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Chapter 1: Problem 13

Show that neither the probability of obtaining the result \(a_{i}\) nor the expectation value \(\langle A\rangle\) is affected by \(|\psi\rangle \rightarrow e^{i \delta}|\psi\rangle\), that is, by an overall phase change for the state \(|\psi\rangle\).

Short Answer

Expert verified
Neither probability nor expectation value changes under global phase change.

Step by step solution

01

Understanding the Phase Change

A phase change of the form \(|\psi\rangle \rightarrow e^{i\delta}|\psi\rangle\) is a change where the state vector \(|\psi\rangle\) is multiplied by a complex number on the unit circle, \(e^{i\delta}\), which only changes the phase but not the magnitude of the state vector.
02

Probability Calculation

The probability of obtaining the result \(a_i\) is given by \(P(a_i) = |\langle a_i|\psi\rangle|^2\). After a phase change \(|\psi\rangle \rightarrow e^{i\delta}|\psi\rangle\), this becomes \(P'(a_i) = |\langle a_i|e^{i\delta}|\psi\rangle|^2 = |e^{i\delta}\langle a_i|\psi\rangle|^2 = |\langle a_i|\psi\rangle|^2\), since \(|e^{i\delta}|=1\). Thus, \(P'(a_i) = P(a_i)\).
03

Expectation Value Calculation

The expectation value of operator \(A\), when the state is \(|\psi\rangle\), is given by \(\langle A\rangle = \langle \psi|A|\psi\rangle\). After the phase change \(|\psi\rangle \rightarrow e^{i\delta}|\psi\rangle\), it becomes \(\langle A'\rangle = \langle e^{i\delta}\psi|A|e^{i\delta}\psi\rangle = e^{-i\delta}\langle \psi|A|\psi\rangle e^{i\delta}\), which simplifies to \(\langle \psi|A|\psi\rangle\) because \(e^{-i\delta}e^{i\delta} = 1\). Hence, \(\langle A'\rangle = \langle A\rangle\).
04

Conclusion

Both the probability \(P(a_i)\) and the expectation value \(\langle A\rangle\) remain unchanged under a global phase change \(|\psi\rangle \rightarrow e^{i\delta}|\psi\rangle\). This invariance is due to the nature of quantum mechanics, where only relative phases affect probabilities and expectation values, but not global phases.

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Key Concepts

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

Phase Invariance
In the fascinating world of quantum mechanics, phase invariance is a fundamental concept. It means that a shift in the phase of a quantum state does not affect observable quantities.
Imagine you have a quantum state \(|\psi\rangle\) and you multiply it by \(e^{i\delta}\), a complex number with magnitude 1. This multiplication changes the phase but not the actual state.

The important takeaway is:
  • Phase changes like this do not alter physical measurements.
  • They preserve quantities like probabilities and expectation values.
This invariance is crucial because it allows for consistent predictions in experiments regardless of the phase shift imposed.
Probability Amplitude
The probability amplitude is key to determining the likelihood of various outcomes in quantum systems. For any state \(|\psi\rangle\), the probability that a certain result \(a_i\) is observed is given by \(|\langle a_i|\psi\rangle|^2\).
This notation means you take the amplitude, \(\langle a_i|\psi\rangle\), and square its magnitude.

With a phase change \(|\psi\rangle \rightarrow e^{i\delta}|\psi\rangle\), the probability remains the same because:
  • The absolute value of a complex number on the unit circle, like \(e^{i\delta}\), is 1.
  • Thus, \(|\langle a_i|e^{i\delta}|\psi\rangle|^2 = |\langle a_i|\psi\rangle|^2\).
Hence, phase does not influence the probability amplitudes themselves.
Expectation Value
The expectation value is a measure of the average outcome of a quantum observable repeated many times. It is calculated by \(\langle A\rangle = \langle \psi|A|\psi\rangle\).
When you introduce a phase change, it transforms as \(|\psi\rangle \rightarrow e^{i\delta}|\psi\rangle\).

Despite this change, the expectation value remains invariant:
  • It becomes \(\langle e^{i\delta}\psi|A|e^{i\delta}\psi\rangle = e^{-i\delta}\langle \psi|A|\psi\rangle e^{i\delta}\).
  • This simplifies back to \(\langle \psi|A|\psi\rangle\) because \(e^{-i\delta}e^{i\delta} = 1\).
This shows how quantum mechanics maintains the expectation values under phase shifts, ensuring predictable outcomes.
Quantum States
Quantum states, represented by vectors like \(|\psi\rangle\), are at the heart of quantum mechanics. These states describe all the information about a quantum system.
Each state can undergo transformations, such as phase shifts, while preserving certain physical properties.

Key characteristics include:
  • States can be superpositions, combining multiple possibilities.
  • Observables derived from these states include probabilities and expectation values.
  • Changes in the state don't influence measurements due to phase invariance.
Understanding quantum states is crucial for grasping how quantum systems behave under various transformations.

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Most popular questions from this chapter

Verify that \(\Delta S_{x}=\sqrt{\left\langle S_{x}^{2}\right\rangle-\left\langle S_{x}\right\rangle^{2}}=0\) for the state \(|+\mathbf{x}\rangle\).

Determine the field gradient of a 50 -cm-long Stern-Gerlach magnet that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at \(T=1500 \mathrm{~K}\). Assume the detector (see Fig. 1.1) is located \(50 \mathrm{~cm}\) from the magnet. Note: While the atoms in the oven have average kinetic energy \(3 k_{\mathrm{B}} T / 2\), the more energetic atoms strike the hole in the oven more frequently. Thus the emitted atoms have average kinetic energy \(2 k_{\mathrm{B}} T\), where \(k_{\mathrm{B}}\) is the Boltzmann constant. The magnetic dipole moment of the silver atom is due to the intrinsic spin of the single electron. Appendix F gives the numerical value of the Bohr magneton, \(e \hbar / 2 m_{e} c\), in a convenient form.

Show for a solid spherical ball of mass \(m\) rotating about an axis through its center with a charge \(q\) uniformly distributed on the surface of the ball that the magnetic moment \(\boldsymbol{\mu}\) is related to the angular momentum \(\mathbf{L}\) by the relation $$ \boldsymbol{\mu}=\frac{5 q}{6 m c} \mathbf{L} $$ Reminder: The factor of \(c\) is a consequence of our using Gaussian units. If you work in SI units, just add the \(c\) in by hand to compare with this result.

It is known that there is a \(36 \%\) probability of obtaining \(S_{z}=\hbar / 2\) and therefore a \(64 \%\) chance of obtaining \(S_{z}=-\hbar / 2\) if a measurement of \(S_{z}\) is carried out on a spin- \(\frac{1}{2}\) particle. In addition, it is known that the probability of finding the particle with \(S_{x}=\hbar / 2\), that is in the state \(|+x\rangle\), is \(50 \%\). Determine the state of the particle as completely as possible from this information.
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