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I saw an awesome derivation of Schrodinger's equation on Wikipedia. Part of it relies on:

We also know that when $t' = t$, we must have the unitary time evolution operator $U(t, t) = 1$. Therefore, expanding the operator $U(t', t)$ for $t'$ close to $t$, we can write $U(t', t) = 1 - iH(t' - t)$, where $H$ is a Hermitian operator. This follows from the fact that the Lie algebra corresponding to the unitary group comprises Hermitian operators. Taking the limit as the time-difference $t' - t$ becomes very small, we obtain Schrodinger's equation.

What is meant by Lie algebra corresponding to the unitary group comprises Hermitian operators in the derivation in that link?

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The short answer is that the idea of group elements that are infinitesimally near the identity element are described by the Lie algebra of the group. If $\mathbb{1}$ is the identity of the unitary group, we can think of infinitesimally nearby elements as $\mathbb{1} + i\epsilon X$, where $\epsilon$ is a small (real) number. The set of all possible $X$ such that $\mathbb{1}+i\epsilon X$ is unitary is the Lie algebra of the unitary group. You can check that in order for $\mathbb{1} + i\epsilon X$ to be unitary, $X$ must be Hermitian. (Simply apply the condition for unitarity, $U^\dagger U = \mathbb{1}$ and ignore second order terms in $\epsilon$.) That's what is meant by "the Lie algebra of the unitary group is the set of Hermitian operators".

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  • $\begingroup$ @PhyEnthusiast Can you write out your work somewhere? $\endgroup$ – knzhou Mar 4 '18 at 10:34
  • $\begingroup$ $(1 + i\epsilon X)^\dagger (1 + i\epsilon X) = 1 \implies (1 - i\epsilon X^\dagger) (1 + i\epsilon X) = 1 \implies 1 - i\epsilon X^\dagger + i\epsilon X + \epsilon^2 X^\dagger X = 1$ forgetting second order terms, we get $$X - X^\dagger = 0.$$ Sorry, I got that wrong twice and got it right only now $\endgroup$ – PhyEnthusiast Mar 4 '18 at 10:40
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Some further information to add to qm-arv's correct answer, some that may help you now, some later on as you get more of a grasp on the subject matter.

A Lie group is essentially a group that is also a manifold, i.e. it can be labelled patchwise by co-ordinates and we also have the further requirement that the group operations (multiplication and inverse) induce continuous functions of these co-ordinates. That is, we can write expressions for the co-ordinates of a product as a function of the co-ordinates of the multiplicands and the expression we get defines a continuous function; likewise for the inverse.

Where this is relevant to the Schrödinger equation is that, for a time-invariant Hamiltonian, the Schrödinger equation describes the evolution of the quantum state of a system when the system is undisturbed; therefore the operator $U(\tau)$ describing evolution through some time interval $\tau$ must be the same whatever the beginning time for the evolution. So if we think about a time interval of length $\tau+s$, it is equivalent to the evolution for time interval $\tau$, followed by the evolution for time interval $s$ (or contrariwise) and we arrive at the conclusion that:

$$U(\tau+s) = U(\tau)\,U(s)\tag{1}$$

That is, the set $\{U(\tau):\,\tau\in\mathbb{R}\}$ of all evolution operators for this undisturbed system for all time intervals must form a group, called, naturally enough, a one parameter group. This is always an Abelian (commutative) group.

We're used to Nature's being continuous in her actions, so it's natural to postulate that $U(\tau)$ is a continuous function of the evolution duration $\tau$.

Now if our quantum system is a finite dimensional one, all the operators in (1) have commuting, finite dimensional matrices. We can manipulate square matrices exactly like scalars, so we can quickly deduce, as one can for real numbers, that the only continuous function that fulfils (1) is a function of the form

$$U(\tau) = \exp(K\,\tau) = \mathrm{id} + K\,\tau + K^2 \frac{\tau^2}{2!}+\cdots;\;\forall \tau\in\mathbb{R}\tag{2}$$

for some constant square matrix $K$. The series is universally convergent for finite dimensional square matrices. Also, for $\tau$ small enough, so that $U(\tau)$ must approach the identity matrix, the matrix logarithm is uniquely defined:

$$\tau\, K = U(\tau)-\mathrm{id} - \frac{(U(\tau)-\mathrm{id})^2}{2} + \frac{(U(\tau)-\mathrm{id})^3}{3}-\cdots;\; \left\|U(\tau)-\mathrm{id}\right\|<1\tag{3}$$

Lie groups always have at least one such one parameter group threading through them; in fact, every member near enough to the identity must be a member of one of these one parameter groups, so another characterization of the Lie algebra $\mathfrak{h}$ in a Lie group $\mathfrak{H}$ is the set of all logarithms of Lie group members $U\in\mathfrak{H}$ such that $\|U-\mathrm{id}\|<1$ together with all scalar multiples of these logarithms. This also leads to an equivalent characterization that the Lie algebra is:

$$\mathfrak{h} = \left\{K|\;\exp(\tau\,K)\in\mathfrak{H}\;\forall\,\tau\in\mathbb{R}\right\}\tag{4}$$

For quantum state vectors, their norms must be conserved by any evolution, so the operators concerned must all be unitary. This means that our logarithms must all be skew-Hermitian, or of the form $K=i\,H$ where $H$ is Hermitian as you have shown and qm-arv's correct answer has stated. In the unitary group, which is compact and connected, every Lie group member is the exponential of a Lie algebra member; this is true for the identity-connected component of all compact Lie groups but some noncompact groups have identity-connected subgroups with members which cannot be written as an exponential of a Lie algebra member. Of course, no element $\gamma$ that is outside the identity connected component of any Lie group can be exponential $\exp(H)$ of a Lie algebra element $H$, because such an element is connected to the identity by the path $\{\exp(t\,H)|\;t\in[0,\,1]\}$.

In the infinite dimensional case, all of this still holds, thanks to the remarkable Stone's Theorem on One Parameter Unitary Groups. This remarkable theorem shows that for any Hilbert space, as long as we talk about one parameter groups where the multiplication in (1) is strongly continuous, there is a one-to-one, onto correspondence between all such groups within the Hilbert space and the (possibly unbounded) self adjoint operators on the Hilbert space. That is, we always have $U(\tau) = \exp(i\,\tau\,H)$ for some, possibly unbounded, self adjoint operator $H$. Even weaker conditions than strong continuity can make this theorem work if the Hilbert space in question is separable, as it always is in quantum mechanics.

Where we are talking about one parameter groups of time evolution operators, the corresponding self adjoint operators are the Hamiltonians defined by isolated quantum state evolutions.

As another example of the use of this powerful theorem in quantum mechanics, we have the strongly continuous group of unitary translation operators defined on the Hilbert space $L^2$ functions by $\tilde{U}(\tau)\, \psi(x) = \psi(x+\tau)$. There is, by the theorem, a self adjoint operator $D$ such that $U(\tau) = \exp(i\,\tau\,D)$.

When we restrict $D$ to smooth functions, $D$ can be written as $D = -i\,\frac{\mathrm{d}}{\mathrm{d} x}$, and we retrieve the Taylor series:

$$\psi(x+\tau) = \psi(x) + i\,\tau\,D\,\psi(x) - \frac{\tau^2}{2!} D^2\,\psi(x) +\cdots = \psi(x) + \tau\,\frac{\mathrm{d}}{\mathrm{d} x} \psi(x) + \frac{\tau^2}{2!}\,\frac{\mathrm{d}^2}{\mathrm{d} x^2} \psi(x) + \cdots\tag{6}$$

elegantly showing the correspondence between the momentum operator $-i\,\frac{\mathrm{d}}{\mathrm{d} x}$ and translation in space.

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The Hermitian part is slightly arbitrary in the sense that the operators could all be multiplied by $i$ and become anti hermitian. Moreover, there is no reason to stick to hermitian operators as $L_+, L_-$ and $L_z$ are a legitimate basis for the Lie algebra $su(2)$ (or, more precisely, its complex extension).

The choice of hermitian operators is convenient because $$ U(\boldsymbol{\alpha})=\exp\left(-i \sum_k\alpha_k H_k\right) \tag{1} $$ is automatically unitary if $H_k$ is Hermitian and the parameters $\alpha_k$, the components of $\boldsymbol{\alpha}$, are real. To prove this simply note that $$ U^{\dagger}(\boldsymbol{\alpha})= \exp\left(i \sum_k \alpha_k H_k\right) $$ since $H^\dagger= H$. Since $\sum_k\alpha_k H_k$ is also Hermitian, call this $W$ and you have $$ U\cdot U^\dagger = e^{-i W}e^{iW}=e^{-i W +i W}=I $$ ($I$ is the $n\times n$ unit matrix, with $n$ the dimension of the matrix representation of $H_k$) as $W$ commutes with itself.

Coming back to the ossibility of writing a unitary matrix using exponentials of non-hermitian operators, the famous example is $SU(2)$, where an $SU(2)$ matrix $R$ can be written “in the antinormal form” $$ R=e^{\xi L_+} e^{\zeta L_z} e^{\eta L_-}\, . $$ Obviously in this example the group transformation is expressed as exponentials of non-hermitian elements which nevertheless span the Lie algebra.

A similar example exists for the Heisenberg-Weyl algebra. You can choose as a basis for the Lie algebra $\hat x, \hat p$ and $I$, or $\hat a, \hat a^\dagger$ and $I$, and express a translation in normal-ordered form or antinormal ordered form. In these last two cases the translation involves exponentiations of non-hermitian operators.

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  • $\begingroup$ This is a good point to point out that one often works with the complexification of $\mathfrak{su}(2)$ and of other unitary group Lie algebras but it might be a little confusing to the OP. I'm not really sure how to word this: perhaps one could emphasize that it is always possible to use a wholly Hermitian Lie algebra exponentiating to unitary evolution operators. Also, as a Lie algebra, it's worth pointing out that the complexified $\mathfrak{su}(2)$ is in fact $\mathfrak{sl}(2,\,\mathbb{C})$ over the reals - exponentiating to quite a different beast. $\endgroup$ – Selene Routley Apr 5 '18 at 9:37
  • $\begingroup$ @WetSavannaAnimalakaRodVance Right... I agree and will think about amending and maybe adding a bit but I fear in this case that more will be less, i.e. more information will not clarify much, especially if one gets into the long grass of complexification. $\endgroup$ – ZeroTheHero Apr 5 '18 at 12:47
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Remember that unitary matrices form a Lie group. If we consider the one-parameter subgroup $\{U(t)\}$, by Lie group theory it turns out that: $$ U(t)=\exp(tK)=\mathrm{id}+Kt + K^2 \frac{t^{2}}{2 !}+\cdots $$ with $K$ in the Lie algebra of unitary matrix, that is well known to be the skew-hermitian matrices.

But we can rewrite the expression above with $H=-iK$: $$ U(t)=\exp(tiH)=\mathrm{id}+iHt - H^2 \frac{t^{2}}{2 !}+\cdots $$

Observe that now $H$ is a hermitian matrix.

Anyway, we can show directly that $H$ is hermitian, using properties of matrix exponential, and without relying in approximations. We have that $$ U(t)^{\dagger}U(t)=I $$

But $$ U(t)^{\dagger}U(t)=\exp(tiH)^{\dagger}\exp(tiH)=\exp(-tiH^{\dagger})\exp(tiH)= $$ $$ =\exp(-tiH^{\dagger}+tiH)=\exp(ti(H-H^{\dagger})) $$ And so, must be $H=H^{\dagger}$

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