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.

## Exercise

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

$$ψ(x)=Ce^{-x^2}$$

### 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}

Therfore,

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

We used the following Gaussian integral.

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

### Find the position standard deviation

Acording to

$$Δx=\sqrt{<x^2>-<x>^2},$$

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}}

Finally,

\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.

## Conclusion

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.