1
$\begingroup$

An annoying fact about my physics textbook (Griffiths' Introduction to Quantum Mechanics) is that it introduces bra-ket notation without telling us how to use it. So I have a two-part question for SE:

  1. In general, how do I write an operator written in functional form (say, the Hamiltonian, $-\frac{\hbar^2}{2m}\nabla^2+V$) as something in the form $\sum|a\rangle\langle b|$?
  2. In particular, one of the questions my professor gave me told me (without explanation) that the Hamiltonian of the finite double well could be written $|L\rangle \langle R| + |R\rangle \langle L|$ for the purposes of calculating tunneling probabilities. How can I reproduce this result?

Details for #2: The text of the question reads:

A box containing a particle is divided into a left and a right compartment by a thin partition. If the particle is known to be on the right side with certainty its state is represented by the state vector $|R\rangle$. This is basically a particle in a box but where we have chosen to neglect spatial variations in either half of the box. ... The particle can tunnel through the partition; this tunneling effect is characterized by the Hamiltonian $H=\Delta (|R\rangle\langle L| + |L\rangle\langle R|)$ where $\Delta$ is a real number with dimensions of energy.

Essentially, I don't understand where that Hamiltonian comes from. I can see that it is something tunnel-y (in the sense that it's doing something with mappings from one state from the other) but not how to rigorously generate it.

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
2
$\begingroup$

  1. If your desired basis is the set ${|n\rangle}$, then the completeness relation tells you: $\hat{O} = \sum_a \sum_b \langle a|\hat{O}|b \rangle |a \rangle \langle b|$. Ideally, we prefer to do this in the orthonormal basis in which the operator $\hat{O}$ is diagonal, in which case this becomes $\hat{O} = \sum_a \langle a|\hat{O}|a \rangle |a \rangle \langle a|$. Then the coefficients of the expansion are just the eigenvalues of the operator and the basis is the set of eigenvectors. This is called the spectral decomposition.

  2. This question is a bit vague. More details would be helpful.

Improve this answer
$\endgroup$
8
  • $\begingroup$ blink Huh. That makes perfect sense. Also, the text of the problem has been added to the question. (Don't worry - even if I wasn't asking how to generate the question in the first place, the set was due two weeks ago.) $\endgroup$
    – linkhyrule5
    Commented Mar 10, 2014 at 11:37
  • $\begingroup$ Wait, I take that back. The set $|n\rangle$ has completely vanished in your second two equations - what happened there? $\endgroup$
    – linkhyrule5
    Commented Mar 10, 2014 at 11:42
  • $\begingroup$ I'm not sure I follow. $n$ is just a variable, it doesn't matter what letter you use. $a$ and $b$ are just dummy indices for summation. $\endgroup$
    – user27578
    Commented Mar 10, 2014 at 12:44
  • $\begingroup$ As for the second question: tunneling between the the wells means that $|L\rangle$ becomes $|R\rangle$ and vice versa. Apply that operator to either state and that's exactly what you get. Though I don't know why your professor is calling it the system's Hamiltonian; it's a simplified time evolution operator. $\endgroup$
    – user27578
    Commented Mar 10, 2014 at 12:55
  • $\begingroup$ shrug It's a thing that we were told to set equal to $E \psi$, so... And I think I see - by "set $|n\rangle$" you mean "the set of vectors $|a\rangle, |b\rangle, |c\rangle\cdots$", correct? $\endgroup$
    – linkhyrule5
    Commented Mar 10, 2014 at 13:35

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.