Show that an expected value of a vacuum state is equal to 1

In summary, the problem states that if the vacuum state is defined from the usual annihilation operators, then the expectation value of the ket will vanish if the last operator is conjugated.
  • #1
mcas
24
5
Homework Statement
Using the following anticommutator relations of fermionic one-particle operators:
[itex]\{\hat{c}_{k\alpha},\hat{c}_{k'\beta} \}= \{ \hat{c}^\dagger_{k\alpha}, \hat{c}^\dagger_{k'\beta} \} = 0[/itex]
[itex]\{\hat{c}_{k\alpha},\hat{c}^\dagger_{k'\beta} \}=\delta_{kk'}\delta_{\alpha\beta}[/itex]

Show that the expected value for a vacuum state [itex]|\phi_0>[/itex] is:

[itex] <\phi_0| \hat{c}_{-k \downarrow} \hat{c}_{k \uparrow}\hat{c}^\dagger_{k \uparrow}\hat{c}_{-k \downarrow}|\phi_0>=1
[/itex]
Relevant Equations
Given in the homework statement
[itex]
\langle \phi_0| \hat{c}_{-k \downarrow} \hat{c}_{k \uparrow}\hat{c}^\dagger_{k \uparrow}\hat{c}_{-k \downarrow}|\phi_0\rangle = \\ \langle \phi_0| - \hat{c}_{k \uparrow} \hat{c}_{-k \downarrow} \hat{c}^\dagger_{k \uparrow}\hat{c}_{-k \downarrow}|\phi_0\rangle = \\ \langle \phi_0| \hat{c}_{k \uparrow} \hat{c}^\dagger_{k \uparrow} \hat{c}_{-k \downarrow}\hat{c}_{-k \downarrow}|\phi_0\rangle = \\ \langle \phi_0|(1- \hat{c}^\dagger_{k \uparrow} \hat{c}_{k \uparrow} ) \hat{c}_{-k \downarrow}\hat{c}_{-k \downarrow}|\phi_0\rangle = \\ \langle \phi_0|\hat{c}_{-k \downarrow}\hat{c}_{-k \downarrow}|\phi_0\rangle - \langle \phi_0|\hat{c}^\dagger_{k \uparrow} \hat{c}_{k \uparrow} \hat{c}_{-k \downarrow}\hat{c}_{-k \downarrow}|\phi_0\rangle
[/itex]

Then I changed [itex]\hat{c}^\dagger_{k \uparrow} \hat{c}_{k \uparrow}[/itex] in the second term to [itex](1- \hat{c}^\dagger_{k \uparrow} \hat{c}_{k \uparrow} )[/itex] and the result was [itex]\langle \phi_0| \hat{c}_{-k \downarrow} \hat{c}_{k \uparrow}\hat{c}^\dagger_{k \uparrow}\hat{c}_{-k \downarrow}|\phi_0\rangle[/itex] which is exactly what I sarted from.

I don't know where to go from this so I would really appreciate any help!
 
Last edited:
Physics news on Phys.org
  • #2
Where did you get this problem from?

Are the ##\hat{c}_{k\alpha}## usual annihilation operators?
How is the ##\ket{\phi_0}## state defined?

If ##\ket{\phi_0}## is the usual vacuum defined from the ##\hat{c}## operators then, if I'm not wrong, by definition ##\hat{c}_{-k \downarrow} \ket{\phi_0}=0## and the whole expectation value vanishes. So, either I'm confused by something, or the problem is wrong, or you have extra information that is crucial for solving the problem.
 
  • Like
Likes mcas and anuttarasammyak
  • #3
Gaussian97 said:
Where did you get this problem from?

Are the ##\hat{c}_{k\alpha}## usual annihilation operators?
How is the ##\ket{\phi_0}## state defined?

If ##\ket{\phi_0}## is the usual vacuum defined from the ##\hat{c}## operators then, if I'm not wrong, by definition ##\hat{c}_{-k \downarrow} \ket{\phi_0}=0## and the whole expectation value vanishes. So, either I'm confused by something, or the problem is wrong, or you have extra information that is crucial for solving the problem.
It's a problem formulated by my professor.
After you post I contacted him and indeed, he made a mistake in the homework statement, as the last operator [itex]\hat{c}_{-k \downarrow}[/itex] should be conjugated for the expected value to be one. In the original statement, the value will be 0.
Thank you!
 

FAQ: Show that an expected value of a vacuum state is equal to 1

What is a vacuum state in physics?

A vacuum state, also known as a ground state, is the lowest energy state of a quantum mechanical system. It is the state in which the system has no particles and no energy fluctuations.

How is the expected value of a vacuum state calculated?

The expected value of a vacuum state is calculated by taking the average of all possible outcomes of a measurement on the vacuum state, weighted by their respective probabilities. This is known as the expectation value or mean value.

Why is the expected value of a vacuum state important?

The expected value of a vacuum state is important because it is a fundamental property of quantum systems. It allows us to make predictions about the behavior of a system in its lowest energy state, which is crucial in understanding the behavior of matter at the quantum level.

How is the expected value of a vacuum state related to the number 1?

In quantum mechanics, the expected value of a vacuum state is equal to the lowest possible energy level, which is represented by the number 1. This is because the vacuum state has no particles and therefore no energy fluctuations, resulting in an expected value of 1.

Can the expected value of a vacuum state ever be different from 1?

No, the expected value of a vacuum state is always equal to 1. This is because the vacuum state represents the lowest possible energy level and has no particles or energy fluctuations, resulting in a constant expected value of 1.

Back
Top