# Interpretation of the vector space in notation bra ket

I have read all of the question related to "bra-ket" but no one seems to take the same perspective as I am going to try to give. I know it might be a rather simple and short question, but I need to understand this to continue studying quantum mechanics (I'm new to this subject). Here's how I understand it:

In my head, or at least how I understood it, it was that previously in classical mechanics, each particle had one physical state with it, but now, in quantum mechanics, you have a lot of quantum states in which the particle can be, each with it's own probability of being there, represented with $$|\psi|^2$$. Up until here, I've understood $$\psi(\vec x, t)$$ as a simple wave function whose square modulus was the density of probability, but now I've been introduced to the $$|\psi \rangle$$ notation, and I'm lacking how to understand what it means. I have understood it so far as a vector? Which components are the possible quantum states of a system and that's why it can be "treated" as a vector? Or what is the physical meaning of that $$|\psi \rangle=\begin{pmatrix}\psi_1\\ \psi_2\end{pmatrix}$$?

You can think of $$\psi(\vec x, t)$$ as being a vector in an infinite dimensional complex vector space (a Hilbert space), where $$|\psi(\vec x, t)|^2$$ is the probability that the particle will be found at location $$\vec x$$ at time $$t$$. You can represent this state as $$|\psi(t)\rangle$$ where $$|\psi(t)\rangle$$ has an infinite number of components, each component being the value of $$\psi(\vec x, t)$$ at a particular location $$\vec x$$, and the locations $$\vec x$$ are the eigenstates of a position operator $$\Psi$$.

Because it is difficult to grasp infinite dimensional vector spaces, bras and kets are usually introduced via finite dimensional examples. The simplest non-trivial vector space would have just one dimension, and so only one eignenstate, but the states in such a system would not be very interesting. The simplest interesting vector space has two dimensions, so we introduce a system that has just two eigenstates; up and down spin is the standard example. If we call these eigenstates $$|1\rangle$$ and $$|2\rangle$$ then the state $$\psi_1 |1\rangle + \psi_2 |2\rangle$$ is represented by the two-component vector $$|\psi \rangle=\begin{pmatrix}\psi_1\\ \psi_2\end{pmatrix}$$. The values $$\psi_1$$ and $$\psi_2$$ depend on our choice of eigenstates, which is equivalent to choosing a basis for the vector space. The notation $$|\psi\rangle$$ emphasises that $$|\psi\rangle$$ is a specific physical state of the system, whereas the values $$\psi_1$$ and $$\psi_2$$ will change if we choose a different basis.

• so the components of $|\psi \rangle$ are just the coefficients of the eigenstates? Isn't it?
– Ivy
Commented Feb 14 at 22:16
• @Ivy Yes, the components of $|\psi\rangle$ are the eigenstates themselves, viz. $\psi_1$ and $\psi_2$. The wave functions are the Fourier coefficients of whatever basis is used to span the Hilbert space. In Gandalf's example, $|\psi\rangle=\langle 1|\psi\rangle|1\rangle +\langle 2|\psi\rangle|2\rangle$, where $\psi_1=\langle 1|\psi\rangle$ and $\psi_2=\langle 2|\psi\rangle$, respectively. Since Gandalf has chosen to represent $|\psi\rangle$ with matrices, then $\langle 1|=(1,0)$ and $\langle 2|=(0,1)$. Remeber, $\langle\cdot|\cdot\rangle$ is an inner-product, and is just a complex number. Commented Feb 14 at 22:43
• @Ivy It is not a bad idea to wait 48 hours before accepting a post, this encourages others to contribute, you never know what insight you may gain from them. Just a thought for the future. BTW welcome to Stack Exchange, if you haven't already heard it from someone! Commented Feb 14 at 22:59

The bra, ket notation for quantum states is a way of expressing the state in the abstract. The quantum state ket $$|\alpha\rangle$$ is a vector in Hilbert space. Here the concept of vector is to be understood in the abstract sense, i.e. in terms of a set of axioms for a vector space as opposed to a directed line segment as the notion can be understood in the classical sense. The key to understanding this sudden move from concrete wave functions to abstract vectors is representation.

The representation one chooses determines how one relates an object defined according to its algebraic properties to some kind of tangible mathematical entity useful for computation. For example, to understand quantum mechanics in a purely formalistic sense, all one needs are Hermitian operators defined on vectors in Hilbert space, the operators can be written with the "hat" notation while the vectors with the bra, ket notation. In this formalistic framework an observable is equivalent to a Hermitian operator with a set of eigenvectors that are assumed to span the entire Hilbert space, thus any state $$|\phi\rangle$$ can be written as: $$|\phi\rangle=\Sigma_{\alpha} |\alpha\rangle\langle\alpha |\phi\rangle .$$ Where we have used the completeness, $$\Sigma_{\alpha}|\alpha\rangle\langle\alpha |=\hat I$$, of the set of eigenvectors of the observable $$\hat A$$ such that: $$\hat A|\alpha\rangle=\alpha |\alpha\rangle.$$Where the $$\alpha$$ are real eigenvalues. Furthermore, in the formalism, one may write the operator $$\hat A$$ by a similar expansion: $$\hat A= \Sigma_{\alpha} \Sigma_{\alpha^\prime}|\alpha\rangle\langle\alpha |\hat A|\alpha^\prime\rangle\langle\alpha^\prime |.$$ The formalism makes it readily apparent that one can represent an operator on Hilbert space as a matrix with elements given by $$\langle\alpha |\hat A|\alpha^\prime\rangle$$, since applying the above expression to $$|\phi\rangle$$, one has: $$\hat A|\phi\rangle=\Sigma_{\alpha^{\prime\prime}}\Sigma_{\alpha^{\prime}}|\alpha^{\prime\prime}\rangle\langle\alpha^{\prime\prime} |\hat A|\alpha^\prime\rangle\langle\alpha^\prime |\big (\Sigma_{\alpha} |\alpha\rangle\langle\alpha |\phi\rangle\big )=\Sigma_{\alpha^{\prime\prime}}\Sigma_{\alpha}|\alpha^{\prime\prime}\rangle\langle\alpha^{\prime\prime} |\hat A |\alpha\rangle\langle\alpha |\phi\rangle,$$ where we have used the orthogonality relation $$\langle\alpha^\prime |\alpha\rangle=\delta_{\alpha^\prime\alpha}$$. Now it is rather clear that this expression gives the result of the action of $$\hat A$$ on $$|\phi\rangle$$ as a ket expressed in the $${|\alpha\rangle}$$ basis, with components: $$\Sigma_{\alpha}\langle\alpha^{\prime\prime} |\hat A |\alpha\rangle\langle\alpha |\phi\rangle.$$ Ostensibly, this quantity corresponds to the rule for multiplying a column vector by a matrix. Thus, it is quite natural to say that an operator on Hilbert space can be represented by matrices and that states may be represented by column matrices. The linear space of Hermitian operators and that of abstract states in Hilbert space are in finite dimensional cases isomorphic to those of matrices. However, matrices do not exhaust the representations of quantum entities. For some e.g., two-state systems, matrices are perhaps the most natural setting for computing, however, what about differential operators, like those encountered in the Hamiltonian operator of the Schrodinger equation?

In the case of differential operators, representation by matrices is not so natural, their eigenspaces are infinite-dimensional for one thing, and their spectrum may even be continuous like that of the free particle Hamiltonian. In this case it is usual to say that a state $$\phi\rangle$$, can be represented by a "wave function" $$\phi(\xi)$$.

So, suppose we have a Hamiltonian $$\hat H(\hat x,\hat p)$$ and we want to express it and a state $$|\psi\rangle$$, in the position basis $$|x\rangle$$. Likely, you already know the answer; in the position basis, the $$\hat x$$ operator is simply the multiply by "$$x$$" operator and the momentum operator is identified with the Hermitian differential operator: $$-i\hbar d/dx$$. So that we can represent the Hamiltonian operator as: $$\hat H=-{\hbar^2\over 2m} {d^2\over dx^2}+V(x).$$ What about a state $$|\psi\rangle$$? Well, we have that $$|\psi\rangle=\int_{-\infty}^{+\infty}|x\rangle\langle x|\psi\rangle\; dx,$$ hence the state is given in terms of the gernelized Fourier expansion where the coefficients are known colloquially as a "wave function":$$\psi(x)=\langle x|\psi\rangle.$$ Suppose that we wish to calculate the modulus, $$\langle\psi |\psi\rangle$$? We get that: $$\langle\psi |\psi\rangle=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\langle\psi|x^\prime\rangle\langle x^\prime |x\rangle\langle x|\psi\rangle\; dx\; dx^\prime=\langle\psi |x\rangle\langle x|\psi\rangle.$$ The integral simplifies due to the fact that $$\langle x^\prime |x\rangle=\delta(x-x^\prime)$$. Thus, although the formalism of linear operators defined on Hilbert spaces allows one to handle any calculation required of quantum mechanics, it is often useful to choose a concrete representation in which to work which is suited to the problem at hand. We need not always work in the position basis, for example, the Schrodinger equation may be expanded in terms of the momentum basis as well, or in to any other complete eigenbasis, so that a general state ket $$|\phi\rangle$$, will be represented as a wave function computed from carrying out the eigenfunction expansion so that: $$\phi(\xi)=\langle \xi|\phi\rangle.$$