Meaning of Bra-ket Notation

In quantum mechanics, use bra vector  \(<a|\) and ket vector \(|a>\). What are the advantages of introducing them?

Meaning of Bra-ket Notation

Represent Complex Inner Product Using a Bra-ket Vector

Express the inner product of two complex functions \(u (x), v (x)\) as \((u (x), v (x)) \), where \(x \) is a real variable. This inner product is expressed by the following integral.

$$\left( u(x),v(x) \right)=\int u(x)^*v(x) dx$$

\(u(x)^*\) is the complex conjugate of \(u(x)\).

In addition, this dot product can be expressed using Bra-Kett vector. In that case, it becomes as follows.

$$\int u(x)^*v(x) dx=<u(x)|v(x)>$$

This is the meaning of the bra-ket vector. That is, the bra-ket vector is introduced because we just wanted to express the inner product easily.

Matrix notation of bra-ket vector

If you want to represent bra-ket vectors like a matrix respectively, you need to make it possible to calculate the inner product of both vectors.

$$<u|=\left( u_1 , u_2 , \ldots \right)$$
$$|v>=\left( \begin{array}{c} v_1 \\ v_2 \\ \vdots  \end{array} \right)$$


\begin{eqnarray} <u|v>&=&\left( u_1 , u_2 , \ldots \right)\left( \begin{array}{c} v_1 \\ v_2 \\ \vdots  \end{array} \right) \\&=&u_1v_1+u_2v_2+\ldots \end{eqnarray}

Why we use bra-ket vectors

Find average value of physical quantity

By using this bra-ket vector, the average value of the physical quantity can be expressed easily.

For example, when the average value of the physical quantity <a> in the wave function \ (Ψ (x, t) \) is represented using bra-ket vectors, the following is obtained.

$$<Ψ|a|Ψ>=\int Ψ^*aΨ dx$$

Changing the physical quantity \(a\) to the position \(x\) or momentum \(p\), you can obtain the average value of the position and momentum respectivity.  Then changing \(a\) to Hamiltonian \(\hat{H}\), you can even obtain the expected value of energy.

Furthermore, if you calculate \(<x^2>\) and \(<p^2>\) in the same way, the standard deviation \(σ=\sqrt{<a^2>-<a>^2}\) also be obtained.


Find the position standard deviation \(Δx\) and momentum standard deviation \(Δp\) in the following wave function \(ψ(x)\).


Normalization of the wave function

First, normalize the wave function and determine the constant \(C\).

\begin{eqnarray} 1&=&\int_{-∞}^∞ |ψ(x)|^2 dx \\&=&\int_{-∞}^∞ ψ(x)^* ψ(x) dx \\&=&\int_{-∞}^{∞} Ce^{-x^2}・Ce^{-x^2} dx\\&=&C^2\int_{-∞}^{∞} e^{-2x^2} dx\\&=&C^2\sqrt{\frac{π}{2}} \end{eqnarray}


$$C=\left( \frac{2}{π} \right)^{\frac{1}{4}}$$

We used the following Gaussian integral.

\int_{-∞}^{∞} e^{-ax^2} dx=\sqrt{\frac{π}{a}}

Find the position standard deviation

Acording to


you should find \(<x^2>\) and \(<x>\) firstly.

\begin{eqnarray} <x>&=&<ψ|x|ψ>\\&=&\int_{-∞}^{∞} ψ^*xψ dx \\&=&\sqrt{\frac{2}{π}}\int_{-∞}^{∞} xe^{-2x^2} dx \end{eqnarray}

Here, \(x\) is an odd function, and \(e^{-2x^2}\) is an even function. Therefore, the integrand \(xe^{-2x^2}\) is an odd function because (odd function) \(\times\) (even function) = (odd function). Integration of an odd function from \(-\infty\) to \(\infty\) results in \(0\), so we can say \(<x>=0\).

\begin{eqnarray} <x^2>&=&<ψ|x^2|ψ>\\&=&\int_{-∞}^{∞} ψ^*x^2ψ dx \\&=&\sqrt{\frac{2}{π}}\int_{-∞}^{∞} x^2e^{-2x^2} dx\\&=&\sqrt{\frac{2}{π}}\frac{\sqrt{π}}{2・2・\sqrt{2}}\\&=&\frac{1}{4} \end{eqnarray}

We used the following Gaussian integral.

\int_{-∞}^{∞} x^2e^{-ax^2} dx=\frac{\sqrt{π}}{2a\sqrt{a}}


\begin{eqnarray} Δx&=&\sqrt{<x^2>-<x>^2}\\&=&\sqrt{\frac{1}{4}-0^2} \\&=&\frac{1}{2} \end{eqnarray}

Find the momentum standard deviation

Next, find the momentum standard deviation \(Δp\). It is almost the same calculation as when \(Δx\).

\begin{eqnarray} <p>&=&<ψ|\hat{p}|ψ>\\&=&\int_{-∞}^{∞} ψ^*\hat{p}ψ dx \\&=&\sqrt{\frac{2}{π} }\int_{-∞}^{∞} e^{-x^2}\left(-iħ\frac{d}{dx} \right)e^{-x^2} dx\\&=&\sqrt{\frac{2}{π} }\int_{-∞}^{∞} e^{-x^2}・(-iħ)・\left(-2xe^{-x^2}\right) dx\\&=&\sqrt{\frac{2}{π} }2iħ\int_{-∞}^{∞}  xe^{-2x^2} dx\\&=&\sqrt{\frac{2}{π} }2iħ・0\\&=&0 \end{eqnarray}
\begin{eqnarray} <p^2>&=&<ψ|\hat{p}^2|ψ>\\&=&\int_{-∞}^{∞} ψ^*\hat{p}^2ψ dx \\&=&\sqrt{\frac{2}{π} }\int_{-∞}^{∞} e^{-x^2}\left(-iħ\frac{d}{dx} \right)^2e^{-x^2} dx\\&=&\sqrt{\frac{2}{π} }\int_{-∞}^{∞} e^{-x^2}・(-iħ)^2・\frac{d}{dx}\left(-2xe^{-x^2}\right) dx\\&=&-ħ^2\sqrt{\frac{2}{π} }\int_{-∞}^{∞} e^{-x^2}・-2\frac{d}{dx}\left(xe^{-x^2}\right) dx\\&=&2ħ^2\sqrt{\frac{2}{π} }\int_{-∞}^{∞} e^{-x^2}・\left(e^{-x^2}+x・(-2xe^{-x^2})\right) dx\\&=&2ħ^2\sqrt{\frac{2}{π} }\int_{-∞}^{∞} \left(e^{-2x^2}-2x^2e^{-2x^2}\right) dx\\&=&2ħ^2\sqrt{\frac{2}{π} }\left( \int_{-∞}^{∞} e^{-2x^2} dx-2\int_{-∞}^{∞}x^2e^{-2x^2} dx \right)\\&=&2ħ^2\sqrt{\frac{2}{π} }\left( \sqrt{\frac{π}{2}}-2\frac{\sqrt{π}}{2・2\sqrt{2}} \right)\\&=&2ħ^2\left( 1-2・\frac{1}{2・2} \right)\\&=&ħ^2 \end{eqnarray}
\begin{eqnarray} Δp&=&\sqrt{<p^2>-<p>^2}\\&=&\sqrt{ħ^2-0^2} \\&=&ħ \end{eqnarray}

Bra-ket vectors and wave function

The wave function in quantum mechanics can be regarded as a complex vector of infinite dimension. Therefore, the wave function \ (φ \) can be written as \ (| φ> \) using a ket vector. Furthermore, there are cases in which quantum numbers included in wave functions are extracted and written like \ (| l, m \!> \).

In the example given in this article, the benefits of the bra-ket vector are small. But in more difficult cases, it will be heavily used. Let’s become familiar with these notations in the moment.


Bra-ket vectors are the way of expressions introduced to make mathematical expressions easier to see. In quantum mechanics, they are used to express wave functions mainly.

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