$O$$ should be unaffected: if $$O|n\rangle = O_n |n\rangle$$ then $$\tilde{O}|\tilde{n}\rangle = O_n |\tilde{n}\rangle$$
3. any expectation value should be invariant $$\langle \psi | O |\psi\rangle = \langle \tilde\psi| \tilde{O} |\tilde\psi\rangle$$

Anything else isn't directly observable in QM. We start by transforming a state $$|\psi\rangle$$ with the prescription $$|\psi\rangle \to |\tilde\psi\rangle = U|\psi\rangle .$$ Since $$U$$ is unitary $$U U^\dagger = 1$$ which immediately yields $$\langle \psi|\psi\rangle \to \langle \psi | U^\dagger U|\psi\rangle = \langle \psi |\psi\rangle$$ Next we demand that the expectation value remains invariant $$\langle \psi | O |\psi\rangle \to \langle \psi | U^\dagger \tilde{O} U |\psi\rangle = \langle \psi | O |\psi\rangle$$ For this to be true in general when the operators don't commute we require that $$O$$ transforms as $$O \to \tilde{O} = U O U^\dagger$$. With this transformation law for operators it is also obvious that the spectrum is unchanged because $$\tilde{O}|\tilde{n}\rangle = U O U^\dagger U |n\rangle = U O |n\rangle = U O_n |n\rangle = O_n U |n\rangle = O_n |\tilde{n}\rangle$$ As mentioned in the other posts the unitary operators form a group, if $$U$$ is unitary so is $$U^{-1} = U^\dagger$$ so it really is somewhat arbitrary how we define our transformation. We are free to start with $$|\psi\rangle \to U^\dagger |\psi\rangle$$. Then we'd also have to change the operator transformation to $$\tilde{O} = U^\dagger O U$$ to remain consistent. There is more to be said about symmetry groups and how they act on a quantum system. One thing I want to point out however is that one can show that you can always find a representation of a symmetry group such that the action on states is given by unitary operators $$U = e^{iA}$$ where $$A=A^\dagger$$ is a hermitian matrix.

If $$U$$ is a symmetry of the Hamiltonian $$H = U H U^\dagger$$ (note that timedependent transformations require a special treatment if we are considering the Hamiltonian) we can rewrite this as $$HU = UH$$ or $$[H, U] = 0$$. This can be used to generate further eigenvalues of the Hamiltonian and classifying them according to their quantum numbers. Again it doesn't matter if we were to write $$H = U^\dagger H U$$ instead, we still arive at $$[H, U] = 0$$ which is really the key point.