# Bras and kets vs. Einstein summation convention

1. Oct 27, 2012

### HJ Farnsworth

Greetings,

This is just an opinion question about notations.

Having learned the basics of bra-ket notation and using the ESC, as far as I can tell, ESC is just plain better, at least when dealing with finite bases. Using bras and kets, you can represent and manipulate states using scalars, vectors (functions), one-forms (dual functions), and operators, but using ESC you can extend this very easily to any kind of tensor.

I haven't done any ESC where the bases involved were countably or uncountably infinite, so maybe that's where bras and kets become more advantageous - but on the other hand, it would surprise me if there's not some simple extension of the rules of the ESC that allows it to account for these types of bases (eg., an index appearing as superscript and subscript implies integration with respect to that index, so ESC becomes EIC).

So basically, it seems to me that bra-ket notation is great for linear algebra, but ESC is great for tensor analysis, which includes linear algebra, and it also works just as well as bras and kets for linear algebra before worrying about higher order tensors. So it's better.

Is there some obvious advantage to bras and kets that isn't occuring to me right now (this thought occured to me about 10 minutes ago, so I haven't really considered it that much)? What are people's opinions?

-HJ Farnsworth

2. Oct 28, 2012

### tom.stoer

In QM you always write down the sum explicitly, i.e.

$$1 = \sum_n|n\rangle\langle n|$$

$$|\psi\rangle = \sum_n\psi_n |n\rangle$$

$$\text{tr} A = \sum_n\langle n|A|n\rangle$$

3. Oct 28, 2012

### Jazzdude

The summation convention is not what you have in mind, you mean Ricci calculus where the summation convention is one tiny part that spares you writing the sums explicitly. You can use the SC also with the Dirac notation, so that's not the important difference.

Ricci goes much farther by defining a set of biorthogonal bases of the tangent and cotangent space. This is very sensible if you have non-cartesian coordinates, but totally superfluous for quantum theory, where the inner product is canonical and no differential geometry on the state space present. Applying Ricci calculus to QT on the Hilbert space would be close to meaningless, but definitely overkill. On the other hand the Dirac notation greatly simplifies inner products, outer products, tensor products, in a way that is very suitable for the structure of quantum theory. The notation is extremely intuitive and allows you to see precisely what kind of object you have written down. Also, multilinear forms play only a side role in quantum theory. The only case where they come in would be the so called Super-operators.

There is one exception to this general picture though. If you formulate quantum theory on its non-commutative phase space it sometimes makes sense to use tools from differential geometry. But even then you gain much more by using the language of differential forms and geometric algebras instead of Ricci calculus, which has some issues with unitary representations.

4. Oct 28, 2012

### andrien

Bra and ket algebra really makes the harmonic oscillator problem easy in terms of creation and destruction operators,which determines the eigenvalues in a very simple way without solving a differential eqn. and these operators have generalized form in quantum field theory.

5. Oct 28, 2012

### dextercioby

Bra's and ket's are concepts in functional analysis, the ESC applies to the part of multilinear algebra which comes with the elementary differential geometry. There's no direct connection between these concepts, in quantum mechanics we always use the SUM symbol, there are no indexed objects ('upstairs' or 'downstairs').

6. Oct 28, 2012

### samalkhaiat

Let $\{ |{}_{a}\rangle \}$ be a set of basis vectors in some index space in $T_{p}(M)$, so that any vector $|A \rangle \in T_{p}(M)$ can be written as
$$|A \rangle = A^{a} \ |{}_{a} \rangle .$$
This is just another way of writing $A = A^{\mu} \ \partial_{\mu}$.
Following the usual idea of linear algebra, we can associate a dual index space (in $T^{*}_{p}(M)$) with each index space in $T_{p}(M)$. So, for a given basis $|{}_{b}\rangle$, we can construct its dual basis $\langle {}^{a}|$ by demanding
$$\langle {}^{a}| {}_{b} \rangle = \delta^{a}_{b}. \ \ \ (1)$$
So, each covector (form) $\langle \omega |$ in the dual space $T^{*}_{p}(M)$ can be represented as
$$\langle \omega | = \langle {}^{a}| \ \omega_{a}.$$
This should remind you with $\omega = \omega_{\nu} \ dx^{\nu}$.
When there is a metric defined on M, we can also write
$$\langle {}_{a}| {}_{b}\rangle = \eta_{ab}, \ \ \ (2)$$
$$\langle {}^{a} | {}^{b}\rangle = \eta^{ab}. \ \ \ (3)$$
Using eq(2), we can show that the tensor operator
$$\Sigma_{ab}= | {}_{a}\rangle \langle {}_{b}| - | {}_{b}\rangle \langle {}_{a}| ,$$
forms a representation of Lorentz algebra. Now consider the matrix element
$$\langle {}^{c}| \Sigma_{ab}| {}_{d} \rangle \equiv ( \Sigma_{ab})^{c}{}_{d} .$$
Using eq(1) and eq(2), we find
$$\langle {}^{c}| \Sigma_{ab}| {}_{d} \rangle = \delta^{c}_{a} \ \eta_{bd} - \delta^{c}_{b} \ \eta_{ad}.$$
We recognise this as the spin matrix associated with the vector representation of Lorentz group. Indeed, under the Lorentz group $SO(1,3)$, the vector $|A \rangle$ transforms according to
$$\langle {}^{c}| \Sigma_{ab}| A \rangle = \langle {}^{c}| \Sigma_{ab}| {}_{d} \rangle \ A^{d} = \delta^{c}_{a} \ A_{b} - \delta^{c}_{b} \ A_{a}.$$
B.S. De Witt extended the above methods to a vector space in which the index takes values in the set $\{a , x \}$; $x \in \mathbb{R}$. In this space, De Witt defines the scalar product by
$$\Phi_{J}\Pi^{J} = \int d^{n}x \ \Phi_{a}(x) \ \Pi^{a}(x).$$
For a more detailed treatment on this see De Witt lectures (The Space-time Approach to Quantum Field Theory) in:
Relativity, Groups and Topology, eds B.S. De Witt and R. Stora (Elsevier Science Publishers B. V., 1984).
Sam

Last edited: Oct 28, 2012