Why is My Calculation of Φ Incorrect?

Click For Summary

Homework Help Overview

The discussion revolves around the calculation of a quantum state represented by <+|Φ> and the normalization of this state. Participants are examining the correctness of a probability calculation involving complex numbers.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the normalization of the quantum state and the implications for calculating probabilities. Questions are raised about the computation of the inner product <+|Φ> and the process of squaring complex numbers. There is also a mention of a potential typo affecting the normalization.

Discussion Status

The discussion is active, with participants exploring different interpretations of the calculations and clarifying the normalization issue. Some guidance has been offered regarding the potential typo and the need for normalization, but no consensus has been reached on the final answer.

Contextual Notes

There is an emphasis on the importance of normalization in quantum mechanics, and participants are navigating the implications of this requirement on their calculations. The original poster's calculation appears to have been based on an unnormalized state, which is a key point of discussion.

gremio594
Messages
7
Reaction score
0
Homework Statement
The state

|ϕ⟩=(1−2√i)/4|0⟩−(3−2i)/4|1⟩
is measured in the Hadamard basis |+⟩, |−⟩. What is the probability to obtain |+⟩ as measurement result?
Relevant Equations
Pψ(v) = |<v|ψ>|^2
I calculated <+|Φ> to be (1-√2i)/4√2 + (3-2i)/4√2. When I squared this I for 16/32 but this is no the right answer.
 
Physics news on Phys.org
First of all, this state is not normalized, so you cannot compute the probability as ##\left|\left<+\right|\left.\phi\right>\right|^2##.
I think probably you have a typo and the term ##2\sqrt{i}## should be ##i\sqrt{2}##, then the state is normalized. But two questions:
1. How did you compute ##\left<+\right|\left.\phi\right>##?
2. How did you square this complex number?
 
Last edited:
Gaussian97 said:
First of all, this state is not normalized, so you cannot compute the probability as ##\left|\left<+\right|\left.\phi\right>\right|^2##.
I think probably you have a typo and the term ##2\sqrt{i}## should be ##i\sqrt{2}##, then the state is normalized. But two questions:
1. How did you compute ##\left<+\right|\left.\phi\right>##?
2. How did you square this complex number?
 
I was not adding the complex numbers together before trying to square them. After doing that I was able to get the right answer
 
What answer did you get finally?
 
Gaussian97 said:
What answer did you get finally?
I think it was (5-2√2)/16
 
Ok, perfect
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K