When you’re reading articles in quantum mechanics, oftentimes you’ll come across what’s known as the Bra-Ket notation. I know how complicated it may seem, but it’s quite simple.
To understand the this notation, you should be familiar with vectors, which we can write as a column matrix (the usual way). For simplicity sake, we’ll be looking at two dimensional vectors like \vec{a} = \begin{pmatrix} x \\ y \end{pmatrix}
Bra-Ket notation can be thought of as another way of expressing these vector quantities. The following is called the ket vector. |a\rangle = \vec{a} = \begin{pmatrix} x \\ y \end{pmatrix}
If we have ket vectors, we can of course have bra vectors. Bra is defined as the Hermitian adjoint of the ket – which in our context of real numbers correspond to transpose of ket. \bra{a} = (\ket{a})^T = \begin{pmatrix} x \\ y \end{pmatrix}^T = \begin{pmatrix} x&y \end{pmatrix}
It’s easy to think of ket vectors as column vectors and bra vectors as row vectors, although extremely simplified. Also remember that these vectors can be considered as matrices.
Doing maths with bra-ket notation
Multiplying by a constant
Just like any vectors or matrices, we can multiply a constant to both ket and bra. To compute this, we simply need to multiply each element with the constant being multiplied. \ket{a}=\begin{pmatrix} 3 \\ 2 \end{pmatrix} \\ 4\ket{a}=4\begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}
simple addition
Just like how we can add up column vectors, we can add ket vectors. If we let \vec{e_1} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} and \vec{e_2} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} , we can represent another vector, say \vec{m}=\begin{pmatrix} 4 \\ 2 \end{pmatrix}, with addition of \vec{e_1}and \vec{e_2}. Let’s try this with column vectors first. \vec{m}=\begin{pmatrix} 4 \\ 2 \end{pmatrix} = 4\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 2\begin{pmatrix} 0 \\ 1 \end{pmatrix} So we can represent them as \vec{m}=4\vec{e_1}+2\vec{e_2} Since each vectors can be represneted through ket vector, we can write the following as well. \ket{m}=4\ket{e_1}+2\ket{e_2}
inner product
You probably come across the following Bra-Ket notations the most.
\braket{a|a}
So what does this mean? Let’s represent both the bra and ket vectors in their component forms – as row and column vectors repectively.
\braket{a|a} = \begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
As you may know, vectors can be thought of as matrices. So, what the above expression tells us to do is to matrix multiply the two matrices \braket{a|a} = \begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = x^2+y^2 This is basically taking the inner product, which we can think of as taking the dot product, of the two vectors. (the dot product analogy is valid as we will be working with real vectors) \braket{a|a}=\begin{pmatrix} x \\ y \end{pmatrix} . \begin{pmatrix} x \\ y \end{pmatrix}
outer product
You may be asking, can we also have expressions like the following? \ket{a}\bra{a}
Like before, let’s represent the vectors into their component forms. \ket{a}\bra{a} =\begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} x&y \end{pmatrix}
We can, again, do the matrix multiplication. \ket{a}\bra{a} =\begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} x&y \end{pmatrix} = \begin{pmatrix} x^2&xy \\ xy&y^2 \end{pmatrix}
This time, we get a matrix, not a number (a scalar). This is called the outer product.
An easy way to remember this is by looking at the orientation of the triangular brakets. If they face inward, they represent inner product, while if they face outward, they represent outer product.
orthonormal vectors
Vectors that have magnitudes of one and are orthogonal to one another is said to be orthonormal vectors. As we should already know, if two vectors are orthogonal, their inner product (dot product) must be zero. If we were to write this with the Bra-Ket notation: let \, \vec{e_1} \, and \, \vec{e_2} \, be \, orthonormal. \\ then \, \braket{e_1|e_2} = 0 One interesting property of orthonormal vectors is that when we add up their outer products, it gives an identity matrix. \ket{e_1}\bra{e_1} + \ket{e_2}\bra{e_2} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1&0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0&1 \end{pmatrix} \\ \; \; =\begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix} + \begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} = I
Quantum mechanics (Just for fun!)
Funnily enough, in quantum mechanics, the inner product is represented with an integral. This is because wavefunctions (or any functions) can be thought of as vectors of infinite dimensions. So for wavefunctions \phi and \psi the inner product is as the following.
\langle \phi | \psi \rangle = \int_{-\infty}^{\infty} \phi^*(x)\,\psi(x)\,dx
you may realise that we are taking the conjugate of \phi. This is because the \phi wavefunction is in a Bra. If we let the original \phi = \ket{\phi}, we can find the bra function by taking the complex conjugate of the \phi.
\bra{\phi} = \ket{\phi}^* = \ket{\phi^*}
To take the inner product, we simply multiply the bra function with the ket function.
\phi^*(x)\,\psi(x)
But since we have infinitely many dimensions in these bra and ket, the inner product turns out of be an integral – as we need to add up all the terms. This is how we end up with \langle \phi | \psi \rangle = \int_{-\infty}^{\infty} \phi^*(x)\,\psi(x)\,dx
In quantum mechanics, we normalise wavefunctions by making “length square” of the wavefunction to equal to one. If we write this in classical notation:
\int_{-\infty}^{\infty} \phi^*(x)\,\phi(x)\,dx = 1
Obviously, writing this out every time we need to express normalisation is cumbersome. We can use the Bra-Ket notation instead to represent the inner product and the normalisation condition:
\langle \phi | \phi \rangle = 1
Thanks for reading my first blog/article. I tried to simplify everything to their core concepts – if you find any errors or have any feedbacks, please don’t hesitate to contact me.
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