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D. J. Griffiths in his Introduction to quantum mechanics says that separable solutions are states of definite total energy. As an argument to this assertion he offers mathematical proof that variance of Hamiltonian is 0 so every member of a sample must share the same value, hence every measurment of a total energy is certain to return the same value of E.

While reading chapter on Linear algebra, I came across paragraph about the trace of matrix. It says that trace, sum of the diagonal elements of matrix, stays unchanged in any basis.

Are those two claims equivalent in formalism of quantum mechanics? Is it even linked?

My thoughts:

Hamiltonian is an operator(linear transformation, matrix), it has same trace in every basis. Sum of all diagonal elements,(that is, sum of all En's associated to some ψn's) will always be the same. If we change the basis, we change set of ψn's, but sum of their corresponding En's will be the same, so total energy E will be the same.

I am not sure if I interconneted it correctly, it seems a bit vague. I'd appreciate any help. Thanks.

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    $\begingroup$ I am pretty sure that the energy you can get terms contributing to energy which are NOT on the diagonal. $\endgroup$ Feb 4, 2017 at 19:04
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    $\begingroup$ 1. The question is much better suited for Physics.SE. 2. When quoting some material from a book, you better mention its edition as well as page numbers. 3. I can hardly follow you thoughts: why on earth do you think there is a connection between those two statements? $\endgroup$
    – Wildcat
    Feb 8, 2017 at 11:12

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Well this is what I can think of.

If $H$ is a matrix and $\psi$ is a vector of length $n$,

$$H \psi = E \psi$$

is an eigenvalue problem. Let's say you got as many solutions (pairs of $\psi, E$) to this problem as there are rows in $H$ (same as dimensions of the vector space, $n$). So solutions are going to be called $\psi_i$ and their corresponding eigenvalues $E_i$. Set

$$V = [\psi_1 \cdots \psi_n]$$

By that I mean that $V$ is a square matrix whose columns are the column vectors $\psi_i$. Matrix multiplication is column-wise:

$$H V = [H \psi_1 \cdots H \psi_n] = [E_1 \psi_1 \cdots E_n \psi_n] \\ = V E$$

where $E$ is a matrix whose diagonal elements are $E_i$. Well then $V^{-1} H V = E$.

A nice property of the trace is its invariance under cyclic permutations,

$$tr(A B C) = tr(C A B) = tr(B C A)$$

Thus

$$\sum_i E_i = tr(E) = tr(V^{-1} H V) = tr(V V^{-1} H) = tr(H)$$

In fact any basis transformation between matrices $A$ and $B$ leads to an equation like $A \Gamma = \Gamma B$, where $\Gamma$ is the matrix representing the basis change. Similarly, if such an equation holds (for a sufficiently well behaved matrix $\Gamma$), we say $A$ and $B$ are similar matrices. This is why the trace is the same in every basis.

And obtaining a complete set of solutions for an eigenvalue problem is equivalent as finding a basis for which the matrix $H$ can be represented as a diagonal matrix $E$.

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    $\begingroup$ But how is this related to separable solutions? $\endgroup$
    – aventurin
    Feb 5, 2017 at 20:48
  • $\begingroup$ In a sense, nothing. My answer makes no assumption about separability, so it is general. On the other hand, the time independent Schrödinger equation, with which I began the answer, is itself the result of a separation of variables. Any solution to this equation is thus already separated, e.g. from its time dependent wave function. $\endgroup$ Feb 6, 2017 at 11:13
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These so-called "separable solutions" are states of definite total energy because they are eigenstates of the Hamiltonian. By definition, a separable solution is an eigenfunction of the Hamiltonian multiplied by the corresponding phase factor, $$ \Psi_i(x, t) = \mathrm{e}^{-i E_i t / \hbar} \psi_i(x) \, , $$ where $$ \hat{H} \psi_i(x) = E_i \psi_i(x) \, . $$ And since $\hat{H}$ is time-independent, such solutions are eigenfunctions of the Hamiltonian as well, $$ \begin{align} \hat{H} \Psi_i(x, t) &= \hat{H} \mathrm{e}^{-i E_i t / \hbar} \psi_i(x) \\ &= \mathrm{e}^{-i E_i t / \hbar} \hat{H} \psi_i(x) \\ &= \mathrm{e}^{-i E_i t / \hbar} E_i \psi_i(x) \\ &= E_i \mathrm{e}^{-i E_i t / \hbar} \psi_i(x) \\ &= E_i \Psi_i(x, t) \, . \end{align} $$ And that is basically it, because any state can be represented by the linear combination of eigenstates of a self-adjoint operator, $$ \Psi(x, t) = c_i \Psi_i(x, t) \, , $$ and, in accordance with the postulates of quantum mechanics, a measurement of energy on a system in a state $\Psi$ can yield value $E_i$ with the probability equal to coefficient in front of the corresponding eigenfunction $\Psi_i$, $$ \Pr(E_i) = c_i \, . $$ So that if a state $\Psi(x, t)$ is one of the Hamiltonian eigenstates $\Psi_j(x, t)$, then all the coefficients $c_i$ are equal to 0 except for $c_j$ which is equal to 1, which means that we will certainly get the corresponding eigenvalue $E_j$ when measuring energy: $\Pr(E_j) = 1$. In words, separable solutions describe states of definite total energy.

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