# Quick questions regarding BRA KET notation

## Main Question or Discussion Point

I'm trying to apply BRA KET notation to my notes on particle physics.

please could someone confirm that the kroneker delta function may be written

$$\delta _{ij} = \left \langle i |j \right \rangle$$

OR would it be written

δij = |i> <j|

I know i and j are indices, so can BRA KET even be used?

Also if you have the Levi-Civita Tensor εijk is it possible to write this in BRA KET form?

In particular take this identity:

$$\varepsilon _{ijk} \varepsilon _{ilm} = \delta _{jl} \delta _{km} - \delta _{jm} \delta _{kl}$$

how would this be written in BRA KET, or am I mixing things up? it would be nice if at all possible to use BRA KET in my particle as I found using it very useful in last semesters quantum!

thanks for any ideas / guidance!

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ShayanJ
Gold Member
You're mixing things up.
In QM, states are represented by kets, which are vectors in an abstract Hilbert space(a vector space with some properties). So $|\phi\rangle , |i \rangle , |\uparrow \rangle ,...$ are abstract vectors.
$\langle i | j \rangle$ is the inner product of vectors $|i \rangle$ and $|j \rangle$. Only if they're part of an orthonormal set of vectors, you can write $\langle i | j \rangle=\delta_{ij}$, otherwise the inner product can be any complex number. So $\langle i | j \rangle=\delta_{ij}$ is not a representation of Kronecker delta but only a way of saying that we have a orthonormal set of vectors.
And about $| i \rangle \langle j |$. This is an operator, something that gets a vector and gives another vector. But its not what Kronecker delta does its not an operator and so $\delta_{ij}=| i \rangle \langle j |$ is wrong too!

Not only that these are not representations of Kronecker delta, but also people usually don't use such representations because $\delta_{ij}=\left\{ \begin{array}{lr} 1 \ \ \ i=j \\ 0 \ \ \ i\neq j \end{array} \right.$ is good enough.

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You're mixing things up.
In QM, states are represented by kets, which are vectors in an abstract Hilbert space(a vector space with some properties). So $|\phi\rangle , |i \rangle , |\uparrow \rangle ,...$ are abstract vectors.
$\langle i | j \rangle$ is the inner product of vectors $|i \rangle$ and $|j \rangle$. Only if they're part of an orthonormal set of vectors, you can write $\langle i | j \rangle=\delta_{ij}$, otherwise the inner product can be any complex number. So $\langle i | j \rangle=\delta_{ij}$ is not a representation of Kronecker delta but only a way of saying that we have a orthonormal set of vectors.
Not only that it is not, in general, a representation of Kronecker delta, but also people usually don't use such representations because $\delta_{ij}=\left\{ \begin{array}{lr} 1 \ \ \ i=j \\ 0 \ \ \ i\neq 0 \end{array} \right.$ is good enough.
And about $| j \rangle \langle i |$. This is an operator, something that gets a vector and gives another vector. But its not what Kronecker delta does its not an operator and so $\delta_{ij}=| j \rangle \langle i |$ is wrong too!
hmm but couldn't you also apply $| j \rangle \langle i |$ sort of like this:

$$x_{i} \delta _{ij} y_{i} = \left \langle x|\delta |y \right \rangle$$

would that be correct?

aside from that, thanks for the reply, was hoping to apply BRA and KET but if it's incorrect then I'll just use standard notation.

ShayanJ
Gold Member
rwooduk said:
$x_i \delta_{ij}y_i=\langle x | \delta | y \rangle$
You mean $\langle x | (| i \rangle \langle j |)| y \rangle=\langle x| i \rangle \langle j | y \rangle=\langle i| x \rangle^* \langle j | y \rangle=x_i^* y_j$?

This is correct only if $| i \rangle$ and $| j \rangle$ are part of an orthonormal set of vectors. But this has nothing to do with Kronecker delta. Because $\delta_{ij}=\left\{ \begin{array}{lr} 1 \ \ \ i=j \\ 0 \ \ \ i\neq j \end{array} \right.$ is a number: 0 or 1. Its just a compact way of writing particular conditionals to choose between 0 and 1 but it still means a number! But $| i \rangle \langle j |$ is an operator.

You mean $\langle x | (| i \rangle \langle j |)| y \rangle=\langle x| i \rangle \langle j | y \rangle=\langle i| x \rangle^* \langle j | y \rangle=x_i^* y_j$?

This is correct only if $| i \rangle$ and $| j \rangle$ are part of an orthonormal set of vectors. But this has nothing to do with Kronecker delta. Because $\delta_{ij}=\left\{ \begin{array}{lr} 1 \ \ \ i=j \\ 0 \ \ \ i\neq j \end{array} \right.$ is a number: 0 or 1. Its just a compact way of writing particular conditionals to choose between 0 and 1 but it still means a number! But $| i \rangle \langle j |$ is an operator.
Yes, that's what I meant, hmm indeed, yes i and j are just numbers not part of an orthonormal set of vectors, ok guess I better start getting used to the upcoming new notation! Thanks very much you have saved me from making errors and a large amount of time!

ShayanJ
Gold Member
i and j are just numbers not part of an orthonormal set of vectors
It seems you misunderstood things. i and j are numbers but $|i\rangle$ and $|j\rangle$ are vectors. Now if we can find a set of vectors $B=\{ | n \rangle \}_{n=1}^\infty$ such that for all pairs of vectors in this set, we have $\langle k | l \rangle =\delta_{k l}$ and $|i\rangle \in B$ and $|j\rangle \in B$ , we can say $|i\rangle$ and $|j\rangle$ are part of an orthonormal set of vectors, otherwise they're just vectors.
If conditions above are satisfied, then any other vector $|x\rangle$ can be written as a linear combination of members of B, i.e. $|x\rangle=\sum_{n=1}^\infty x_n |n\rangle$ where $x_n$ are complex numbers. Then we can write $\langle i |x\rangle=\sum_{n=1}^\infty x_n \langle i|n\rangle=\sum_{n=1}^\infty x_n \delta_{in}=x_i$.

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It seems you misunderstood things. i and j are numbers but $|i\rangle$ and $|j\rangle$ are vectors. Now if we can find a set of vectors $B=\{ | n \rangle \}_{n=1}^\infty$ such that for all pairs of vectors in this set, we have $\langle k | l \rangle =\delta_{k l}$ and $|i\rangle \in B$ and $|j\rangle \in B$ , we can say $|i\rangle$ and $|j\rangle$ are part of an orthonormal set of vectors, otherwise they're just vectors.
If conditions above are satisfied, then any other vector $|x\rangle$ can be written as a linear combination of members of B, i.e. $|x\rangle=\sum_{n=1}^\infty x_n |n\rangle$ where $x_n$ are complex numbers. Then we can write $\langle i |x\rangle=\sum_{n=1}^\infty x_n \langle i|n\rangle=\sum_{n=1}^\infty x_n \delta_{in}=x_i$.

thanks, yes i misunderstood, thats much clearer! thanks again.