# Why should multiplication of a ket vector by a complex number change only its "direction"?

Dirac argues on page 17 of his book, The Principles of Quantum Mechanics, that multiplication of a ket by a complex number shouldn't change the state this ket represents. But then concludes:

Thus a state is represented by the direction of a ket vector, and any length one may assign to the ket vector is irrelevant.

Bearing in mind that he has only stated that we need to make a generalization of vectors in a space of an infinite number of dimensions, how is he able to conclude this?

• Length is irrelevant because the theory gives relative probabilities. Moreover, when dealing with operators and vector spaces, it is important to remember that if $\textbf{v}$ is an eigenvector, so is $a\textbf{v}$ for any $a \in F$ (for the field under which the vector space is defined). Regardless of what a ket represents, it is impossible to distinguish between eigenvectors of the different lengths. Not sure if I've answered why a ket's length doesn't matter a priori, but hopefully this serves as some motivation. Commented Aug 19, 2014 at 22:50
• @Nathan I'm looking for an answer in the spirit of Dirac talking vaguely about vectors in an infinite dimensional space up to this point. I guess this will become clearer later on in the book, but Dirac seems to suggest this statement is obvious for physical/mathematical reasons. Commented Aug 19, 2014 at 23:03
• I haven't read Dirac's book, so I can't answer exactly how he concludes that a ket's size is irrelevant. If I were to guess, however, I would continue to support my eigenvector hypothesis. In QM, observable quantities are represented by operators in Hilbert space. Their eigenstates (and linear combinations thereof) are the only allowed states that the state vector may assume (since they span the space). I would assume that this is the reason why a ket's length does not matter (since it is a linear combination of these eigenvectors). Commented Aug 19, 2014 at 23:08
• I'm a bit confused what is given here - do we know anything else than states are vectors in a Hilbert space yet? Your introductory sentence seems to imply that we also know that scalar multiplication won't change the state, but then, the conclusion that length is irrelevant follows immediately. Commented Aug 19, 2014 at 23:14
• @ACuriousMind A vector multiplied by a negative real number changes its direction, yes? How are we to interpret multiplying it by a complex number? Commented Aug 19, 2014 at 23:24

Obviously I cannot know what Dirac is thinking, but I think it is just that his direction does not correspond exactly to your direction.

We "know", just as Dirac does, that quantum states are members of some Hilbert space $\mathcal{H}$. We also know that scalar multiplication should not change the state, so $\lvert \psi \rangle$ and $\lambda \lvert \psi \rangle$ should represent the same state for $\lambda \in \mathbb{C} / \{0\}$.

The ray in the Hilbert space (or the direction) given by a single vector $\lvert \psi \rangle$ of a state is $R_\psi = \{\lambda \lvert \psi \rangle \vert \lambda \in \mathbb{C} / \{0\}\}$. Dirac wants us, since all $\lvert \phi \rangle \in R_\psi$ represent the same state, to declare the equivalence relation $\lvert \psi \rangle \sim \lvert \phi \rangle \text{ iff } \lvert\phi\rangle \in R_\psi$ on $\mathcal{H}$ and obtain the projective Hilbert space as $\mathbb{P}\mathcal{H} := \mathcal{H} / \sim$.

It should be noted that, in "intuitive" terms, the complex ray $R_\psi$ is like a punctured plane created by taking the naive real ray $r_\psi := \{c\lvert \psi \rangle\ \vert c \in \mathbb{R^+}\}$ and rotating it around the origin as $R_\psi = \bigcup_{k \in [0,2\pi)} (\mathrm{e}^{\mathrm{i}k} r_\psi)$

Stated in a simpler way, kets are the generalisation of vectors to complex and potentially continuous/infinite dimension space (Hilbert spaces). Yet you can keep in mind the image of a vector to begin with.

When you multiply a vector by a nonzero real number, its direction does not change. The same is valid for vectors from a Hilbert space. If you multiply them by a nonzero complex number, their "direction" remains the same.

Regarding the change in physical information, it is unchanged too for the following reason. In state $|\psi>$, the expectation value of quantity $A$ is $\frac{<\psi|A|\psi>}{<\psi|\psi>}$. If you multiply $|\psi>$ by any nonzero complex number, you can check that the expectation value does not change. It's the same physics.

A physical state is defined by the density matrix, so, if you define the density matrix by :

$\rho = \frac{|\psi\rangle \langle \psi|}{\langle \psi| \psi\rangle}$

it is easy to see that any multiplication by a complex number does not change the density matrix, so does not change the physical state.

It is probably what Dirac means, while I have not the book.