# Real symmetric matrix in Wigner's theorem

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?

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

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.

"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.

• 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? Feb 27, 2019 at 23:02

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.