1
$\begingroup$

A consequence of Wigner's theorem is that if a Hamiltonian matrix obeys time reversal symmetry then it is real-symmetric. It seems to me that for this to make sense then "real symmetric" should be a basis independent notion, i.e. if $\{ | i \rangle \}_{i=1}^n$ is an orthonormal basis and $H$ a time-symmetric Hamiltonian then $\langle{i} |H | j \rangle \in \mathbb{R}$ for all i and j. But if we have another basis related by a unitary matrix $U$ I don't immediately see how it follows that the matrix is also real in the new basis, $\langle{i}| U^\dagger H U | j \rangle \in \mathbb{R}$.

Is it the case that "real symmetric" is a basis independent property and if so how would one prove it?

EDIT: the answers below point out that real symmetric is not a basis independent property so let me modify the question a bit: given that real symmetric is not basis independent property, how can this be reconciled with Wigner's theorem which says that T-symmetric Hamiltonians are real-symmetric? After all, isn't QM supposed to be done on abstract Hilbert spaces without reference to a basis?

$\endgroup$

3 Answers 3

1
$\begingroup$

$\let\d=\delta \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle} \def\mxelm#1#2#3{\bra#1#2\ket#3} \def\hx{\hat x} \def\hp{\hat p} \def\PD#1#2{{\partial #1 \over \partial #2}}$

how can this be reconciled with Wigner's theorem which says that T-symmetric Hamiltonians are real-symmetric?

The answer is simple: Wigner's theorem doesn't say that. It says $$T\,H\,T^{-1} = H \qquad T\,H = H\,T \tag1$$ with $T$ antiunitary.

Consider the simplest possible system: a one-dimensional particle. The basic observables are position $x$ and momentum $p$. All other observables are built as functions thereof. To define $T$ it's necessary and sufficient to say as it acts on $x$ and on $p$: $$T\,\hx\,T^{-1} = \hx \qquad T\,\hp\,T^{-1} = -\hp.$$ Note that because of antiunitarity commutation relation is conserved. Starting from $$[\hx,\hp] = i \hbar\,I$$ and applying $T \ldots T^{-1}$ we obtain on the left $$T\,[\hx,\hp]\,T^{-1} = [\hx,-\hp] = -i \hbar\,I$$ and on the right $$T\,(i \hbar\,I)\,T^{-1} = -i \hbar\,I.$$

To speak of a Hamiltonian matrix you need first of all a representation. Let's take the Hilbert space $L^2(\Bbb R)$ and a basis made of real orthonormal functions $u_n(x)$. Then $$\hx \mapsto x \qquad \hp \mapsto -i \hbar\,\PD{}x.$$ It's easy to verify that in this representation $$T \mapsto K \quad\rm (complex\ conjugation).$$

Let's check eq. (1) on a base function $u_m(x)$: $$\eqalign{K\,H\,u_m(x) &= \bigl(H\,u_m(x)\bigr)^* = \Bigl(\sum_n H_{mn} u_n(x)\Bigr)^* \cr &= \sum_n H_{mn\vphantom0}^* u_n(x)} \tag2$$ (remember $u_n$ is real). $$H\,K\,u_m(x) = H\,u_m(x) = \sum_n H_{mn} u_n(x).\tag3$$ (2) and (3) coincide for all $m$ if and only if all $H_{mn}$'s are real.

$\endgroup$
1
$\begingroup$

"Real symmetric" is not a basis-independent property in general.

Consider an arbitrary Hermitian matrix with complex elements.

Its eigenvalues are all real, therefore there is a basis (the eigenvectors) which transforms in into a real symmetric (in fact, diagonal) matrix. But you can easily construct a small (2x2) example where a real, non-diagonal, symmetric matrix is transformed into a Hermitian matrix.

On the other hand, if you start with an arbitrary real symmetric matrix, why would you want to choose a basis to make it Hermitian but not real symmetric? If you choose a real orthogonal matrix $U$ instead of a unitary matrix, the transformed matrix will still be real symmetric.

One fundamental difference between a Hermitian and a real symmetric matrix is that the eigenvectors of a real symmetric matrix are also real, but the eigenvectors of a Hermitian matrix are not. Given that fact, it seems strange to want to transform a real symmetric matrix into a Hermitian one, even though you can.

$\endgroup$
1
  • 1
    $\begingroup$ Do you not see my issue? QM is supposed to be done on abstract Hilbert spaces without reference to a basis. So do you not see how requiring a Hamiltonian to have real matrix elements is problematic if it's a basis dependent notion? $\endgroup$ Feb 27, 2019 at 23:02
-1
$\begingroup$

Matrices are not basis independent since they are the linear transformations on the real vector space, $R^n$. However, unlike all other vector spaces that we come across in nature, this vector space is special because it comes equipped with a natural basis, aka the standard basis vectors $e_i = (0,0, .., 1, 0, ...,0)$, where The 1 is in the ith position.

The correct notion of a basis independent transformation is any linear transformation between vector spaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.