Schrodinger solution spin half particles

In summary: OK, let's be a little more careful. Suppose we take ##\psi = e^{-i\omega t/2} \begin{pmatrix}1\\1\end{pmatrix}##. Then ##\psi## satisfies the Schrodinger equation$$\hat{H} \psi = \begin{pmatrix}\frac{\hbar \omega}{2}&0\\0&-\frac{\hbar \omega}{2}\end{pmatrix} e^{-i\omega t/2} \begin{pmatrix}1\\1\end{pmatrix} = i\hbar \frac{d\psi}{dt} = i \frac{\hbar \omega
  • #1
crowlma
8
0

Homework Statement


The evolution of a particular spin-half particle is given by the Hamiltonian [itex]\hat{H} = \omega\hat{S}_{z},[/itex] where [itex]\hat{S}_{z}[/itex] is the spin projection operator.
a) Show that [itex]\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{-i\frac{\omega}{2}t}\\
e^{i\frac{\omega}{2}t}
\end{pmatrix} [/itex] is a solution to the Schrodinger equation.
b) Calculate [itex]<\hat{S}_{x}>[/itex] as a function of time with respect to this state.

We are told
[itex]\hat{S}_{z} = \frac{\bar{h}}{2}\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}, [/itex][itex]\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}[/itex]

Homework Equations


det([itex]\hat{S}_{z} - λI)=0[/itex]


The Attempt at a Solution


This was a previous exam example - we went over it in class but I got a little bit lost. I know it has to do with eigenvalues and eigenvectors, and I can get up to a certain point but then I get stuck, and I've no clue about where to start for b, we didn't get time to do that in class.

I get that det([itex]\hat{S}_{z} - λI)=0[/itex], and I know that ([itex]\hat{S}_{z} - λI) = \begin{pmatrix}
\frac{\bar{h}}{2}-λ&0\\
0&-\frac{\bar{h}}{2}-λ
\end{pmatrix}.[/itex] Substituting this into det([itex]\hat{S}_{z} - λI)=0[/itex] gives [itex]\frac{-\bar{h}^{2}}{4}+λ^{2}=0.[/itex] Solving for λ gives [itex]λ=\frac{\bar{h}}{2}[/itex] or [itex]λ=-\frac{\bar{h}}{2}[/itex]. Then I substitute this back in, so that if [itex]λ=\frac{\bar{h}}{2}[/itex] then [itex] \begin{pmatrix}
0&0\\
0&-\bar{h}
\end{pmatrix} Z=0[/itex] where Z is some vector. Also if [itex]λ=-\frac{\bar{h}}{2}[/itex] then [itex] \begin{pmatrix}
\bar{h}&0\\
0&0
\end{pmatrix} Z=0[/itex] where Z is some vector. This is where I get stuck - I can't seem to solve for vector Z, which takes the form of \begin{pmatrix}a\\b\end{pmatrix}, without just ending up with zeros. Not sure how to find a meaningful value of Z. And even if I had one, not sure how to bring that around to prove [itex]\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{-i\frac{\omega}{2}t}\\
e^{i\frac{\omega}{2}t}
\end{pmatrix} [/itex] is a solution to the Schrodinger equation.
 
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  • #2
crowlma said:
Then I substitute this back in, so that if [itex]λ=\frac{\bar{h}}{2}[/itex] then [itex] \begin{pmatrix}
0&0\\
0&-\bar{h}
\end{pmatrix} Z=0[/itex] where Z is some vector. Also if [itex]λ=-\frac{\bar{h}}{2}[/itex] then [itex] \begin{pmatrix}
\bar{h}&0\\
0&0
\end{pmatrix} Z=0[/itex] where Z is some vector. This is where I get stuck - I can't seem to solve for vector Z, which takes the form of \begin{pmatrix}a\\b\end{pmatrix}, without just ending up with zeros. Not sure how to find a meaningful value of Z.

Let's take one more step and write, for ##\lambda = \hbar/2##,

$$ \begin{pmatrix}
0&0\\
0&-\bar{h}
\end{pmatrix} Z = \begin{pmatrix}
0&0\\
0&-\bar{h}
\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} =\begin{pmatrix}
0&0\\
0&-\bar{h} b
\end{pmatrix} =0.$$

This equation implies that ##b=0##, but ##a## is not constrained. So an eigenvector of ##\hat{S}_z## with eigenvalue ##\lambda = + \hbar## is a multiple of

$$ \chi_+ = \begin{pmatrix}1\\0\end{pmatrix}.$$

Similarly, an eigenvector with eigenvalue ##\lambda = + \hbar## is a multiple of (it'll help if you fill in the steps for yourself)

$$ \chi_- = \begin{pmatrix}0\\1\end{pmatrix}.$$


These unit vectors are probably familiar to you from the discussion in class. Eigenvectors of ##\hat{H}## are also multiples of these, but the corresponding eigenvalue also involves a factor of ##\omega##.

And even if I had one, not sure how to bring that around to prove [itex]\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{-i\frac{\omega}{2}t}\\
e^{i\frac{\omega}{2}t}
\end{pmatrix} [/itex] is a solution to the Schrodinger equation.

The time-dependent Schrodinger equation (TDSE) is

$$ \hat{H} \psi = i\hbar \frac{\partial \psi}{\partial t}.$$

The idea is to write ##\psi ## as a linear combination of the eigenvectors of ##\hat{H}## that we found above:

$$ \psi(t) = a(t) \chi_+ + b(t) \chi_- = \begin{pmatrix}
a(t)\\
b(t)
\end{pmatrix}.$$

When ##\hat{H}## acts on ##\chi_\pm##, we can substitute in the appropriate eigenvalue. The TDSE will then become a pair of independent equations for the functions ##a(t),b(t)##. You should see if you can make progress on the problem from this and post back if you're still having trouble,
 
  • #3
Okay so, from there I got to [itex]\hat{H} = ω\hat{S}_{z}[/itex] so then [itex]ω\hat{S}_{z} \begin{pmatrix}
a(t)\\
b(t)
\end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex] so then
[itex] \begin{pmatrix}
\frac{ω \bar{h}}{2}&0\\
0&-\frac{ω \bar{h}}{2}
\end{pmatrix} \begin{pmatrix}
a(t)\\
b(t)
\end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex]
From here I can multiply it out to [itex] \begin{pmatrix}
\frac{iω}{2}a(t)\\
-\frac{iω}{2}b(t)
\end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex] and so I can solve for [itex]\psi = \begin{pmatrix}
e^{\frac{iωt}{2}}\\
e^{-\frac{iωt}{2}}
\end{pmatrix} [/itex]

This looks much closer to the final result, but I can't seem to work out a) why my signs appear to be around the wrong way and b) where the [itex]\frac{1}{\sqrt{2}}[/itex] comes into play. :S
 
  • #4
crowlma said:
Okay so, from there I got to [itex]\hat{H} = ω\hat{S}_{z}[/itex] so then [itex]ω\hat{S}_{z} \begin{pmatrix}
a(t)\\
b(t)
\end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex] so then
[itex] \begin{pmatrix}
\frac{ω \bar{h}}{2}&0\\
0&-\frac{ω \bar{h}}{2}
\end{pmatrix} \begin{pmatrix}
a(t)\\
b(t)
\end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex]
From here I can multiply it out to [itex] \begin{pmatrix}
\frac{iω}{2}a(t)\\
-\frac{iω}{2}b(t)
\end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex] and so I can solve for [itex]\psi = \begin{pmatrix}
e^{\frac{iωt}{2}}\\
e^{-\frac{iωt}{2}}
\end{pmatrix} [/itex]

You've introduced an extra factor of ##i## in the next to last equation. It should be

$$\begin{pmatrix}
\frac{ω}{2}a(t)\\
-\frac{ω}{2}b(t)
\end{pmatrix} = i\hbar\frac{d\psi}{dt}$$

This looks much closer to the final result, but I can't seem to work out a) why my signs appear to be around the wrong way and b) where the [itex]\frac{1}{\sqrt{2}}[/itex] comes into play. :S

The [itex]\frac{1}{\sqrt{2}}[/itex] was introduced to normalize the wavefunction.
 
  • #5
Ah ok, the normalisation makes sense. Forgot about that. Also makes sense about the extra i - I stuffed up typing in the equation, it should have read:
[itex] \begin{pmatrix}
\frac{iω}{2}a(t)\\
-\frac{iω}{2}b(t)\\
\end{pmatrix} = \frac{\delta\psi}{\delta t} [/itex]
Still struggling to work out how I get to:
[itex] \upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{-\frac{iωt}{2}}\\
e^{\frac{iωt}{2}}\\
\end{pmatrix} [/itex]

I seem to keep arriving at:
[itex] \upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{\frac{iωt}{2}}\\
e^{-\frac{iωt}{2}}\\
\end{pmatrix} [/itex] instead? Does the change in signs occur at some point in the normalisation?
 
  • #6
crowlma said:
Ah ok, the normalisation makes sense. Forgot about that. Also makes sense about the extra i - I stuffed up typing in the equation, it should have read:
[itex] \begin{pmatrix}
\frac{iω}{2}a(t)\\
-\frac{iω}{2}b(t)\\
\end{pmatrix} = \frac{\delta\psi}{\delta t} [/itex]
Still struggling to work out how I get to:
[itex] \upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{-\frac{iωt}{2}}\\
e^{\frac{iωt}{2}}\\
\end{pmatrix} [/itex]

I seem to keep arriving at:
[itex] \upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
e^{\frac{iωt}{2}}\\
e^{-\frac{iωt}{2}}\\
\end{pmatrix} [/itex] instead? Does the change in signs occur at some point in the normalisation?

When you moved the ##i## from one side of the equation to the other, you missed the minus sign coming from ##1/i = - i##. You should have

[itex] \begin{pmatrix}
-\frac{iω}{2}a(t)\\
\frac{iω}{2}b(t)\\
\end{pmatrix} = \frac{\delta\psi}{\delta t} [/itex]
 
  • #7
The first part all makes sense but now I'm stuck on part b) - calculate [itex]<\hat{S}_{x}>[/itex].
[itex]\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}
[/itex]
So the expectation value = [itex]\int^{\infty}_{-\infty} \psi*<\hat{S}_{x}>\psi dx[/itex] so I get to [itex]\int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix}
e^{i\frac{w}{2}t}\\
e^{-i\frac{w}{2}t}
\end{pmatrix}\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}\begin{pmatrix}
e^{-i\frac{w}{2}t}\\
e^{i\frac{w}{2}t}
\end{pmatrix} [/itex].
From here I get a bit stuck because if I multiply [itex]<\hat{S}_{x}>[/itex] by [itex] \psi [/itex] then I end up with [itex]\int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix}
e^{i\frac{w}{2}t}\\
e^{-i\frac{w}{2}t}
\end{pmatrix}\begin{pmatrix}
e^{i\frac{w}{2}t}\\
e^{-i\frac{w}{2}t}
\end{pmatrix} [/itex] which comes out to
[itex] \int^{\infty}_{-\infty}\frac{\bar{h}}{2}\psi*^{2}[/itex]. Is this the final answer? I feel like there should be a way to simplify, am I mixing up the calculations somewhere? If I can get to [itex] \int^{\infty}_{-\infty} \psi\psi*[/itex] then I know this cancels out to 1 since the solution is normalised.
 
  • #8
When the wavefunctions are spinors or vectors, instead of ##\psi^*## appearing in the inner product, we must have the adjoint wavefunction ##\psi^\dagger##. In addition to taking the complex conjugate, the adjoint also takes the transpose of the object: ##\psi^\dagger = (\psi^*)^T = (\psi^T)^*##. This is because the inner product ##\langle \psi | \chi \rangle## must be a c-number.
 
  • #9
Ok, so I've followed this through and I end up with [itex]\frac{1}{2}\frac{\bar{h}}{2}[/itex][itex]\int^{\infty}_{-\infty} \begin{pmatrix}
e^{-i\frac{ω}{2}t}t&e^{i\frac{ω}{2}t}
\end{pmatrix}\begin{pmatrix}
e^{i\frac{ω}{2}t}\\
e^{-i\frac{ω}{2}t}
\end{pmatrix}[/itex] which then works out to be [itex]\frac{\bar{h}}{2}[/itex] ?
 
  • #10
In post #7, you had

$$\psi^* = \begin{pmatrix}
e^{i\frac{w}{2}t}\\
e^{-i\frac{w}{2}t}
\end{pmatrix}, $$

which is correct. When we take the transpose of this, we have

$$ \psi^\dagger = \begin{pmatrix}
e^{i\frac{w}{2}t} &
e^{-i\frac{w}{2}t}
\end{pmatrix} .$$

When you put this back in the expectation value, you should write the result in terms of trig functions to verify that the expectation value is real.
 
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1. What is Schrodinger's solution for spin half particles?

Schrodinger's solution for spin half particles is a quantum mechanical equation that describes the behavior of particles with a spin of 1/2, such as electrons. It is based on the Schrodinger equation and takes into account the spin of the particle as an additional degree of freedom.

2. How does Schrodinger's solution for spin half particles differ from the classical model?

Unlike the classical model, which only considers the position and momentum of particles, Schrodinger's solution also takes into account the spin of the particle. This allows for a more accurate description of the behavior of particles at the quantum level.

3. What is the significance of Schrodinger's solution for spin half particles?

Schrodinger's solution has been crucial in understanding the behavior of quantum particles, particularly in the field of quantum mechanics. It has also played a major role in the development of technologies such as transistors and lasers.

4. Can Schrodinger's solution be applied to all spin half particles?

Yes, Schrodinger's solution can be applied to all spin half particles, including electrons, protons, and neutrons. It is a fundamental equation in quantum mechanics and is applicable to all spin 1/2 particles.

5. Are there any limitations to Schrodinger's solution for spin half particles?

While Schrodinger's solution is a powerful tool for understanding spin half particles, it does have its limitations. It does not take into account the effects of relativity, and it is not applicable to particles with spins greater than 1/2.

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