Quantum states are unit vectors... with respect to which norm?

Question

The most general definition of a quantum state I found is (rephrasing the definition from Wikipedia)

Quantum states are represented by a ray in a finite- or infinite-dimensional Hilbert space over the complex numbers.

Moreover, we know that in order to have a useful representation we need to ensure that the vector representing the quantum state is a unit vector.

But in the definition above, they don't precise the norm (or the scalar product) associated with the Hilbert space considered. At first glance I though that the norm was not really important, but I realised yesterday that the norm was everywhere chosen to be the Euclidian norm (2-norm). Even the bra-ket notation seems to be made specifically for the euclidian norm.

My question: Why is the Euclidian norm used everywhere? Why not using an other norm? Does the Euclidian norm has useful properties that can be used in quantum mechanics that others don't?

Answer 1

Born's rule states that $|\psi(x)|^2 = P(x)$ which is the probability of finding the quantum system in the state $|x\rangle$ after a measurement. We need the sum (or integral!) over all $x$ to be 1:

\begin{align} \sum_x P_x &= \sum_x |\psi_x|^2 = 1,\\ \int P(x)dx &= \int |\psi(x)|^2 dx= 1. \end{align}

Neither of these are valid norms because they are not homogenous. You can make them homogenous simply by doing the square root:

\begin{align} \sqrt{\sum_x |\psi_x|^2} = 1,\\ \sqrt{\int |\psi(x)|^2dx} = 1. \end{align}

and you may recognize this as the Euclidean norm and a generalization of the Euclidean norm to a non-discrete domain. We could also use a different norm:

\begin{align} \sqrt{\sum_x \psi_x A \psi_x^*} = 1,\\ \sqrt{\int \psi(x)A\psi^*(x)} = 1, \end{align}

for some positive definite matrix/function A.

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However a $p$-norm with $p>2$ would not be as useful because for example:

\begin{align} \sqrt[5]{\sum_x |\psi_x|^5} \\ \end{align}

does not have to be 1.

In this way the Euclidean norm is special because 2 is the power in Born's rule, which is one of the postulates of quantum mechanics.

Answer 2

Some terminology seems a little bit jumbled here. Quantum states are represented (within a finite dimensional Hilbert space) by complex vectors of length 1, where length is measured by the Euclidean norm. They are not unitary, because unitary is a classification of a matrix, not a vector.

Quantum states are changed/evolved according to some matrix. Given that quantum states have length 1, it turns out to be necessary and sufficient that the maps of pure states to pure states are described by unitary matrices. These are the only matrices that preserve the (Euclidean) norm.

It is certainly a valid question "could we use a different ($p$) norm for our quantum states?" If you then classify the operations that map normalised states to normalised states, they are incredibly limited. If $p\neq 2$, the only valid operations are permutation matrices (with different phases on each element). Physics would be a whole lot more boring.

A good way to get a feel for this is to try drawing a 2D set of axes. Draw on it the shapes corresponding to the set of points of length 1 under different $p$-norms. $p=2$ gives you the circle, $p=1$ gives you a diamond, and $p\rightarrow\infty$ gives a square. What operations can you do that map the shape onto itself? For the circle, it's any rotation. For anything else, it's just rotations by multiples of $\pi/2$. The following comes from Wikipedia:

If you want more details, you might want to look here.

Answer 3

More mathematically, because $\mathbb{R}^n$ with an $L^p$ norm is a Hilbert space only for $p=2$.

Answer 4

An elegant argument can be derived by asking which theories can we build which are described by vectors $\vec v = (v_1,\dots,v_N)$, where the allowed transformations are linear maps $\vec v\to L\vec v$, probabilities are given by some norm, and probabilities must be preserved by those maps.

It turns out that there are basically only three options:

1. Deterministic theories. Then we don't need those vectors, since we are always in one specific state, i.e. the vectors are $(0,1,0,0,0)$ and the like, and the $L$'s are only permutations.

2. Classical probabilistic theories. Here, we use the $1$-norm and stochastic maps. The $v_i$ are probabilities.

3. Quantum mechanics. Here, we use the $2$-norm and unitary transformations. The $v_i$ are amplitudes.

These are the only possibilities. For other norms no interesting transformations exist.

If you want a more detailed and nice explanation of this, Scott Aaronson's "Quantum Computing since Democritus" has a Lecture on this, as well as a paper.

Answer 5

The other answers addressed why $p=2$ in terms of which $L^p$ space to use, but not the weighting.

You could put in a Hermitian positive definite matrix $M_{ij}$ so that that the inner product is $\sum x_i^* M_{ij} y_j$. But that doesn't gain you much. This is because you might as well change variables. For ease, consider the case when $M$ is diagonal. with the diagonal case that would be interpreting $M_{ii} \mid x_i \mid^2$ as a probability instead of $\mid x_i \mid^2$. $M_{ii}>0$ so why not just change variables to $\tilde{x}_i = \sqrt{M_{ii}} x_i$. You can think of this as $L^2$ functions on the space of $n$ points where each point is weighted by $M_{ii}$.

For the continuous 1 variable case, yes you could use $L^2 (\mathbb{R} , w(x) dx)$ as well. $w(x)$ just reweights the lengths. That's still a perfectly good Hilbert space. But the problem is that translation $x \to x+a$ was supposed to be a symmetry and $w(x)$ breaks that. So might as well not use $w(x)$. For some purposes, that symmetry is not present, so you do have a $w(x) \neq 1$.

In some cases it is useful not to move to standard form. It shuffles around how you do some calculations. For example, if you're doing some numerics, then you can reduce your errors by this sort of reshuffling to avoid really small or large numbers that your machine finds difficult.

A tricky thing is to make sure you keep track of when you changed your variables and when you didn't. You don't want to get confused between changing to the standard inner product doing some unitary and changing variables back vs trying to do that in one step. You are likely to drop factors of $\sqrt{M_{ii}}$ etc by mistake, so be careful.

Answer 6

The Euclidean norm on an $n$-dimensional space, as defined here, is not the only norm used for quantum states.

A quantum state doesn't have to be defined on an n-dimensional Hilbert space, for example the quantum states for a 1D harmonic oscillator are functions $\psi_i(x)$ whose ortho-normality is defined by:

$$ \int \psi_i(x)\psi_j^*(x)dx. $$

If $i=j$ we get:

$$ \int |\psi(x)|^2dx = \int P(x)dx = 1, $$

because the total probability must be 1.
If $i\ne j$, we get 0, meaning that the functions are orthogonal.

The Euclidean norm, as defined in the link I gave, is more for quantum states on discrete variables where $n$ is some countable number. In the above case, $n$ (which is the number of possible values that $x$ can be) is uncountable, so the norm doesn't fit into the definition given for a Euclidean norm on an $n$-dimensional pace.

We could also apply a square root operator to the above norm, and still we'd have the required property that $\int P(x)dx=1$, and the Euclidean norm can then be thought of as a special case of this norm though, for the case where $x$ can only be chosen from some countable number of values. The reason why we use the above norm in quantum mechanics is because it guarantees that the probability function $P(x)$ integrates to 1, which is a mathematical law based on the definition of probability. If you had some other norm which can guarantee that all laws of probability theory are satisfied, you would be able to use that norm too.