Probability of getting specific states -- Quantum Mechanics

In summary: So, you cannot determine the probabilities without knowing the state.Yes, it doesn't give you the information about the state. So, you cannot determine the probabilities without knowing the state.In summary, the conversation discusses a 4x4 Hamiltonian with eigenkets and the probabilities of an electron being in different states. However, without knowing the state of the electron, it is not possible to determine the probabilities of it being in specific states.
  • #1
Thomas Brady
12
0
I'm pretty new to quantum, so I'm pretty sure I'm missing something basic here. I've got a 4x4 Hamiltonian with eigenkets $$\psi_{U} = 1/(\sqrt 2) (\psi_{1up} \pm \psi_{2up})$$ and $$\psi_{D} = 1/(\sqrt 2) (\psi_{1down} \pm \psi_{2down})$$ The only difference between the two states is the spin as signified by the subscripts up (U) and down (D). The plus states have the eigenvalue ##E_0 - t## and the minus states have the eigenvalue ##E_0 + t##. Knowing this, how can I say what the probability is of an electron being in any of the ##\psi_{1up}##, ##\psi_{1down}##, ##\psi_{2up}##, and ##\psi_{2down}## states, without knowing the probability it is in ##\psi_{U}## or ##\psi_{D}##?

Just to be clear there are of course 4 eigenkets, but the difference between each of the plus and minus eigenkets is the spin
 
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  • #2
Thomas Brady said:
Knowing this, how can I say what the probability is of an electron being in any of the ψ1up\psi_{1up}, ψ1down\psi_{1down}, ψ2up\psi_{2up}, and ψ2down\psi_{2down} states, without knowing the probability it is in ψU\psi_{U} or ψD\psi_{D}?
The basis sets ##\psi_u^+,\psi_u^-,\psi_d^+,\psi_u^-## and ##\psi_{1\textrm {up}},\psi_{1\textrm {down}},\psi_{2\textrm {up}},\psi_{2\textrm {down}}## are connected by some transformation matrix. Thus, a given state of an electron can be equivalently expressed in either set. If you don't know the expansion coefficients, which represents the probability being found in the corresponding basis, in one of the sets, you cannot determine the coefficients in the other set.
 
  • #3
blue_leaf77 said:
The basis sets ##\psi_u^+,\psi_u^-,\psi_d^+,\psi_u^-## and ##\psi_{1\textrm {up}},\psi_{1\textrm {down}},\psi_{2\textrm {up}},\psi_{2\textrm {down}}## are connected by some transformation matrix. Thus, a given state of an electron can be equivalently expressed in either set. If you don't know the expansion coefficients, which represents the probability being found in the corresponding basis, in one of the sets, you cannot determine the coefficients in the other set.

So, essentially I cannot determine the probabilities, with the given information
 
  • #4
Thomas Brady said:
with the given information
The given information? You mean the equations you posted there? Those vectors you gave there are merely the eigenvectors of the Hamiltonian. An electron can have arbitrary wavefunction which is some linear combination of those 4 vectors. In order to know the probabilities you have to know the state the electron is in.
 
  • #5
blue_leaf77 said:
The given information? You mean the equations you posted there? Those vectors you gave there are merely the eigenvectors of the Hamiltonian. An electron can have arbitrary wavefunction which is some linear combination of those 4 vectors. In order to know the probabilities you have to know the state the electron is in.

Right, and with the problem I was given it does not appear to specify which state it is in.
 
  • #6
Thomas Brady said:
Right, and with the problem I was given it does not appear to specify which state it is in.
Yes, it doesn't give you the information about the state.
 

Related to Probability of getting specific states -- Quantum Mechanics

What is the probability of getting a specific state in Quantum Mechanics?

The probability of getting a specific state in Quantum Mechanics is given by the square of the amplitude of the state in the wave function. This is known as the Born rule and is a fundamental principle in quantum mechanics.

How is the probability of getting a specific state calculated in Quantum Mechanics?

The probability of getting a specific state is calculated by taking the absolute value of the amplitude of the state in the wave function and squaring it. This gives the probability of observing that particular state when a measurement is made.

Can the probability of getting a specific state be greater than 1 in Quantum Mechanics?

No, the probability of getting a specific state cannot be greater than 1 in Quantum Mechanics. This is because the sum of all possible probabilities must equal to 1, as there is a 100% chance of observing a state when a measurement is made.

What is the relationship between the probability of getting a specific state and the uncertainty principle in Quantum Mechanics?

The uncertainty principle in quantum mechanics states that it is impossible to know both the position and momentum of a particle with certainty. This is related to the probability of getting a specific state as the more precisely we know the state of a particle, the less certain we are about its position and momentum.

How does the probability of getting a specific state change over time in Quantum Mechanics?

In Quantum Mechanics, the probability of getting a specific state can change over time due to the wave function evolving according to the Schrodinger equation. The probability of observing a specific state can also be affected by measurements and interactions with other particles.

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