Bra-Ket Notation

Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.

Updated by Alexander Fufaev on 03/20/2022

Table of contents

Wave function as a vector in Hilbert space Here you will learn how the wave function from quantum mechanics can be understood as an infinite-dimensional vector living in Hilbert space.
Bra and Ket vectors Here you will learn how bra and ket vectors are defined and how they are related to each other.
Scalar Product and Inner Product in Bra-Ket notation Here you will learn how to use bra-ket notation to write up the overlap integral as the scalar product of bra-ket vectors.
Tensor product (density matrix)

Wave function as a vector in Hilbert space

Consider any one-dimensional wave function $ \mathit{\Psi}(x)$ describing a quantum mechanical particle. The value of the wave function, for example at location $ \class{red}{x_1} $ is $ \mathit{\Psi}(\class{red}{x_1}) $, at location $\class{green}{x_2}$ the function value is $ \mathit{\Psi}(\class{green}{x_2}) $, at location $\class{blue}{x_3}$ it is $ \mathit{\Psi}(\class{blue}{x_3}) $ and so on. You can assign a function value to each $x$ value in this way. We can then represent the whole function values as a list of values. We can think of this list of values as a column vector $ \mathit{\Psi}$ that lives in an abstract space. The vector then has the components:

$$ \begin{align} \begin{bmatrix} \mathit{\Psi}(\class{red}{x_1}) \\ \mathit{\Psi}(\class{green}{x_2}) \\ \mathit{\Psi}(\class{blue}{x_3}) \\ . \\ . \\ . \end{bmatrix} \end{align} $$

This vector is spanned by the coordinates $ \mathit{\Psi}(\class{red}{x_1}) $, $ \mathit{\Psi}(\class{green}{x_2}) $, $ \mathit{\Psi}(\class{blue}{x_3}) $. We can even illustrate this approximated vector (see Ilustration 1, right):

Get this illustration

Left: Real wave function $\Psi(x)$ and its three example function values. Right: Three function values span an approximate coordinate system in which the wave function $\Psi$ is understood as a vector.

As soon as we add an additional function value $ \mathit{\Psi}(x_4) $, the space becomes four-dimensional and can no longer be represented graphically. The principle of how a function can be understood as a vector is hopefully now clear. In the case of the wave function, we call the vector 2 the state vector.

Theoretically there are of course infinitely many $ x $ values. Therefore there are also infinitely many associated function values $ \mathit{\Psi}(x) $. If there are infinitely many function values $ \mathit{\Psi}(x) $, the space in which the vector $ \mathit{\Psi}$ lives is infinite-dimensional. Remember, this is not an infinite-dimensional position space, but an abstract space, as shown in Illustration 1. This abstract space in which quantum mechanical state vectors live is called Hilbert space. In general, this is an infinite-dimensional vector space. But it can also be finite-dimensional, as for example in the case of spin states.

If you approximate the wave function for example for numerical calculations with the computer, by finitely many function values (otherwise it does not work), the $ \mathit{\Psi}$-vector 1 will have finitely many components. It is clear that if you choose a larger $n$, the representation of the wave function as a vector becomes more accurate:

$$ \begin{align} \begin{bmatrix} \mathit{\Psi}(\class{red}{x_1}) \\ \mathit{\Psi}(\class{green}{x_2}) \\ \mathit{\Psi}(\class{blue}{x_3}) \\ . \\ . \\ . \\ \mathit{\Psi}(x_n) \end{bmatrix} \end{align} $$

If the Hilbert space is finite-dimensional, then a $ \mathit{\Psi}$ vector, such as 2, with $n$ components, lives in an $n$ dimensional Hilbert space.

Bra and Ket vectors

So, as you learned, we can represent a quantum mechanical particle in two ways:

as a wave function
as a state vector

In order to distinguish the description of the particle as a state vector from the description as a wave function, we write the state vector 1 as follows:

Ket vector

$$ \begin{align} |\mathit{\Psi}\rangle ~=~ \begin{bmatrix} \mathit{\Psi}(\class{red}{x_1}) \\ \mathit{\Psi}(\class{green}{x_2}) \\ \mathit{\Psi}(\class{blue}{x_3}) \\ . \\ . \\ . \end{bmatrix} \end{align} $$

Wave function $\mathit{\Psi}(x)$ represented as a column vector is called ket vector $|\mathit{\Psi}\rangle$. It doesn't matter what you write between $ | ~~ \rangle $. For example, you could have written $|\mathit{\Psi}(x)\rangle$. The main thing is that it is clear from the ket notation which quantum mechanical system the ket vector 3 represents.

So when you see the ket notation $|\mathit{\Psi}\rangle$, then you know that it means the representation of the particle state as a state vector.
On the other hand, if you see $\mathit{\Psi}(x)$, then you know that it means the representation of the particle state as a wave function.

The vector $|\mathit{\Psi}\rangle^\dagger$ adjoint to the ket vector is called bra vector. The symbol '$\dagger$' is pronounced as 'dagger'. For a clever, compact notation, we write the bra vector with an inverted arrow: $|\mathit{\Psi}\rangle^\dagger ~:=~ \langle\mathit{\Psi}|$. Note that 'adjoint' is sometimes also called 'Hermitian adjoint'.

To get the bra vector $ |\mathit{\Psi}\rangle $ adjoint to the ket vector $ \langle\mathit{\Psi}| $, you need to do two things:

Transpose the ket vector 3. This makes it a row vector:
Transposed ket vector
Formula anchor $$ \begin{align} \left[ \mathit{\Psi}(\class{red}{x_1}),~ \mathit{\Psi}(\class{green}{x_2}),~ \mathit{\Psi}(\class{blue}{x_3}),~~... \right] \end{align} $$
Complex-conjugate the ket vector 3. This operation 'adds asterisks' to the components.

Bra Vector

$$ \begin{align} \langle\mathit{\Psi}| ~=~ \left[ \mathit{\Psi}(\class{red}{x_1})^*,~ \mathit{\Psi}(\class{green}{x_2})^*,~ \mathit{\Psi}(\class{blue}{x_3})^*,~~... \right] \end{align} $$

What are Bra-Ket vectors?

The wave function $\mathit{\Psi}$ in the vector representation corresponds to the ket vector $|\mathit{\Psi}\rangle$ and the row vector $\langle\mathit{\Psi}|$ adjoint to the ket vector is the bra vector.

Since we have interpreted the wave function $\mathit{\Psi}$ as a ket vector $| \mathit{\Psi} \rangle$, we can calculate with it practically in the same way as with usual vectors you know from mathematics. For example, we can form a scalar product or a tensor product between the bra-ket vectors. The thing that is probably new to you is that the components of the vector can be complex and the number of components can be infinite. In this case, you would have a vector with infinitely many complex components. Practically, of course, that is, in the numerical representation of a ket vector, it always has finitely many components. However, even in finite-dimensional Hilbert spaces, which are just as important in quantum mechanics, the number of components of the bra-ket vectors is finite. The spin state vectors for example, live in a two-dimensional Hilbert space. Thus, the Bra and Ket vectors would have only two components.

Scalar Product and Inner Product in Bra-Ket notation

You can form the scalar product of the bra-ket vectors. By the way, in abstract infinite dimensional space, scalar product is called inner product. In finite dimensional Hilbert space, the scalar product between the Bra and Ket vectors looks like this:

$$ \begin{align} \langle\mathit{\Psi} | \mathit{\Psi} \rangle ~=~ \left[ \mathit{\Psi}_{\class{red}{1}}^*,~ \mathit{\Psi}_{\class{green}{2}}^*,~...~, \mathit{\Psi}_n^* \right] ~ \begin{bmatrix} \mathit{\Psi}_{\class{red}{1}} \\ \mathit{\Psi}_{\class{green}{2}} \\ . \\ . \\ \mathit{\Psi}_n \end{bmatrix} \end{align} $$

Here you can see one of the many examples of why notating the bra vector with a reversed arrow is convenient. The notation is suitable for scalar products in finite as well as inner products in infinite Hilbert spaces. Here we do not need to include the scalar product point and a vertical line. We write $\langle\mathit{\Psi} | \mathit{\Psi} \rangle$ instead of $ \langle\mathit{\Psi} | ~\cdot~ | \mathit{\Psi} \rangle $.

Of course, you can also form the scalar product between two different state vectors:

$$ \begin{align} \langle\mathit{\Phi} | \mathit{\Psi} \rangle ~=~ \left[ \mathit{\Phi}_{\class{red}{1}}^*,~ \mathit{\Phi}_{\class{green}{2}}^*,~...~, \mathit{\Phi}_n^* \right] ~ \begin{bmatrix} \mathit{\Psi}_{\class{red}{1}} \\ \mathit{\Psi}_{\class{green}{2}} \\ . \\ . \\ \mathit{\Psi}_n \end{bmatrix} \end{align} $$

You can multiply out the vectors in 7 just as you do with the usual matrix multiplication:

$$ \begin{align} \langle\mathit{\Phi} | \mathit{\Psi} \rangle ~=~ \mathit{\Phi}_{\class{red}{1}}^* \, \mathit{\Psi}_{\class{red}{1}} ~+~ \mathit{\Phi}_{\class{green}{2}}^* \, \mathit{\Psi}_{\class{green}{2}} ~+~ ... ~+~ \mathit{\Phi}_n^* \, \mathit{\Psi}_n \end{align} $$

You can write Eq. 8 even shorter with a sum sign (if the dimension of the Hilbert space is infinite, then the sum is of course only an approximation):

Scalar product of two states

$$ \begin{align} \langle\mathit{\Phi} | \mathit{\Psi} \rangle ~=~ \underset{i=1}{\overset{n}{\boxed{+}}} ~ \mathit{\Phi}_i^* \, \mathit{\Psi}_i \end{align} $$

Here $n$ is the dimension of the Hilbert space, i.e. the number of components of a state vector living in this Hilbert space.

If we take two normalized and orthogonal states $ \mathit{\Psi}_i $ and $ \mathit{\Psi}_j $ and give them indices rather than different letters, then the scalar product gives either 0 or 1 (imagine the scalar product of two basis vectors).

Scalar product of two different orthonormal states ($i \neq j$) yields 0.
Scalar product of two equal, orthonormal states ($i = j$) yields 1.

This orthogonality of states can be combined into a single equation with a Kronecker delta and Einstein sum convention:

Scalar product of two orthonormal states

$$ \begin{align} \langle\mathit{\Psi}_i | \mathit{\Psi}_j \rangle ~=~ \delta_{ij} \end{align} $$

We can make the illustrative transition to an infinite dimensional Hilbert space as follows: We scale the scalar product 9 with $\Delta x$ and let $\Delta x \rightarrow 0 $ go to zero. Then $\Delta x$ changes into $\text{d}x $ and the sum sign into an integral sign, which corresponds to a continuous sum:

$$ \begin{align} \langle\mathit{\Psi} | \mathit{\Psi} \rangle ~=~ \int \mathit{\Psi}^*\mathit{\Psi} \, \text{d}x \end{align} $$

The integral 9 is sometimes called the overlap integral because, like the scalar product, it indicates how much the two states $ \langle\mathit{\Psi} | $ and $| \mathit{\Psi} \rangle$ overlap.

Of course, we can calculate the overlap between two different states. Bra and Ket vector do not have to describe the same state:

Inner product of two states

$$ \begin{align} \langle\mathit{\Phi} | \mathit{\Psi} \rangle ~=~ \int \mathit{\Phi}^*\mathit{\Psi} \, \text{d}x \end{align} $$

What does the scalar product (inner product) say illustratively?

The scalar product and inner product $\langle\mathit{\Phi} | \mathit{\Psi} \rangle$ is a number that measures how much two states $\mathit{\Phi}$ and $\mathit{\Psi} $ overlap.

Tensor product (density matrix)

Another important operation between bra-ket vectors is the tensor product: $|\mathit{\Phi} \rangle ~\otimes~ \langle\mathit{\Psi} |$. We can omit the tensor sign '$\otimes$' because it is immediately clear from the bra-ket notation that it is not a scalar product (the bra and ket vectors are swapped).

The result of the tensor product is a matrix. In the case of the tensor product of two quantum mechanical states this matrix is called density matrix or base-independent density operator:

Density matrix

$$ \begin{align} |\mathit{\Phi} \rangle \langle\mathit{\Psi} | ~=~ \begin{bmatrix} \mathit{\Phi}_{\class{red}{1}} \, \mathit{\Psi}^*_{\class{red}{1}} & \mathit{\Phi}_{\class{red}{1}} \, \mathit{\Psi}^*_{\class{green}{2}} & ... \\ \mathit{\Phi}_{\class{green}{2}} \, \mathit{\Psi}^*_{\class{red}{1}} & \mathit{\Phi}_{\class{green}{2}} \, \mathit{\Psi}^*_{\class{green}{2}} & ... \\ ... & ... & \mathit{\Phi}_n \, \mathit{\Psi}^*_n \end{bmatrix} \end{align} $$

If the state $ |\mathit{\Psi} \rangle $ is normalized and we form the tensor product $|\mathit{\Psi} \rangle \langle\mathit{\Psi} | $, then this product is called a projector or projection matrix. If we apply the projection matrix $|\mathit{\Psi} \rangle \langle\mathit{\Psi} | $ to a ket vector $ |\mathit{\Phi} \rangle $ (imagine a matrix which is applied to a vector): $|\mathit{\Psi} \rangle \langle\mathit{\Psi} | \, \mathit{\Phi} \rangle $, then we get a new ket vector giving the projection of $ |\mathit{\Phi} \rangle $ onto $ |\mathit{\Psi} \rangle $. Thus, we can find out how much of the state $ |\mathit{\Phi} \rangle $ is in the state $ |\mathit{\Psi} \rangle $.

Another important application of the projection matrix is the insertion of identity. Consider a set $ \{|\Psi_i\rangle \} $ of state vectors $|\Psi_i\rangle$ forming an orthonormal basis. That is, these state vectors are orthogonal to each other, normalized, and they span the Hilbert space in which they live. The sum of the projection matrices of these basis vectors gives a $n$ x $n$ unit matrix:

$$ \begin{align} \underset{i=1}{\overset{n}{\boxed{+}}} \, |\mathit{\Psi}_i \rangle \langle\mathit{\Psi}_i | ~=~ \mathbb{1} \end{align} $$

We can apply this unit matrix to any state $ |\mathit{\Phi} \rangle $:

'Insertion of the identity' for a finite base

$$ \begin{align} |\mathit{\Phi} \rangle ~&=~ \mathbb{1} \, |\mathit{\Phi} \rangle \\\\
~&=~ \underset{i=1}{\overset{n}{\boxed{+}}} \, |\mathit{\Psi}_i \rangle \langle\mathit{\Psi}_i | \mathit{\Phi} \rangle \end{align} $$

Here the scalar product $\langle\mathit{\Psi}_i | \mathit{\Phi} \rangle$ picks out the $i$th component, of the state $| \mathit{\Phi} \rangle$. Let us call the $i$th component for short as $\langle\mathit{\Psi}_i | \mathit{\Phi} \rangle := \varphi_i $:

$$ \begin{align} |\mathit{\Phi} \rangle ~=~ \underset{i=1}{\overset{n}{\boxed{+}}} ~ \varphi_i \, |\mathit{\Psi}_i \rangle \end{align} $$

With the trick 'insertion of the identity' we are able to represent a state $|\mathit{\Phi} \rangle$ in the basis $ \{|\Psi_i\rangle \} $.

An infinite-dimensional Hilbert space is spanned by infinitely many basis states. To translate the 'insertion of the identity' (Eq. 15) to an infinite Hilbert space, we scale 15 with $\Delta x$ and let it go to zero. Then $\delta x$ turns into an infinitesimal interval $\text{d}x$ and the sum sign becomes an integral sign:

$$ \begin{align} \int \text{d}x \, |\mathit{\Psi} \rangle \langle\mathit{\Psi} | ~=~ \mathbb{1} \end{align} $$

This unit matrix now has infinitely many columns and rows. With Eq. 17 we are now able to represent a ket vector $ | \Phi \rangle $ living in an infinite-dimensional Hilbert space:

'Insertion of the identity' for an infinite base

$$ \begin{align} |\mathit{\Phi} \rangle ~&=~ \mathbb{1} \, |\mathit{\Phi} \rangle \\\\
~&=~ \int \text{d}x \, |\mathit{\Psi} \rangle \langle\mathit{\Psi} | \mathit{\Phi} \rangle \end{align} $$

With this basic knowledge, you should now be able to understand the equations of quantum mechanics in bra-ket notation.

Log into an avatar

Unlock upgrades