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A unitary linear transformation is one for which \(\hat{U}^{\dagger} \hat{U}=1 .\) (a) Show that unitary transformations preserve inner products, in the sense that \(\langle\hat{U} \alpha \mid \hat{U} \beta\rangle=\langle\alpha \mid \beta\rangle\), for all vectors \(|\alpha\rangle,|\beta\rangle .\) (b) Show that the eigenvalues of a unitary transformation have modulus 1 . (c) Show that the eigenvectors of a unitary transformation belonging to distinct eigenvalues are orthogonal.

Short Answer

Expert verified
(a) Unitary operators preserve inner products. (b) Eigenvalues have modulus 1. (c) Eigenvectors for distinct eigenvalues are orthogonal.

Step by step solution

01

Understand Unitary Transformations

A unitary linear transformation \( \hat{U} \) is defined such that its Hermitian conjugate \( \hat{U}^{\dagger} \) multiplied by \( \hat{U} \) equals the identity operator: \( \hat{U}^{\dagger} \hat{U} = 1 \). This implies that \( \hat{U} \hat{U}^{\dagger} = 1 \) as well, since the inverse equals the Hermitian adjoint for unitary operators.
02

Show Preservation of Inner Products

To show that unitary transformations preserve inner products, consider two vectors \(|\alpha\rangle\) and \(|\beta\rangle\): \[ \langle \hat{U} \alpha | \hat{U} \beta \rangle = (\hat{U} |\alpha\rangle)^{\dagger} (\hat{U} |\beta\rangle) = \langle \alpha| \hat{U}^{\dagger} \hat{U} | \beta \rangle = \langle \alpha | \beta \rangle \]This shows that the inner product remains unchanged by the transformation.
03

Determine the Modulus of Eigenvalues

Given \( \hat{U} |\lambda\rangle = \lambda |\lambda\rangle \) with \( |\lambda\rangle \) an eigenvector and \( \lambda \) an eigenvalue, apply the unitary condition:\[ \hat{U}^{\dagger} \hat{U} |\lambda\rangle = |\lambda\rangle \]Thus, \( \hat{U}^{\dagger} \lambda |\lambda\rangle = \lambda^{*} \lambda |\lambda\rangle = |\lambda\rangle \). This implies \(|\lambda|^2 = 1\), so \(|\lambda| = 1\).
04

Show Orthogonality of Eigenvectors

If \( |\lambda\rangle \) and \( |\mu\rangle \) are eigenvectors belonging to different eigenvalues \( \lambda \) and \( \mu \), assume \( \langle \lambda | \mu \rangle eq 0 \):\[ \hat{U} |\lambda\rangle = \lambda |\lambda\rangle, \quad \hat{U} |\mu\rangle = \mu |\mu\rangle \]Calculate inner products: \[ \lambda \langle \lambda | \mu \rangle = \langle \lambda | \hat{U} | \mu \rangle = \mu \langle \lambda | \mu \rangle \]Since \( \lambda eq \mu \), it leads to \( \langle \lambda | \mu \rangle = 0 \), showing they are orthogonal.

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Key Concepts

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

Inner Product Preservation
Unitary transformations hold a special place in linear algebra, particularly due to their ability to preserve the inner product of vectors. An inner product is a way to multiply two vectors and obtain a scalar, often reflecting a kind of 'dot product'. The preservation of this inner product under unitary transformations ensures that the length (magnitude) and the angle between vectors remain unchanged.
This means if you have two vectors \(|\alpha\rangle\) and \(|\beta\rangle\), and you apply a unitary transformation \(\hat{U}\) to both, the inner product
  • \(\langle \hat{U}\alpha | \hat{U}\beta \rangle\)
equals the original inner product \(\langle \alpha | \beta \rangle\). This relationship, shown by \(\langle \alpha| \hat{U}^{\dagger} \hat{U} | \beta \rangle = \langle \alpha | \beta \rangle\), guarantees that the transformation does not alter the intrinsic geometric properties of the vector space.
This property is crucial in quantum mechanics and various other fields where unitary transformations are commonly used, as it ensures that physical properties, like probabilities, remain consistent before and after transformations.
Eigenvalues Modulus
The eigenvalues of a unitary transformation all have a modulus (absolute value) of 1, which is an important property reflecting stability in transformations. For a given eigenvalue \(\lambda\) associated with an eigenvector \(|\lambda\rangle\), the condition imposed by unitary transformations is:
  • \(\hat{U} |\lambda\rangle = \lambda |\lambda\rangle\)
Through the operation of the Hermitian conjugate, \(\hat{U}^{\dagger}\lambda|\lambda\rangle = \lambda^{*} \lambda |\lambda\rangle = |\lambda\rangle\), we deduce that \(|\lambda|^2 = 1\).
Thus, \(|\lambda| = 1\). This fact ensures that unitary transformations preserve the energy or 'intensity' of the eigenvectors, which is essential in many practical applications, such as keeping signal integrity in quantum computing and wave mechanics.
This property also contributes to the unitary transformation's ability to maintain the length of vectors as it avoids any stretching or shrinking, ensuring vectors map to a space of equivalent dimensions and lengths.
Orthogonality of Eigenvectors
Orthogonality among eigenvectors of a unitary transformation is a fascinating concept that plays a critical role in simplifying problems in mathematics and physics. When eigenvectors correspond to distinct eigenvalues, they are orthogonal to each other. This means that their inner product is zero.
  • If two eigenvectors \(|\lambda\rangle\) and \(|\mu\rangle\) belong to different eigenvalues \(\lambda\) and \(\mu\), their inner product \(\langle \lambda | \mu \rangle \) equals zero.
This arises from:
  • \(\lambda \langle \lambda | \mu \rangle = \mu \langle \lambda | \mu \rangle\)
where differing eigenvalues \(\lambda eq \mu\) suggest \(\langle \lambda | \mu \rangle = 0\).
Orthogonality is a vital feature because it ensures that the eigenvectors can be used as a basis for the vector space. This, in turn, simplifies many mathematical problems because each component can be handled independently. In contexts like quantum mechanics, this orthogonality leads to the superposition principle as states are expressed as linear combinations of orthogonal basis states.

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Most popular questions from this chapter

The \(2 \times 2\) matrix representing a rotation of the \(x y\) -plane is $$\mathbf{T}=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right)$$ Show that (except for certain special angles-what are they?) this matrix has no real eigenvalues. (This reflects the geometrical fact that no vector in the plane is carried into itself under such a rotation; contrast rotations in three dimensions.) This matrix does, however, have complex eigenvalues and eigenvectors. Find them. Construct a matrix \(S\) which diagonalizes \(T\). Perform the similarity transformation \(\left(\mathbf{S T S}^{-1}\right)\) explicitly, and show that it reduces \(\mathbf{T}\) to diagonal form.

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.

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 \(p=\pm n \pi \hbar / a\).

Show that $$\langle x\rangle=\int \Phi^{*}\left(-\frac{\hbar}{i} \frac{\partial}{\partial p}\right) \Phi d p .$$ where \(\Phi(p, t)\) is the momentum-space wave function. In general, $$\langle Q(x, p, t)\rangle=\left\\{\begin{array}{ll} \int \Psi^{*} \hat{Q}\left(x, \frac{\hbar}{i} \frac{\partial}{\partial x}, t\right) \Psi d x, & \text { in position space; } \\ \int \Phi^{*} \hat{Q}\left(-\frac{h}{\partial p}, p, t\right) \Phi d p, & \text { in momentum space. } \end{array}\right.$$

Functions of matrices are defined by their Taylor series expansions; for example, $$e^{M}=\mathbf{1}+\mathbf{M}+\frac{1}{2} \mathbf{M}^{2}+\frac{1}{3 !} \mathbf{M}^{3}+\cdots$$ (a) Find \(\exp (\mathbf{M})\), if $$\mathbf{M}=\left(\begin{array}{lll} 0 & 1 & 3 \\\0 & 0 & 4 \\\0 & 0 & 0\end{array}\right) ; \quad \text { (ii) } \mathbf{M}=\left(\begin{array}{cc}0 & \theta \\\\-\theta & 0\end{array}\right) .$$(b) Show that if \(\mathbf{M}\) is diagonalizable, then $$\operatorname{det}\left(e^{\mathrm{M}}\right)=e^{\mathrm{Tr}(\mathrm{M})}$$ (This is actually true even if \(\mathbf{M}\) is \(n o t\) diagonalizable, but it's harder to prove in the general case.) (c) Show that if the matrices \(\mathbf{M}\) and \(\mathbf{N}\) commute, then $$e^{\mathbf{M}+\mathbf{N}}=e^{\mathbf{M}} e^{\mathbf{N}}$$ Prove (with the simplest counterexample you can think up) that Equation \(3.154\) is \(n\) ot true, in general, for noncommuting matrices. (d) If \(\mathrm{H}\) is Hermitian, show that \(e^{i \mathbf{H}}\) is unitary.

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