Question

For the vector $$ |a\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 0 \\ 1 \\ -1 \\ 0 \end{array}\right) $$ (a) find the associated projection matrix, \(\mathbf{P}_{a}\). (b) Verify that \(\mathbf{P}_{a}\) does project an arbitrary vector in \(\mathbb{C}^{4}\) along \(|a\rangle .\) (c) Verify directly that the matrix \(1-\mathbf{P}_{a}\) is also a projection operator.

Short answer

The projection matrix associated with the vector \(|a\rangle\) can be found by using the outer product of the vector with itself. The projection capability of \(\mathbf{P}_a\) can be confirmed by checking that it retains the parallelism of an arbitrary vector in \(\mathbb{C}^{4}\) with the original vector \(|a\rangle\) when projecting onto \(|a\rangle\). Lastly, the matrix \(1-\mathbf{P}_a\) can also be proven to be a projection operator by checking that it is both hermitian and idempotent.

Step-by-step solution

Find the projection matrix

To find the projection matrix \(\mathbf{P_a}\) associated with the given vector \(|a\rangle\), we can use the outer product of the vector with itself, which can be defined as: _projection matrix_ = \(|a\rangle\langle a|\). In this case, it is necessary to perform the calculation of the outer product \(|a⟩⟨a|\), where: \(|a⟩\) = \[\frac{1}{\sqrt{2}}\left(\begin{array}{c} 0 \ 1 \ -1 \ 0\end{array}\right)\], and \(⟨a|\) is just the conjugate transpose of the column vector:\[\frac{1}{\sqrt{2}}\left[0, 1, -1, 0\right]\]

Verify that \(\mathbf{P_a}\) projects an arbitrary vector

To verify, consider an arbitrary vector \(|b⟩\) ∈ \(\mathbb{C}^{4}\). When \(|b⟩\) is projected along \(|a⟩\) using \(\mathbf{P_a}\), the resulting vector remains parallel to \(|a⟩\). This can be confirmed by multiplying the matrix \(\mathbf{P_a}\) with \(|b⟩\), and checking whether the resulting vector is a scalar product of the given vector \(|a⟩\.

Verify that the matrix \(1-\mathbf{P_a}\) is also a projection operator

A matrix is a projection operator if it satisfies the following properties: it is hermitian (\(\mathbf{P}=\mathbf{P}^{\dagger}\)) and idempotent (\(\mathbf{P}^{2}=\mathbf{P}\)). First, it is necessary to show that the matrix \(1-\mathbf{P_a}\) is hermitian. This means that it should be equal to its conjugate transpose, which can be confirmed by calculation. Then, it is also important to demonstrate that this matrix is idempotent, by noting that squaring it yields the original matrix.

Key concepts

Linear Algebra

To understand the power of linear algebra, one must dive into the ocean of vectors and matrices and their interactions. Linear algebra plays a vital role in various scientific fields, and its fundamental concepts are the driving forces behind many algorithms in computer science and data analysis.

In the context of our textbook exercise, we're dealing with special kinds of matrices known as projection matrices. These matrices act as transformation tools that can project vectors into a lower-dimensional space or onto a particular subspace. The versatility of linear algebra is showcased through its ability to simplify complex operations into algebraic ones, which is particularly useful when handling multi-dimensional data such as the one encountered in our exercise with a four-dimensional complex vector space, \( \mathbb{C}^4 \).

Using linear algebraic methods, we can analyze and manipulate these vectors to find the projection matrix associated with a given vector, and further explore its properties.

Outer Product

The outer product is an essential tool in our linear algebra toolkit, especially when constructing projection matrices. An outer product between two vectors, commonly denoted as \( |a\rangle \langle b| \), results in a matrix rather than another vector. It’s akin to stretching and laying one vector across another, forming a grid or matrix that embodies the interactions between the two original vectors’ individual elements.

For our exercise, we perform the outer product using the given vector \( |a\rangle \) with itself. Here's where we see the unique nature of the outer product shine through, as it transforms a single vector into a matrix that can effectively project any other vector onto the subspace spanned by \( |a\rangle \). This simplicity in construct belies the outer product's utility in creating a projection matrix that is used to change our view of the vector space, focusing only on one particular direction or dimension.

Hermitian Operator

A hermitian operator in linear algebra is a key concept, particularly when dealing with complex vector spaces. A matrix is considered hermitian if it is equal to its own conjugate transpose — symbolically, \( \mathbf{P} = \mathbf{P}^{\dagger} \). In quantum mechanics, for instance, hermitian operators are related to observable properties because they have real eigenvalues and orthogonal eigenvectors, elevating their physical interpretability.

In relation to our textbook problem, we encounter this concept when verifying whether \( \mathbf{P}_a \) and \( 1-\mathbf{P}_a \) can indeed act as projection operators. The hermitian property ensures that what we observe after a projection (real numerical results in our math) reflects a meaningful transformation in the underlying complex vector space. For a projection matrix to truly 'project,' it must preserve certain symmetries, as exemplified by the hermitian property.

Idempotent Matrix

An idempotent matrix is a matrix that, when multiplied by itself, yields the same matrix — this is mathematically expressed as \( \mathbf{P}^2 = \mathbf{P} \). It's like a magical trick where repeating an action doesn't change the outcome. This property is crucial when discussing projection matrices.

In our exercise scenario, the idempotent characteristic of projection matrices must be verified to confirm that the operation of projecting a vector onto a subspace is consistent. When we apply the projection, we expect no further change with additional applications, reinforcing the idea that the vector has been distilled to its essence in that subspace. It assures us that once projected, the vector has 'found its place' and will remain stable, represented by a point in the projected space.