0
$\begingroup$

I'm reading Dirac's book about QM. I reached the chapter called "representations" where Dirac introduces how can bras, kets, and observables be decomposed using a base. I have found issues understanding what does Dirac's notation means and how he justifies it. He says:

We may suppose the basic bras to be labelled by one or more parameters, $\lambda_1, \lambda_2, \lambda_3, \cdots \lambda_u $ each of which may take on certain numerical values, the basic bras can then be written as $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u |$ and the representative of $|a \rangle$ will be written as $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u | a \rangle$. This representative will now consist of a set of numbers, one for each set of values that $\lambda_1, \lambda_2, \lambda_3, \cdots \lambda_u $ may have in their respective domains.

First of all I don't understand what is the connection between those $\lambda$s parameters and the $\langle \lambda |$s eigenbras. Then I'm not sure about what does $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u | a \rangle$ mean: is it a compact notation for $\langle \lambda_1 | a \rangle, \langle \lambda_2 | a \rangle, \langle \lambda_3 | a \rangle, \cdots, \langle \lambda_u | a \rangle$ ? And if this is right, is $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u |$ a shorthand for $\langle \lambda_1 |, \langle \lambda_2 |, \langle \lambda_3 |, \cdots, \langle \lambda_u |$? And last, how can Dirac's talk about domain, when the $\lambda$s as I understood aren't functions?

$\endgroup$
1
  • 1
    $\begingroup$ The eigenvalues are used as labelled for the bras. A single bra (not a tensor product) can be labelled by multiple eigenvalues. $\endgroup$ Commented Mar 16, 2021 at 21:23

2 Answers 2

5
$\begingroup$

I will try to translate the following passage by Dirac that you quote in a more familiar language, with minimal changes and few explanatory injections.

Broke

We may suppose the basic bras to be labelled by one or more parameters, $\lambda_1, \lambda_2, \lambda_3, \cdots \lambda_u $, each of which may take on certain numerical values, the basic bras can then be written as $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u |$ and the representative of $|a \rangle$ will be written as $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u | a \rangle$. This representative will now consist of a set of numbers, one for each set of values that $\lambda_1, \lambda_2, \lambda_3, \cdots \lambda_u $ may have in their respective domains.

Woke

We may suppose the basis bras of a Hilbert space to be labelled by one or more parameters, $\lambda_1, \lambda_2, \lambda_3, \cdots, \lambda_u $, each of which may take on certain numerical values. In particular, let's imagine the set of operators $\{O_1, O_2,\cdots, O_u\}$ to form a largest set of commuting operators on the said Hilbert space. Then, $\lambda_1,\lambda_2,\cdots,\lambda_u $ can be imagined to be the eigenvalues of the operators $O_1, O_2,\cdots, O_u$ respectively. Each basis bra can then be written as $\langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u |$. We are using the eigenvalues of the largest set of commuting operators as labels for the basis bras because we are constructing the basis bras as the eigenbras of the said operators. Now, the wavefunction of $|\psi \rangle$ in this basis will be written as $\psi(\lambda_1,\lambda_2,\cdots,\lambda_u) = \langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u | \psi \rangle$. This wavefunction's value will be a complex number, one complex number for each valid (see below for the "valid") combination of the values that $\lambda_1, \lambda_2, \lambda_3, \cdots \lambda_u $ may take.

Few points of clarification:

  • Notice that $\psi(\lambda_1,\lambda_2,\cdots,\lambda_u) = \langle \lambda_1 \lambda_2 \lambda_3 \cdots \lambda_u | \psi \rangle$ is a single (complex) number. The full description of the state $\vert\psi\rangle$ would entail specifying $d$ of these numbers where $d$ is the dimensionality of the Hilbert space.
  • Relatedly, $\langle \lambda_1\lambda_2\cdots\lambda_u\vert$ is a single bra. In particular, as I mentioned, the bras are being labeled by the eigenvalues of the largest set of commuting operators whose eigenbras they are. For example, for the particle spin, the bras would be of the form $\langle lm\vert$ where $l$ is such that $\langle lm\vert L^2 = \langle lm\vert l^2$ and $m$ is such that $\langle lm\vert L_z = \langle lm\vert m$. In other words, we are just naming the shared eigenbra of $L^2$ and $L_z$, characterized by its total angular momentum $l$ and its $z-$spin $m$, as $\langle lm\vert$.
  • The domains of $\lambda_i$ is simply the set of eigenvalues of the operator $O_i$. More broadly, the domain of a parameter, in physics, is understood to be the set of values that it can take.
  • Finally, notice that it need not be the case that if $\lambda_1$ can take $N_1$ values and $\lambda_2$ can take $N_2$ values then the total number of basic bras $\langle \lambda_1\lambda_2\vert$ would be $N_1N_2$. It may be but it may not be. In particular, you might not necessarily vary $\lambda_i$s independently to generate one eigenbra for each variation. For example, for a spin, $l$ can take integer values and $m$ can take integer and half-integer values -- however, there is no basis bra $\langle l=1, m=3/2\vert$ because $m$ can only take values from $-l $ to $l$ for a given $l$.
$\endgroup$
3
  • $\begingroup$ Hi, thanks for your answer and your time. You were very clear, but I don't really understand why would you take more than one observable to make a representation, would a single one be sufficent for a representation? $\endgroup$
    – Luke__
    Commented Mar 17, 2021 at 7:57
  • $\begingroup$ @LucaMattioni That's a good question. The basic idea is that you need as many unique eigenvalues as the dimensionality of the Hilbert space, and it is not always the case that a single physically meaningful operator will be non-degenerate in all its eigenvalues. For example, for a spin $1/2$ particle on a line, the physically meaningful operators are $x,p,\ \mathrm{and}\ S_z$. No single one of them is non-degenerate: every eigenstate of $x$ is two-fold degenerate, corresponding to the two eigenvalues of the spin. [...] $\endgroup$
    – user87745
    Commented Mar 17, 2021 at 14:06
  • $\begingroup$ [...] The same story goes for $p$, each of its eigenvalues is two-fold degenerate corresponding to the two eigenvalues of the spin. So, you either need $x,S_z$ or you need $p,S_z$. However, for spin$-0$ particle, simply $x$ would suffice. So, if the physically meaningful operators that you work with are degenerate in their eigenvalues then you need to find a bunch of them who together pin down a one-dimensional subspace of the Hilbert space when their eigenvalues are specified. $\endgroup$
    – user87745
    Commented Mar 17, 2021 at 14:08
1
$\begingroup$

$⟨λ_1λ_2λ_3⋯λ_u|$ is one basic bra. It is just a way to order the base vector in a specific order. $⟨λ_1λ_2λ_3⋯λ_u|a⟩$ is the bra-ket product (scalar product) between $⟨λ_1λ_2λ_3⋯λ_u|$ and $|a⟩$. For example, if u=2 and the domain of the λs are {1,2,3} and {1,2} than you can describe any 6 different basic bras in this way: $$⟨11|,⟨12|,⟨21|,⟨22|,⟨31|,⟨12|$$

$\endgroup$

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.