Quick questions regarding BRA KET notation

[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

δ_ij = |i> <j|

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

Also if you have the Levi-Civita Tensor ε_ijk 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!

 

[ShayanJ]

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.

 

[rwooduk]

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]

$x_i \delta_{ij}y_i=\langle x | \delta | y \rangle$

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]

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]

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 $\langle i |x\rangle=\sum_{n=1}^\infty x_n \langle i|n\rangle=\sum_{n=1}^\infty x_n \delta_{in}=x_i$.

 

[rwooduk]

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.