# Bra-ket notation in a 3 dimensional system

Recently we started to learn quantum mechanics in a three-dimensional system. However, I am very confused from the beginning. Firstly, professor told us that for a position vector $$\pmb{x}={}^t(x,y,z)$$ the ket vector in a 3-dimensional system can be written as $$|\pmb{x}\rangle = |x\rangle |y\rangle |z\rangle$$. What even is this? I know that $$\langle a|b\rangle$$ represents an inner product but I don't even know what $$|x\rangle |y\rangle |z\rangle$$ means. Also we were told that $$\hat{\pmb{x}}|\pmb{x}\rangle = \pmb{x}|\pmb{x}\rangle$$ holds. Is this something that is provable from $$\hat{x}|x\rangle=x|x\rangle$$, $$\hat{y}|x\rangle=y|y\rangle$$,$$\hat{z}|z\rangle=z|z\rangle$$ and $$|\pmb{x}\rangle = |x\rangle |y\rangle |z\rangle$$ or just a definition?

• Start looking up Direct Product. Oct 2, 2019 at 12:58
• How could it be direct product?Is the order match? Oct 2, 2019 at 15:46
• Have you encountered bra-ket notation in 1D or have you been doing everything with wavefunctions? I.e. does the equation $\psi(x)=\langle x | \psi \rangle$ make sense to you? Oct 2, 2019 at 15:53
• Yes, I learned bra-ket notation in 1d. I know that $\langle x |\psi \rangle$ is a wave function. Oct 2, 2019 at 23:45
• @jacob1729 Above comment is for you. Mao, make sure to tag users if you want them to be notified of your comment. Oct 3, 2019 at 10:00

(NB: Some of this isn't completely rigorous, particularly regarding infinite dimensional vector spaces, but hopefully it's not misleading.)

Broadly speaking, the state of a quantum system can be described as being a superposition of the possible classical configurations. So a classical bit is a system with two possible states $$\{A,B\}$$ and the quantum analogue is any vector in the vector space spanned by two vectors which we conventionally label $$|A\rangle,|B\rangle$$.

The classical configurations of a particle are its coordinates $$(x,y,z)$$. Following the above we assign to each possible configuration a state vector $$|(x,y,z)\rangle$$. There are some subtleties to this but this answer will just ignore them and pretend this is really a basis. This particular notation is a bit clumsy so I'll write it as $$|x,y,z\rangle$$ from now on. From this perspective that $$|x,y,z\rangle$$ is the state where a particle will definitely be found at position $$(x,y,z)$$, and an arbitrary state is given as a linear superposition of all of these `basis' states.

Let's think back to the classical description of a particle. The configuration is completely given by three real numbers which can be varied independently. This means that we can consider the system not as one particle moving in three dimensions, but a collection of three particles each moving in one dimensions. (This logic may be familiar from Lagrangian mechanics where a system of $$N$$ particles is often treated as one particle moving in $$3N$$ dimensions.) So instead of describing the configuration $$(x,y,z)$$ by the state $$|x,y,z\rangle$$ we could instead use the description $$|\text{Particle 1 at }x,\text{Particle 2 at }y,\text{Particle 1 at }z\rangle$$. Because these properties have nothing to do with each other it can be convenient to separate them out:

$$|x,y,z\rangle \sim |\text{Particle 1 at }x\rangle|\text{Particle 2 at }y\rangle|\text{Particle 3 at }z\rangle$$

Or even more briefly, $$|x,y,z\rangle \sim |x\rangle |y\rangle |z\rangle$$. Formally this listing of kets one after another is representing the tensor product but physically it just means that you're considering the system as composed of three subsystems and listing their states. Each factor in this expression can be considered to be a vector in its own Hilbert space equivalent to that of a 1 dimensional particle. As such we can introduce operators on this vector space by combining the 1D ones. That is, consider three operators $$A,B,C$$ acting on 1D states, then we can form the combination:

$$(A\otimes B\otimes C)|x\rangle|y\rangle|z\rangle = (A|x\rangle)(B|y\rangle)(C|z\rangle)$$

where $$\otimes$$ symbol represents a tensor product (really we should be writing $$|x\rangle \otimes |y\rangle$$ etc, but that's very longwinded).

As such there is an '$$x$$-component operator' given by $$\hat{x}=x\otimes \mathbb{1} \otimes \mathbb{1}$$, a '$$y$$-component operator' given by $$\hat{y}=\mathbb{1} \otimes y \otimes \mathbb{1}$$ and so on.

Now, it remains to explain what the equation:

$$\hat{\pmb{x}}|\pmb{x}\rangle = \pmb{x}|\pmb{x}\rangle$$

is supposed to mean. After all, the vector space is a complex vector space, and it's not clear what it means to multiply a state vector by a Euclidean vector. What it means is that we consider an ordered triple of ordinary operators $$\hat{\pmb{x}}=(\hat{x},\hat{y},\hat{z})$$ then when we act on the vector $$|\pmb{x}\rangle=|x,y,z\rangle$$ we get an ordered triple of vectors as a result. In this case its obvious the result is:

$$\hat{\pmb{x}}|\pmb{x}\rangle=(x|\pmb{x}\rangle,y|\pmb{x}\rangle,z|\pmb{x}\rangle)=(x,y,z)|\pmb{x}\rangle$$

the last of which we interpret as $$\pmb{x}|\pmb{x}\rangle$$.