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In a simple example, Most of the Hamiltonians are talked about on its eigenbasis or a basis that can be transformed from the eigenbasis. With this, the eigenvalues do not change even if on a different basis, but the Eigenvector will change depending on which basis. (See example on 1.6 in https://www2.ph.ed.ac.uk/~gja/qp/qp1.pdf)

Question:
But in practice, most of the Hamiltonians cannot be solved analytically. If I use a basis which is not an eigenbasis $\{|\psi^{non-e}_{n}\rangle\}$ of the Hamiltonian $H$, what will happen?

For example, On a two-body system, what if I use the Gaussian form wave function ($\psi_{nlm}^{Gaussian}\propto e^{-\alpha r^2}$) on a Hamiltonian with potential in Coulomb form $V_{ij}=\frac{A}{r}$ (which the exact solution of Coulomb potential is proportional to $\psi_{nlm}^{Coulomb}\propto e^{-\beta r}$). (Or more practically, a Cornell form $V_{ij}=\frac{A}{r}+Br$)

Can I get the matrix element of $H$, Eigenvalues of $H$ on this basis $\{|\psi^{non-e}\rangle\}$?, and will the Eigenvalues be useful?

Edit 1: A similar thing about this question should be the variational method, $$\bar{H}=\frac{\langle \psi_{trial}|H|\psi_{trial}\rangle}{\langle\psi_{trail}|\psi_{trail}\rangle}.$$ But the variational method is valid only for the ground state. Well, in this case, it is a set of basis function which is similar to the trail function. $$\bar{H}=\begin{pmatrix}\bar{H}_{ij}&\cdots\\\vdots&\ddots\end{pmatrix},~where~\bar{H}_{ij}=\langle \psi^{trial}_i|H|\psi^{trial}_j\rangle. $$ Is it possible?

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In an arbitrary basis the Hamiltonian is simply a non-diagonal matrix, and the eigenvalues represent the energy levels. Let me give you a much much simpler example: consider the Hamiltonian describing a $1/2$ fermion in a magnetic field along $x$ axis $$ H = - \gamma S_x $$ where $S_x$ is the spin x operator. Now the Hilbert space is two-dimensional and if you take as a basis the eigenstates of the spin z operator $S_z$, and you call them $\left| \uparrow \right\rangle$ and $\left| \downarrow \right\rangle$, then the spin x operator, and hence the Hamiltonian are not diagonal $$ S_x = \frac{\hbar}{2} \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right). $$ The eigenvalues and eigenvectors of $H$ then are $$ E_0 = - \gamma \hbar/2 \;\;\;\; , \;\;\;\; \left| \rightarrow \right\rangle = \frac{1}{\sqrt{2}} \left( \left| \uparrow \right\rangle + \left| \downarrow \right\rangle \right) $$ $$ E_1 = \gamma \hbar/2 \;\;\;\; , \;\;\;\; \left| \leftarrow \right\rangle = \frac{1}{\sqrt{2}} \left( \left| \uparrow \right\rangle - \left| \downarrow \right\rangle \right) $$.

If instead you decide from the beginning to use the eigenstates of $S_x$ (i.e. $\left| \rightarrow \right\rangle $ and $\left| \leftarrow \right\rangle $ ) as a basis for the Hilbert space, then the spin x operator and hence the Hamiltonian are diagonal:

$$ S_x = \frac{\hbar}{2} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right). $$

with the same eigenvalues and eigenvectors that coincide with the basis elements.

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  • $\begingroup$ Thank you, but well, that is a simple case. As I mentioned the basis of $\{|\leftarrow\rangle, |\rightarrow\rangle\}$ can be completely transformed into basis $\{|\uparrow\rangle, |\downarrow\rangle\}$. But for this case, two bases cannot be completely transformed into each other. $\endgroup$ Feb 13, 2021 at 15:44
  • $\begingroup$ This is very general: all the accessible states of the system form an Hilbert space, which is a vector space. Chosing a basis means by definition that any element of the space can be written as a linear combination of the basis, but there are two techincal complications in your problem: 1) the Hilbert space is infinite dimensional 2) all the states depend on a continuum label, which is the coordinate $r$. So if you want to use numerics you have to fix this problems, but in principle the eigenvalues of $H$ are the energy levels and the eigenstates are linear combinations of the basis $\endgroup$
    – Matteo
    Feb 13, 2021 at 16:02

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