# Quick questions regarding BRA KET notation

• rwooduk
That is, 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 for all vectors ## | x \rangle ##, we can write ## |x\rangle =\sum_{n=1}^{\infty} x_n | n \rangle ## where ## x_n =\langle n | x \rangle ##, we can say that ## B ## is a basis of the Hilbert space and ## | x \rangle ## can be represented as a linear combination of basis vectors ## | x \rangle

#### rwooduk

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!

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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• rwooduk
Shyan said:
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.

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.

• rwooduk
Shyan said:
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!

rwooduk said:
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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• rwooduk
Shyan said:
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, that's much clearer! thanks again.