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Chapter 3: Problem 1

Consider the ordinary vectors in three dimensions \(\left(a_{x} \hat{\imath}+a_{y} \hat{\jmath}+a_{z} \hat{k}\right)\) with complex components. (a) Does the subset of all vectors with \(a_{z}=0\) constitute a vector space? If so, what is its dimension; if not, why not? (b) What about the subset of all vectors whose \(z\) component is 1 ? (c) How about the subset of vectors whose components are all equal?

Short Answer

Expert verified
(a) Yes, dimension = 2; (b) No, it doesn't form a vector space; (c) Yes, dimension = 1.

Step by step solution

01

Checking Subspace Conditions for Part (a)

To determine if the subset where \(a_{z} = 0\) forms a vector space, we must check if it satisfies the vector space conditions: closure under addition and scalar multiplication. Let's consider two vectors \( \mathbf{u} = (a_{x1}, a_{y1}, 0) \) and \( \mathbf{v} = (a_{x2}, a_{y2}, 0) \) in this subset. Their sum is \( \mathbf{u} + \mathbf{v} = (a_{x1} + a_{x2}, a_{y1} + a_{y2}, 0) \), which is also in the subset (keeps \(a_{z}=0\)). Scalar multiplication \(c \mathbf{u} = (ca_{x1}, ca_{y1}, 0)\) remains in the subset for any scalar \(c\). Thus, it is a vector space.
02

Dimension of the Subspace (a)

For the vector space in step 1, since the condition \(a_{z}=0\) reduces the freedom of the vector by one dimension, and we can still freely choose \(a_{x}\) and \(a_{y}\), this vector space is 2-dimensional. A basis is \( \{(1, 0, 0), (0, 1, 0)\} \).
03

Assessing Vector Space Conditions for Part (b)

The subset where \(a_{z} = 1\) is checked for closure under addition. Consider two vectors \( \mathbf{u} = (a_{x1}, a_{y1}, 1) \) and \( \mathbf{v} = (a_{x2}, a_{y2}, 1) \); their sum is \( \mathbf{u} + \mathbf{v} = (a_{x1} + a_{x2}, a_{y1} + a_{y2}, 2) \), not meeting \(a_{z} = 1\). Also, scalar multiplication \(c \mathbf{u} = (ca_{x1}, ca_{y1}, c)\) doesn't yield \(a_{z} = 1\) unless \(c = 1\). Hence, it doesn't satisfy vector space conditions.
04

Analyzing Subspace Conditions for Part (c)

For vectors where components are all equal \( (a, a, a) \), consider closure under addition. For \( \mathbf{u} = (a, a, a) \) and \( \mathbf{v} = (b, b, b) \), their sum \( \mathbf{u} + \mathbf{v} = (a+b, a+b, a+b) \) remains in the subset. For scalar multiplication \(c \mathbf{u} = (ca, ca, ca)\) also keeps all components equal. Thus, it satisfies vector space conditions, forming a 1-dimensional space.
05

Finding Dimension for Subspace (c)

Since the condition specifies all components must be equal, every vector can be expressed as \(a(1, 1, 1)\) where \(a\) is scalar. Thus, the dimension is 1 and a basis is \( \{(1, 1, 1)\} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Subspace
In the realm of vector spaces, a **vector subspace** is a subset that, at itself, is a vector space. To determine if a subset qualifies, it must satisfy two fundamental properties: closure under addition and closure under scalar multiplication.

**Closure under Addition** means that if you add any two vectors from the subset, the resulting vector must also be in the subset. For instance, if we consider vectors in three dimensions with the condition that the third component, \(a_z\), is zero, then for any two vectors \(\mathbf{u} = (a_{x1}, a_{y1}, 0)\) and \(\mathbf{v} = (a_{x2}, a_{y2}, 0)\), their sum is \(\mathbf{u} + \mathbf{v} = (a_{x1} + a_{x2}, a_{y1} + a_{y2}, 0)\), which still maintains \(a_z = 0\).

**Closure under Scalar Multiplication** requires that multiplying any vector in the subset by any scalar results in a vector that is also within the subset. So, if we take any vector like \(\mathbf{u}\) and multiply it by a scalar \(c\), the result \(c\mathbf{u} = (ca_{x1}, ca_{y1}, 0)\) keeps \(a_z = 0\).

In summary, if a subset satisfies these two properties, it forms a vector subspace and thus adheres to the overall structure of a vector space.
Dimension of a Vector Space
The **dimension of a vector space** is an essential concept that denotes the number of vectors in a basis鈥攅ssentially, the minimum number of vectors required to span the entire space. For instance, consider the space formed by all vectors such that \(a_z = 0\) in three dimensions. Here, the space can be spanned by two vectors, \(\{(1, 0, 0), (0, 1, 0)\}\). Therefore, the dimension is 2.

Moreover, the dimension offers insight into the "degrees of freedom" or independent directions that a vector within the space may have. In another case, where vectors have all equal components like \((a, a, a)\), the space only needs one basis vector, \((1, 1, 1)\), to represent any vector. Thus, this space is one-dimensional.

Identifying the dimension gives us an intuitive grasp of how vast the space is and how vectors within it might interact and be represented.
Closure and Scalar Multiplication
**Closure and scalar multiplication** are fundamental operations that determine the structure and integrity of any vector space or subspace.

For closure under vector addition, whenever you add two vectors from a set, the result must also reside within the same set.
For example, in a subspace where a condition such as "all components are equal" is provided, if \(\mathbf{u} = (a, a, a)\) and \(\mathbf{v} = (b, b, b)\) are in the set, their sum \(\mathbf{u} + \mathbf{v} = (a+b, a+b, a+b)\) must also be in the set. This ensures that the vector space structure remains uninterrupted.

**Scalar multiplication** refers to multiplying a vector by a scalar (a real or complex number) and having the result still belong to the same set. If we take \(\mathbf{u} = (a, a, a)\) and multiply it by scalar \(c\), the resulting vector \(c\mathbf{u} = (ca, ca, ca)\) retains the vector's inherent property (e.g., equal components).

This process assures that the vector scaling aligns with the space's conditions, allowing the set to reliably form a subspace maintaining its mathematical consistency.

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Most popular questions from this chapter

In the usual basis \((\hat{i}, \hat{\jmath}, \hat{k})\), construct the matrix \(\mathbf{T}_{x}\) representing a rotation through angle \(\theta\) about the \(x\) -axis, and the matrix \(\mathbf{T}_{y}\) representing a rotation through angle \(\theta\) about the \(y\) -axis. Suppose now we change bases, to \(\hat{\imath}=\hat{\jmath}, \hat{\jmath}=\) \(-\hat{\imath}, \hat{k}=\hat{k}\). Construct the matrix \(\mathbf{S}\) that effects this change of basis, and check that \(\mathbf{S T}_{x} \mathbf{S}^{-1}\) and \(\mathbf{S T}_{y} \mathbf{S}^{-1}\) are what you would expect.

Dirac proposed to peel apart the bracket notation for an inner product, \(\langle\alpha \mid \beta\rangle\), into two pieces, which he called bra \((\langle\alpha|)\) and ket \((|\beta\rangle)\). The latter is a vector, but what exactly is the former? It's a linear function of vectors, in the sense that when it hits a vector (to its right) it yields a (complex) number- product. \({ }^{31}\) (When an operator hits a vector, it delivers another vector; when a bra hits a vector, it delivers a number.) Actually, the collection of all bras constitutes another vector space - the so-called dual space. The license to treat bras as separate entities in their own right allows for some powerful and pretty notation (though I shall not exploit it further in this book), For example, if \(|\alpha\rangle\) is a normalized vector, the operator $$\hat{P} \equiv|\alpha\rangle\langle\alpha|$$ picks out the component of any other vector that "lies along" \(|\alpha\rangle\) : $$\hat{P}|\beta\rangle=(\alpha|\beta\rangle|\alpha\rangle ;$$ we call it the projection operator onto the one-dimensional subspace spanned by \(|\alpha\rangle .\) (a) Show that \(\hat{P}^{2}=\hat{P}\). Determine the eigenvalues of \(\hat{P}\), and characterize its eigenvectors. (b) Suppose \(\left|e_{j}\right\rangle\) is an orthonormal basis for an \(n\) -dimensional vector space. Show that $$\sum_{j=1}^{n}\left|e_{j}\right\rangle\left\langle e_{j}\right|=\mathbf{1}$$ This is the tidiest statement of completeness.the inner (c) Let \(\hat{Q}\) be an operator with a complete set of orthonormal eigenvectors: $$\hat{Q}\left|e_{j}\right\rangle=\lambda_{j}\left|e_{j}\right\rangle \quad(j=1,2,3, \ldots n)$$ Show that \(\hat{Q}\) can be written in terms of its spectral decomposition: $$\hat{Q}=\sum_{j=1}^{n} \lambda_{j}\left|e_{j}\right\rangle\left\langle e_{j}\right|$$ Hint: An operator is characterized by its action on all possible vectors, so what you must show is that $$\hat{Q}|\alpha\rangle=\left\\{\sum_{j=1}^{n} \lambda_{j}\left|e_{j}\right\rangle\left\langlee_{j}\right|\right\\}|\alpha\rangle$$ for any vector \(|\alpha\rangle\).

Suppose you start out with a basis \(\left.\left(\left|e_{1}\right\rangle, \mid e_{2}\right), \ldots,\left|e_{n}\right\rangle\right)\) that is not orthonormal. The Gram-Schmidt procedure is a systematic ritual for generating from it an orthonormal basis \(\left.\left(\left|e_{1}^{\prime}\right\rangle, \mid e_{2}^{\prime}\right), \ldots,\left|e_{n}^{\prime}\right\rangle\right)\). It goes like this: (i) Normalize the first basis vector (divide by its norm): (ii) Find the projection of the second vector along the first, and subtract it off: $$\left|e_{2}\right\rangle-\left\langle e_{1}^{\prime} \mid e_{2}\right\rangle\left|e_{1}^{\prime}\right\rangle .$$ This vector is orthogonal to \(\left|e_{1}^{\prime}\right\rangle\); normalize it to get \(\left|e_{2}^{\prime}\right\rangle\). (iii) Subtract from \(\left|e_{3}\right\rangle\) its projections along \(\left|e_{1}^{\prime}\right\rangle\) and \(\left|e_{2}^{\prime}\right\rangle\) : $$\left|e_{3}\right\rangle-\left\langle e_{1}^{\prime} \mid e_{3}\right\rangle\left|e_{1}^{\prime}\right\rangle-\left\langle e_{2}^{\prime} \mid e_{3}\right\rangle\left|e_{2}^{\prime}\right\rangle .$$ This is orthogonal to \(\left|e_{1}^{\prime}\right\rangle\) and \(\left|e_{2}^{\prime}\right\rangle ;\) normalize it to get \(\left|e_{3}^{\prime}\right\rangle .\) And so on. Use the Gram-Schmidt procedure to orthonormalize the three-space basis \(\left|e_{1}\right\rangle=(1+i) \hat{t}+(1) \hat{\jmath}+(i) \hat{k},\left|e_{2}\right\rangle=(i) \hat{\imath}+(3) \hat{\jmath}+(1) \hat{k},\left|e_{3}\right\rangle=(0) \hat{i}+(28) \hat{\jmath}+(0) \hat{k}\) $$\left|e_{1}^{\prime}\right\rangle=\frac{\left|e_{1}\right\rangle}{\left\|e_{1}\right\|}$$

Prove that \(\operatorname{Tr}\left(\mathbf{T}_{1} \mathbf{T}_{2}\right)=\operatorname{Tr}\left(\mathbf{T}_{2} \mathbf{T}_{1}\right.\) ). It follows immediately that \(\operatorname{Tr}\left(\mathbf{T}_{1} \mathbf{T}_{2} \mathbf{T}_{3}\right)=\operatorname{Tr}\left(\mathbf{T}_{2} \mathbf{T}_{3} \mathbf{T}_{1}\right)\), but is it the case that \(\operatorname{Tr}\left(\mathbf{T}_{1} \mathbf{T}_{2} \mathbf{T}_{3}\right)=\operatorname{Tr}\left(\mathbf{T}_{2} \mathbf{T}_{1} \mathbf{T}_{3}\right)\), in gen- cral? Prove it, or disprove it. Hint: The best disproof is always a counterexampleand the simpler the better!

Find the momentum-space wave function \(\Phi_{n}(p, t)\) for the \(n\) th stationary state of the infinite square well. Construct \(\left|\Phi_{n}\right|^{2}\) (it's simplest to write separate formulas for odd and even \(n\) ). Show that \(\left|\Phi_{n}\right|^{2}\) is finite at \(p=\pm n \pi \hbar / a\).
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