# Wave function as a ket vector in a Hilbert space

There's something I don't understand: I've learned that quantum wave functions can be described as a "ket vector" in an abstract vector space called Hilbert space. The position wave function, for example, used to express the probability of finding the particle at a point, can be described as a vector in an infinite dimensional Hilbert space. But we also have the wave function used to describe spin ("spinor"), and this wave function exists in a 2-dimensional Hilbert space. So my question is, what is the relationship between these two different wave functions? I mean, both are portrayed as a representation of the state of a particle, but they are clearly not the same thing. I've also heard that the wave function contains everything that there is to know about the particle, but I'm like, "which wave function?" I would be really thankful if someone could help!

I've learned that quantum wave functions can be described as a "ket vector" in an abstract vector space called Hilbert space. The position wave function, for example, used to express the probability of finding the particle at a point, can be described as a vector in an infinite dimensional Hilbert space.

It seems you are talking about the position wave function $$\psi(\vec{r}). \tag{1}$$ Yes, this function is a member of an $$\infty$$-dimensional Hilbert space because there are infinitely many positions $$\vec{r}$$. However, this kind of wave function can only represent a spinless particle, but it cannot describe a particle with spin (e.g. an electron).

But we also have the wave function used to describe spin ("spinor"), and this wave function exists in a 2-dimensional Hilbert space.

Just to clarify: a spinor is a "vector" consisting of 2 complex numbers (without any dependency on position $$\vec{r}$$), like $$\begin{pmatrix}\psi_+ \\ \psi_-\end{pmatrix}. \tag{2}$$ So this spinor is a member of a 2-dimensional Hilbert space. This 2-component spinor can be visualized as a flag pole with a flag.

When rotating a spinor its 2 components transform in a well-defined way. For more see An introduction to spinors (especially pages 2 to 5) by Andrew Steane.

So my question is, what is the relationship between these two different wave functions?

The actual wave function of an electron (or any other spin $$\frac{1}{2}$$ particle for that matter) is the tensor product of (1) and (2) above. $$\begin{pmatrix}\psi_+(\vec{r}) \\ \psi_-(\vec{r})\end{pmatrix} \tag{3}$$ So this function is a member of an $$\infty\times 2$$-dimensional Hilbert space. $$\psi_+(\vec{r})$$ is the probability-amplitude of the particle being at position $$\vec{r}$$ and having spin-up. Likewise $$\psi_-(\vec{r})$$ is the probability-amplitude of the particle being at position $$\vec{r}$$ and having spin-down.

I mean, both are portrayed as a representation of the state of a particle, but they are clearly not the same thing. I've also heard that the wave function contains everything that there is to know about the particle, but I'm like, "which wave function?"

It is the spinor wave function given in (3) which contains everything knowable about the particle.

• Wow, that is truly amazing. So the full state of a particle is a ket vector in some Hilbert space, and this vector contains information about everything you can know about the particle, such as spin, position, momentum, energy etc.? So in that sense, the wave function tells us everything there is to know about the particle? Jun 18, 2020 at 13:42
• @FelisSuper Yes, you are correct. Jun 18, 2020 at 14:54
• Small technical comment: it should be the tensor product, I think, not the cartesian one (which is called the direct sum, when dealing with vector/Hilbert spaces). Jun 18, 2020 at 19:52
• @coconut Thanks, it seems you are right. I fixed my answer. Jun 18, 2020 at 20:01
• @JamalS Agreed. I've linked an introductory-level article. Jun 22, 2020 at 10:02

So let's start without spin. You can extract the wavefunction from the 'ket vector' by taking the inner product with the $$|x\rangle$$ state. The $$|x\rangle$$ ket represents a state with definite position where a particle is localised entirely on $$x$$. This is not a physical state (you can't normalise it) but still a useful tool. The wavefunction is then extracted like $$\psi(x)=\langle x|\psi\rangle$$ This might seem weird when you've never seen it written this way but it makes a lot of things more clear. The $$|x\rangle$$ states form an orthonormal basis: $$\langle x|y\rangle=\delta(x-y)\\ \int dx|x\rangle\langle x|=\mathbb{1}$$ and to convince you this is right you can calculate the inner product of the wavefunction: \begin{align}\langle\psi|\psi\rangle&=\langle\psi|\left(\int dx|x\rangle\langle x|\right)|\psi\rangle\\ &=\int dx\langle\psi|x\rangle\langle x|\psi\rangle\\ &=\int dx\ \psi^*(x)\psi(x) \end{align} To extend this to spin we consider the state $$|x,\alpha\rangle$$. This is a state with position $$x$$ and spin $$\alpha$$. For a spin 1/2 particle $$\alpha$$ could be up and down: $$\alpha=\{\uparrow,\downarrow\}$$. For the wavefunction this means $$\psi_\alpha(x)=\langle x,\alpha|\psi\rangle$$ We can collect the $$\alpha$$ components in a column vector. For the spin 1/2 particle: $$\begin{pmatrix}\psi_\uparrow(x)\\ \psi_\downarrow(x)\end{pmatrix}$$ The spin states also form an orthonormal basis. So for completion the inner product becomes $$\langle\psi|\psi\rangle=\sum_\alpha\int dx\ \psi_\alpha^*(x)\psi_\alpha(x)$$

This was a little more than you asked for but I hope it's more clear this way.

• Yes, thank you so much for this answer, it is these kinds of answers (those that go a bit more in depth) that I like the most. It was also very clear, and I think I understood everything. Jun 19, 2020 at 8:09

Yes, the terminology is sometimes a little bit sloppy. The Hilbert space is actually the product of the infinite dimensional Hilbert space defined on $$\mathbb R^3$$ and two dimensional spinor space (or in relativistic qm, 4-dimensional space of Dirac spinors). My recommendation is to ignore terminology, and focus on mathematical structure. The wave function can be restricted to either space, and that is what people talk about. But when you say "the wave function contains everything that there is to know about the particle" this refers to the full wave function, not its restriction to either position or spin space,

• Thank you, I think I get it now. Jun 19, 2020 at 8:06