What is unique about the bra in Dirac bra-ket notation?

[Kashmir]

It's said that every ket has a unique bra. For any vector $|v> ∈ V$ there is a unique bra $<v| ∈ V*$.

I'm not sure what that means. What is unique? Can anyone please help me understand.
Thank you

 

[jedishrfu]

I found this reference that may be helpful

https://courses.cs.washington.edu/courses/cse599d/06wi/lecturenotes2.pdf

 

[Kashmir]

I found this reference that may be helpful

https://courses.cs.washington.edu/courses/cse599d/06wi/lecturenotes2.pdf

thank you but the article discusses in the light of column vectors . I wanted something more general.

 

[vanhees71]

Well, for physicists it's almost always ok not to worry about these mathematical subtleties. One of my math professors (an expert on Lie-group theory) once said that the physicists get away with this, because the separable Hilbert space is almost as harmless as a finite-dimensional unitary vector space (of course he emphasized the word "almost"). So it's good to be at least aware of the corresponding subtleties (in this example the distinction between bound and unbound linear forms on Hilbert space) to avoid mistakes. For a nice pedagogical introduction to the subtleties, but here more concerning self-adjoint operators and their spectra, see

https://arxiv.org/abs/quant-ph/9907069

 

[PeterDonis]

the article discusses in the light of column vectors . I wanted something more general.

The statement the article makes can be generalized from the column vector case. Just note that the bra $\bra{v}$ corresponding to the ket $\ket{v}$ always satisfies $\langle v \vert v \rangle = |v|^2$, i.e., the inner product of the two is the squared norm of $\ket{v}$. There will be one unique bra that has that property in any Hilbert space, not just the special case discussed in the article; that unique bra is the bra that corresponds to the ket $\ket{v}$. To put it another way, bras are linear maps from kets to scalars; the bra $\bra{v}$ is the unique bra that maps the ket $\ket{v}$ to the scalar that is its squared norm.

 

[Kashmir]

The statement the article makes can be generalized from the column vector case. Just note that the bra $\bra{v}$ corresponding to the ket $\ket{v}$ always satisfies $\langle v \vert v \rangle = |v|^2$, i.e., the inner product of the two is the squared norm of $\ket{v}$. There will be one unique bra that has that property in any Hilbert space, not just the special case discussed in the article; that unique bra is the bra that corresponds to the ket $\ket{v}$. To put it another way, bras are linear maps from kets to scalars; the bra $\bra{v}$ is the unique bra that maps the ket $\ket{v}$ to the scalar that is its squared norm.

If you don't mind Can I ask how prove it someway? Could you please give me at least a hint. Thank you :)

 

[PeterDonis]

Can I ask how prove it someway?

Bras are defined by the scalars that they map kets to. It's impossible to have multiple bras that all map the same ket to the same scalar. So there can only be one bra that maps a given ket to any scalar, including the scalar that is that ket's squared norm.

 

[PeroK]

It's said that every ket has a unique bra. For any vector $|v> ∈ V$ there is a unique bra $<v| ∈ V*$.

I'm not sure what that means. What is unique? Can anyone please help me understand.
Thank you

Uniqueness is a mathematical property whereby two things with a given property (or set of properties) must be the same. In this case: $$\text{If} \ \forall u \in V: \langle v_1| u \rangle = \langle v_2| u \rangle, \ \text{then} \ v_1 = v_2$$That means that two different kets cannot generate the same bra.

 

[PeroK]

If you don't mind Can I ask how prove it someway? Could you please give me at least a hint. Thank you :)

Using the above and the properties of the inner product we have:$$\langle v_1| u \rangle = \langle v_2| u \rangle \ \Rightarrow \ \langle v_1 - v_2|u \rangle = 0$$Now, choosing $u = v_1 - v_2$:$$\langle v_1 - v_2|v_1 - v_2 \rangle = 0 \ \Rightarrow \ v_1 - v_2 = 0 \ \Rightarrow \ v_1 = v_2$$This proves that bras are unique.

 

[vanhees71]

Bras are defined by the scalars that they map kets to. It's impossible to have multiple bras that all map the same ket to the same scalar. So there can only be one bra that maps a given ket to any scalar, including the scalar that is that ket's squared norm.

You can prove it directly from the Hilbert-space axioms. Technically one must be a bit careful, because everything I say refers to bound linear forms, i.e., those that are continuous in the sense of the norm of the Hilbert space.

So let $L: H \rightarrow \mathbb{C}$ be such a linear form. Further the separable Hilbert space (there is only one up to isomorphism) has a complete orthonormal set $|u_n \rangle$ ($n \in \mathbb{N}$), $\langle u_j|u_k \rangle=\delta_{jk}$, i.e., for each vector $|\psi \rangle$ there's a unique series $\psi_n$ such that
$$|\psi \rangle=\sum_{n=1}^{\infty} u_n |\psi_n \rangle$$
and
$$\psi_n =\langle u_n|\psi \rangle.$$
Now one can define another series $L_n$ via
$$L_n^*=L(|u_n \rangle).$$
Since $L$ is continuous for all $|\psi \rangle$ you then have
$$L(|\psi \rangle)=L(\sum_{n=1}^{\infty} \psi_n |u_n \rangle=\sum_{n=1}^{\infty} \psi_n L(u_n \rangle)=\sum_{n=1}^{\infty} \psi_n L_n^*.$$
Now we define
$$|L \rangle=\sum_{n=1}^{\infty} L_n |u_n \rangle.$$
Then we have
$$\langle u_n|L \rangle=\langle L|u_n \rangle^*=L_n.$$
From this we get, using the completeness of the $|u_n \rangle$,
$$\langle L|\psi \rangle=\sum_{n=1}^{\infty} \langle L|u_n \rangle \langle u_n \psi \rangle=\sum_n L_n^* \psi_n=L(|\psi \rangle).$$
This proves the existence of a vector $|L \rangle$ such that $\langle L|\psi \rangle=L(|\psi \rangle)$.

For the uniqueness assume that also $|L' \rangle$ has this property, but then for all vectors $|\psi \rangle$
$$\langle L' |\psi \rangle-\langle L|\psi \rangle=L(|\psi \rangle)-L(|\psi \rangle)=0.$$
Thus $|L'-L \rangle$ is a vector for which $\langle L'-L|\psi \rangle=0$ for all $\psi$ and thus also for $|\psi \rangle=|L \rangle-|L'\rangle$. So we find
$$\langle L-L'|L-L' \rangle=\|L-L'\|^2=0 \; \Rightarrow \; |L \rangle=|L' \rangle,$$
i.e., $|L \rangle$ is the unique vector with the property that $L(|\psi \rangle)=\langle L|\psi \rangle$ for all $|\psi \rangle \in H$.

 

[PeroK]

So let $L: H \rightarrow \mathbb{C}$ be such a linear form. Further the separable Hilbert space (there is only one up to isomorphism) has a complete orthonormal set $|u_n \rangle$ ($n \in \mathbb{N}$), $\langle u_j|u_k \rangle=\delta_{jk}$, i.e., for each vector $|\psi \rangle$ there's a unique series $\psi_n$ such that
$$|\psi \rangle=\sum_{n=1}^{\infty} u_n |\psi_n \rangle$$
and
$$\psi_n =\langle u_n|\psi \rangle.$$
Now one can define another series $L_n$ via
$$L_n^*=L(|u_n \rangle).$$
Since $L$ is continuous for all $|\psi \rangle$ you then have
$$L(|\psi \rangle)=L(\sum_{n=1}^{\infty} \psi_n |u_n \rangle=\sum_{n=1}^{\infty} \psi_n L(u_n \rangle)=\sum_{n=1}^{\infty} \psi_n L_n^*.$$
Now we define
$$|L \rangle=\sum_{n=1}^{\infty} L_n |u_n \rangle.$$
Then we have
$$\langle u_n|L \rangle=\langle L|u_n \rangle^*=L_n.$$
From this we get, using the completeness of the $|u_n \rangle$,
$$\langle L|\psi \rangle=\sum_{n=1}^{\infty} \langle L|u_n \rangle \langle u_n \psi \rangle=\sum_n L_n^* \psi_n=L(|\psi \rangle).$$
This proves the existence of a vector $|L \rangle$ such that $\langle L|\psi \rangle=L(|\psi \rangle)$.

... which shows that the set of kets is the dual space of $H$. I.e. the only bounded linear functionals (for a separable Hilbert space) are the kets.

 

[vanhees71]

The point is that if $H^*$ is the "topological dual" (i.e., the set of the continuous/bounded linear functionals), then (and only then) you have $H=H^*$.

What you also need are unbound functionals like "position eigenvectors", which however live in the dual space of the domain of the position operator as a self-adjoint operator, and this domain $D$ is a proper dense subspace of $H$, i.e., $D^*$ is larger than $H$. In the bra-ket notation the bras can denote such "generalized eigenvectors, normalizable only to a $\delta$-distribution, i.e., $\langle x|x' \rangle=\delta(x-x')$.

 

[Kashmir]

Using the above and the properties of the inner product we have:$$\langle v_1| u \rangle = \langle v_2| u \rangle \ \Rightarrow \ \langle v_1 - v_2|u \rangle = 0$$Now, choosing $u = v_1 - v_2$:$$\langle v_1 - v_2|v_1 - v_2 \rangle = 0 \ \Rightarrow \ v_1 - v_2 = 0 \ \Rightarrow \ v_1 = v_2$$This proves that bras are unique.

Which property was used here $$\langle v_1| u \rangle = \langle v_2| u \rangle \ \Rightarrow \ \langle v_1 - v_2|u \rangle = 0$$

 

[PeroK]

Which property was used here $$\langle v_1| u \rangle = \langle v_2| u \rangle \ \Rightarrow \ \langle v_1 - v_2|u \rangle = 0$$

Linearity. For a complex inner product and using the physics convention of conjugate linearity in the first argument we have: $$\langle u_1 + u_2| v_1 + v_2 \rangle = \langle u_1| v_1 \rangle + \langle u_1| v_2 \rangle + \langle u_2| v_1 \rangle + \langle u_2| v_2 \rangle $$ $$\langle \alpha u| \beta v \rangle = \alpha^*\beta \langle u| v \rangle$$ $$\langle u| v \rangle = \langle v| u \rangle^*$$

 

[Kashmir]

Linearity. For a complex inner product and using the physics convention of conjugate linearity in the first argument we have: $$\langle u_1 + u_2| v_1 + v_2 \rangle = \langle u_1| v_1 \rangle + \langle u_1| v_2 \rangle + \langle u_2| v_1 \rangle + \langle u_2| v_2 \rangle $$ $$\langle \alpha u| \beta v \rangle = \alpha^*\beta \langle u| v \rangle$$ $$\langle u| v \rangle = \langle v| u \rangle^*$$

Thankyou so much. I found a book by ballentine, it's also clear in his approach. Thank you very much Sir.