Quantum Mechanics: Math Interlude

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Any ket vector \(|\psi⟩\) can be multiplied by a number \(z\) (which, in general can be real or complex) to produce a new vector \(|\phi⟩\):


In general, \(z=x+iy\). Sometimes the number \(z\) will just be a real number with no imaginary part, but in general it can have both a real and imaginary part. The complex conjugate of the number \(z\) is represented by \(z^*\) and is obtained by changing the sign of the imaginary part of \(z\); so \(z^*=x-iy\). Let’s look at some examples of complex numbers and their complex conjugates. Suppose that \z\) is any real number with no imaginary part: \(z=x+(i·0)=x\). The complex conjugate of any real number is \(z^*=x-(i·0)=x\). In other words taking the complex conjugate \(z^*\) of any real number \(z=x\) just gives the real number back. Suppose, however, that \(z=x+iy\) is any complex number and we take the complex conjugate twice. Let’s see what we get. \(z^*\) is just \(z^*=x-iy\) (a new complex number). If we take the “complex conjugate of the complex conjugate,” we have \((z^*)^*=x+iy=z\). For any complex number \(z\),\(z^*=z\) .

If we multiply any complex number \(z\) by its complex conjugate \(z^*\), we’ll get


For any complex number \(z\) the product \(z^*z\) is always a real number that is greater than or equal to zero. This product is called the modulus squared and, in a very rough sense, represents the length squared of a vector in a complex space. We can write the modulus squared as \(|z|^2=zz^*\). From fig. # we can also see that any complex number can be represented by \(z=x+iy=rcosθ+irsinθ=re^{iθ}\). The complex conjugate of this is \(z^*=x-iy=rcosθ+irsin(-θ)=re^{-iθ}\). The modulus squared of any vector in the complex plain is given by \(|z|^2=zz^*=(re^{iθ})(re^{iθ}=r^2\). If \(|z|^2=r^2=1\) and hence \(|z|=r=1\), then the magnitude of the complex vector is 1 and the vector is called normalized.

Any vector \(|A⟩\) can be expressed as a column vector: \(\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}\). To multiply \(|A⟩\) by any number \(z\) we simply multiply each of the components of the column vector by \(z\) to get \(z|A⟩=z\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}=\begin{bmatrix}zA_1 \\⋮ \\zA_N\end{bmatrix}\). We can add two complex vectors \(|A⟩\) and \(|B⟩\) to get a new complex vector \(|C⟩\). Each of the new components of \(|C⟩\) is obtained by adding the components of \(|A⟩\) and \(|B⟩\) to get \(|A⟩+|B⟩=\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}+\begin{bmatrix}B_1 \\⋮ \\B_N\end{bmatrix}=\begin{bmatrix}A_1+B_1 \\⋮ \\A_N+B_N\end{bmatrix}=|C⟩\). For every ket vector \(|A⟩=\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}\) there is an associated bra vector which is the complex conjugate of \(|A⟩\) and is given by \(⟨A|=\begin{bmatrix}A_1^* &... &A_N^*\end{bmatrix}\). The inner product between any two vectors \(|A⟩\) and \(|B⟩\) is written as \(⟨B|A⟩\). The outer product between any two vectors \(|A⟩\) and \(|B⟩\) is written as \(|A⟩⟨B|\). The rule for taking the inner product between any two such vectors is

$$⟨B|A⟩=\begin{bmatrix}B_1^* &... &B_N^*\end{bmatrix}\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}=B_1^*A_1\text{+ ... +}B_N^*A_N.$$

Whenever you take the inner product of a vector \(|A⟩\) with itself you get

$$⟨A|A⟩=\begin{bmatrix}A_1^* &... &A_N^*\end{bmatrix}\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}=A_1^*A_1\text{+ ... +}A_N^*A_N.$$

We learned earlier that the product between any number \(z\) (which can be a real number but is in general a complex number) and its complex conjugate \(z^*\) (written as ) is always a real number that is greater than or equal to zero. This means that each of the terms \(A_i^*A_i\) is greater than or equal to zero and, therefore, will always equal a real number greater than or equal to zero.

Suppose we have some arbitrary matrix \(\textbf{M}\) whose elements are given by

$$\textbf{M}=\begin{bmatrix}m_{11} &... &m_{N1} \\⋮ &\ddots &⋮ \\m_{1N} &... &m_{NN}\end{bmatrix}.$$

To find the transpose of this matrix (written as \(\textbf{M}^T\)) we interchange the order of the two lower indices of each element. (Another way of thinking about it is that we “reflect” each element about the diagonal.) When we do this we get

$$\textbf{M}^T=\begin{bmatrix}m_{11} &... &m_{1N} \\⋮ &\ddots &⋮ \\m_{N1} &... &m_{NN}\end{bmatrix}.$$

The Hermitian conjugate of a matrix (represented by \(\textbf{M}^†\)) is obtained by first taking the transpose of the matrix and then taking the complex conjugate of each element to get

$$\textbf{M}^†=\begin{bmatrix}m_{11}^* &... &m_{1N^*} \\⋮ &\ddots &⋮ \\m_{N1}^* &... &m_{NN}^*\end{bmatrix}.$$

We represent observables/measurable as linear Hermitian operators. In our electron spin example, the observable/measurable is given by the linear Hermitian operator \(\hat{σ}_r\).