# I Dirac notation and calculations

#### electrogeek

Hello everyone,

I'm stuck on the question which I have provided below to do with Dirac notation:

In these questions |a>, |b> and |c> can be taken to form an orthonormal basis set
Consider the state |ξ> = α(|a> − 2|b> + |c>). What value of α makes |ξ> a normalised state?

I'm brand new to Dirac notation so I'm not too sure what to do? I was thinking that in order for |ξ> to be normalised, then (|ξ> )^2 = 1. But I don't know whether you can do (|ξ>)(|ξ>)? I've only ever seen something like <a|a> which equals 1 if you do <a|a> or 0 if you do <b|c>.

If you can square a ket, then would I be right in saying α is 1/ sqrt (3) because expressions like |a>|a> would be 1 and |a>|b> would be 0?

Any help will be greatly appreciated!

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#### PeroK

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Hello everyone,

I'm stuck on the question which I have provided below to do with Dirac notation:

In these questions |a>, |b> and |c> can be taken to form an orthonormal basis set
Consider the state |ξ> = α(|a> − 2|b> + |c>). What value of α makes |ξ> a normalised state?

I'm brand new to Dirac notation so I'm not too sure what to do? I was thinking that in order for |ξ> to be normalised, then (|ξ> )^2 = 1. But I don't know whether you can do (|ξ>)(|ξ>)? I've only ever seen something like <a|a> which equals 1 if you do <a|a> or 0 if you do <b|c>.

Any help will be greatly appreciated!
Whether you're using Dirac notation of not, a state vector is normalised if its inner product with itself is 1. In Dirac notation this is

$\langle \xi|\xi \rangle = 1$

#### electrogeek

Whether you're using Dirac notation of not, a state vector is normalised if its inner product with itself is 1. In Dirac notation this is

$\langle \xi|\xi \rangle = 1$
Ah okay thank you. Are there any rules for calculating <ξ|? I know it's the hermitian conjugate but I don't really know how'd you calculate it for something like this. I know that if you treat kets like column vectors then bras are like row vectors bit that's it.

#### PeroK

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Ah okay thank you. Are there any rules for calculating <a|? I know it's the hermitian conjugate but I don't really know how'd you calculate it for something like this. I know that if you treat kets like column vectors then bras are like row vectors bit that's it.
The basic rule for bras is that:

If $| \xi \rangle = \alpha |a \rangle + \beta | b \rangle$

then $\langle \xi |= \langle a| \alpha^* + \langle b| \beta^*$

#### electrogeek

The basic rule for bras is that:

If $| \xi \rangle = \alpha |a \rangle + \beta | b \rangle$

then $\langle \xi |= \langle a| \alpha^* + \langle b| \beta^*$
Ah okay. Thanks for the help. I'll give the question a go now.

#### electrogeek I've just had a bit of free time to give the question a go and I got alpha to be 1/sqrt(6). I've attached my workings above - is this right?

#### PeroK

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I've just had a bit of free time to give the question a go and I got alpha to be 1/sqrt(6). I've attached my workings above - is this right?
Yes. Note that, in simple terms you have a vector expressed as $\alpha (1, -2, 1)$ in an orthonormal basis. The magnitude of this vector is $\alpha \sqrt{6}$. Hence, $\alpha = 1/\sqrt{6}$ for normalisation.

It's good to go through all the steps in Dirac notation, but it's only the usual linear algebra in the end.

• Nugatory

#### electrogeek

Yes. Note that, in simple terms you have a vector expressed as $\alpha (1, -2, 1)$ in an orthonormal basis. The magnitude of this vector is $\alpha \sqrt{6}$. Hence, $\alpha = 1/\sqrt{6}$ for normalisation.

It's good to go through all the steps in Dirac notation, but it's only the usual linear algebra in the end.
Ah brilliant! Thanks for the help. Hopefully this will put me in a good position for the rest of the questions.

#### PeroK

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2018 Award
Ah brilliant! Thanks for the help. Hopefully this will put me in a good position for the rest of the questions.
Just a minor additional point. This all assumes $\alpha$ is real and positive, which is the convention for normalisation constants. But, of course, any complex number with modulus $1/\sqrt{6}$ would do.

#### tnich

Homework Helper
Hello everyone,

I'm stuck on the question which I have provided below to do with Dirac notation:

In these questions |a>, |b> and |c> can be taken to form an orthonormal basis set
Consider the state |ξ> = α(|a> − 2|b> + |c>). What value of α makes |ξ> a normalised state?

I'm brand new to Dirac notation so I'm not too sure what to do? I was thinking that in order for |ξ> to be normalised, then (|ξ> )^2 = 1. But I don't know whether you can do (|ξ>)(|ξ>)? I've only ever seen something like <a|a> which equals 1 if you do <a|a> or 0 if you do <b|c>.

If you can square a ket, then would I be right in saying α is 1/ sqrt (3) because expressions like |a>|a> would be 1 and |a>|b> would be 0?

Any help will be greatly appreciated!
You have the right idea. If you think of $| a\rangle$ and $|b\rangle$ as vectors in Hilbert space, then $\langle a | b\rangle$ is their dot product. So multiply out $\langle \xi | \xi \rangle$ and see what you get.