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| {{lead missing|date=October 2013}}
| | Hello, dear friend! I am Vivien. I am delighted that I can join to the whole world. I live in Netherlands, in the ZH region. I dream to check out the different countries, to obtain acquainted with intriguing individuals. |
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| ==Introduction==
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| Translation operator acts on an [[eigenvector]] of [[generalized coordinate]] operator, <math> \mathbf{\hat Q} </math> (e.g. [[position operator]]<ref>Chapter-2,Volume-1, Claude Cohen-Tannoudji, Bernard Diu, Franck Laloë</ref> ) and constructs another eigenvector of <math> \mathbf{\hat Q} </math> , with any real eigen value.
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| Mathematically, it is defined as
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| :<math>\mathbf{\hat T} (\lambda) = \exp\left(\frac{-i\lambda\mathbf{\hat P}}{\hbar}\right) </math>
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| where <math> \mathbf{\hat P} </math> is the conjugate [[momentum operator]].
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| Hermitian conjugate of Translation operator,
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| '''
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| :<math>\mathbf{\hat T^ \dagger} (\lambda) =\mathbf{\hat T^{-1}}(\lambda) = \mathbf{\hat T} (-\lambda) </math>
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| So,
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|
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| :<math>\mathbf{\hat T^ \dagger} (\lambda) \mathbf{\hat T}(\lambda)=\mathbf{I} </math>'''
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| Thus, Translation operator is a [[unitary operator]]. '''(''' This can also be seen from the fact that <math> \mathbf{\hat P} </math> is a [[Hermitian operator]]. ''')'''
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| ==A. Operation on the Eigenvectors of <math> \mathbf{\hat Q} </math> ==
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| Let <math>|q\rangle</math> be a non-zero eigenvector of <math> \mathbf{\hat Q} </math> with eigenvalue <math> q </math> .
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| Consider the [[commutator]] <math> [\mathbf{\hat Q,\hat T}(\lambda)] </math>
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| : <math> [\mathbf{\hat Q,\hat T}(\lambda)] = i\hbar \left(\frac{-i\lambda}{\hbar}\right) \exp\left(\frac{-i\lambda\mathbf{\hat P}}{\hbar}\right) = \lambda \mathbf{\hat T}(\lambda)</math> , '''(''' using <math> [\mathbf{\hat Q,\hat P}] = i\hbar </math> and [[Taylor expansion]]''')'''
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| This relation can also be written as,
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| :<math> \mathbf{\hat Q \hat T}(\lambda) = \mathbf{\hat T}(\lambda) (\mathbf{\hat Q} + \lambda) </math>
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| Applying this equation to the vector <math> |q\rangle </math>,
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| :<math> \mathbf{\hat Q \hat T}(\lambda)|q\rangle = \mathbf{\hat T}(\lambda) (\mathbf{\hat Q} + \lambda)|q\rangle = \mathbf{\hat T}(\lambda) (q + \lambda)|q\rangle = (q + \lambda)\mathbf{\hat T}(\lambda)|q\rangle </math>
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| This equation shows that <math> \mathbf{\hat T}(\lambda)|q\rangle </math> is another non-zero eigenvector of <math> \mathbf{\hat Q} </math> ,with an eigen value of <math> q+\lambda </math> . '''(''' <math> \mathbf{\hat T}(\lambda)|q\rangle </math> is non-zero because <math> \mathbf{\hat T}(\lambda) </math> is unitary ''')''' . Since <math> \lambda </math> can take any real value, the spectrum of <math> \mathbf{\hat Q} </math> is therefore a continuous spectrum, composed of all possible values on the real axis.
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| ==B. Properties of Translation Operator==
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| '''<big>1.</big>''' Translation Operator is a Unitary operator.
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| '''<big>Proof:</big>'''
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| :<math>\mathbf{\hat T^ \dagger} (\lambda) =\left(\exp\left(\frac{-i\lambda\mathbf{\hat P}}{\hbar}\right)\right)^ \dagger = \mathbf{\hat T^{-1}}(\lambda) = \mathbf{\hat T} (-\lambda) </math>
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| So,
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| :<math> \mathbf{\hat T^\dagger} (\lambda) \mathbf{\hat T} (\lambda) = \mathbf{\hat T^{-1}} (\lambda) \mathbf{\hat T} (\lambda) = \mathbf{I} </math>
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| '''<big>2.</big>''' <math> \mathbf{\hat T} (\lambda) \mathbf{\hat T} (\mu)= \mathbf{\hat T} (\lambda+\mu) </math>
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| '''<big>3.</big>''' <math> \langle q| \mathbf{ \hat T}(\lambda) = \langle q - \lambda| </math>
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| '''<big>Proof:</big>'''
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| : <math> \mathbf{ \hat T}(\lambda)|q\rangle = |q + \lambda\rangle </math>
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| Its [[Hermitian adjoint|adjoint]] expression is:
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| : <math> \langle q| \mathbf{ \hat T^ \dagger}(\lambda) = \langle q + \lambda| </math>
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| Using property 1,
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| : <math> \langle q| \mathbf{ \hat T}(-\lambda) = \langle q + \lambda| </math>
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| Replacing <big>λ</big> by <big>-λ</big>,
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| :<math> \langle q| \mathbf{ \hat T}(\lambda) = \langle q - \lambda| </math>
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| In the next sections <math> \mathbf{\hat Q} </math> will represent position operator and <math> \mathbf{\hat P} </math> will represent momentum operator.
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| ==C. Action of <math> \mathbf{\hat T}(\lambda) </math> on the Wave function in the { <math> |q\rangle </math> } Representation==
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| Since <math> \mathbf{\hat Q} </math> is an [[observable]], the set of its eigenvectors <big>{</big> <math> |q\rangle </math> <big>} </big> constitutes a basis of [[state space]].<ref>Page no.-108, Chapter-2,Volume-1, Claude Cohen-Tannoudji, Bernard Diu, Franck Laloë</ref> It is possible to charactrize each [[bra-ket notation|ket]] by its " [[wave function]] in the <big>{</big> <math> |q\rangle </math> <big>}</big> representation":
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| : <math> \psi(q) = \langle q|\psi\rangle </math><ref>Page no. 68, Section 1.10, R. Shankar, Principles of Quantum Mechanics</ref>
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| Consider the wave function associated with the ket <math> \mathbf{\hat Q}|\psi\rangle </math> in the <big>{</big> <math> |q\rangle </math> <big>}</big> representation":
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| Using the fact that <math> \mathbf{\hat Q} </math> is Hermitian and <math>|q\rangle</math> is the eigenvector of <math> \mathbf{\hat Q} </math> with eigenvalue '''<big>q</big>''', we get
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| : <math> \langle q|\mathbf{\hat Q}|\psi\rangle = q\langle q|\psi\rangle = q\psi(q) </math>
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| The action of <math> \mathbf{\hat Q} </math> in the <big>{</big> <math> |q\rangle </math> <big>}</big> representation is therefore simply a multiplication by '''<big>q</big>'''.
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| The wave function associated with the ket <math> \mathbf{\hat T}(\lambda)|\psi\rangle </math> in the <big>{</big> <math> |q\rangle </math> <big>}</big> representation is written as:
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| : <math> \langle q|\mathbf{\hat T}(\lambda)|\psi\rangle = \langle q-\lambda|\psi\rangle = \psi(q-\lambda) </math>
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| (Using property 3)
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| The action of the operator <math> \mathbf{\hat T}(\lambda) </math> in the <big>{</big> <math> |q\rangle </math> <big>}</big> representation is therefore a translation of the wave function over a distance <big>λ</big> parallel to the q-axis. For this reason, <math> \mathbf{\hat T}(\lambda) </math> is called the '''translation operator'''.
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| '''For example,''' if <math> \psi(x) \sim \exp(-x^2) </math> is a Gaussian wave function peaked at the origin, <math> \psi(x-\varepsilon) \sim \exp(-(x-\varepsilon)^2) </math> is an identical Gaussian wave function peaked at <big>x=ε</big>. Thus the wavefunction <math> \psi_\varepsilon(x) </math> is obtained by translating (without distortion) the wave function <math> \psi(x) </math> by an amount <big>ε</big> to the right.
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| ==D. Action of <math> \mathbf{\hat P} </math> on the Wave function in the { <math> |q\rangle </math> } Representation using Translation operator==
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| When <big>ε</big> is an infinitely small quantity, we have:
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| : <math> \mathbf{\hat T}(-\varepsilon) = \exp\left(\frac{i\varepsilon\mathbf{\hat P}}{\hbar}\right) = I + i\frac{\varepsilon}{\hbar}\mathbf{\hat P} + O(\varepsilon^2) </math>
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| Consequently,
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| : <math> \langle q|\mathbf{\hat T}(-\varepsilon)|\psi\rangle = \psi(q) + i\frac{\varepsilon}{\hbar}\langle q|\mathbf{\hat P}|\psi\rangle + O(\varepsilon^2) </math>
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| Using the result of section C, we can write
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| : <math> \langle q|\mathbf{\hat T}(-\varepsilon)|\psi\rangle = \psi(q+\varepsilon) </math>
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| Comparison of the last two equations shows that:
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| : <math> \psi(q+\varepsilon) = \psi(q) + i\frac{\varepsilon}{\hbar}\langle q|\mathbf{\hat P}|\psi\rangle + O(\varepsilon^2) </math>
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| It follows that:
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| : <math> \langle q|\mathbf{\hat P}|\psi\rangle = \frac{\hbar}{i}\lim_{\varepsilon\rightarrow 0} \frac{\psi(q+\varepsilon)- \psi(q)}{\varepsilon} = \frac{\hbar}{i}\frac{d}{dq}\psi(q) </math>
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| The action of <math> \mathbf{\hat P} </math> in the <big>{</big> <math> |q\rangle </math> <big>}</big> representation is therefore that of <math> \frac{\hbar}{i}\frac{d}{dq}\psi(q) </math>.
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| ==E. Expectation values of Position and Momentum in the Translated state==
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| Consider a single particle in one dimension. <math> \mathbf{\hat Q} </math> will be replaced by <math> \mathbf{\hat X} </math> , representing the position operator for a particle in one dimension. Unlike classical mechanics, in Quantum mechanics, a particle neither has a well defined position nor a well defined momentum. In the quantum formulation the [[expectation value]]s<ref>Page no. 127, Section 4.2, R. Shankar, Principles of Quantum Mechanics</ref> play the role of the classical variables. Therefore, in quantum mechanics, a shift in a particle's position is denoted by:
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| : <math> \langle\mathbf{\hat X}\rangle \rightarrow \langle\mathbf{\hat X}\rangle + \varepsilon </math>
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| When the translation operator operator acts on a state <math> |\psi\rangle </math>, it gets modified into a translated state, <math> |\psi_\varepsilon\rangle </math> .
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| : <math> \mathbf{\hat T}(\varepsilon)|\psi\rangle = |\psi_\varepsilon\rangle </math>
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| Expectation value of position in the translated state is calculated as:
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| : <math>\begin{align} \langle\psi_{\varepsilon}|\mathbf{\hat X}|\psi_\varepsilon\rangle & = \int_{-\infty}^{\infty} \, x \, |\psi_{\varepsilon}(x)|^2 \, dx \\
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| & = \int_{-\infty}^{\infty} \, x \, |\psi(x-\varepsilon)|^2 \, dx \end{align} </math>
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| Substituting x-ε = x',
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| : <math> \begin{align} \langle\psi_{\varepsilon}|\mathbf{\hat X}|\psi_\varepsilon\rangle & = \int_{-\infty}^{\infty} \,( x'+\varepsilon) \, |\psi(x')|^2 \, dx'\\
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| & = \int_{-\infty}^{\infty} \, x' \, |\psi(x')|^2 \, dx' + \varepsilon \\
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| & = \langle\psi|\mathbf{\hat X}|\psi\rangle + \varepsilon \end{align} </math>
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| Therefore, when a translation operator, <math> \mathbf{\hat T} (\varepsilon) </math> acts on a state of a particle, it shifts the expectation value of its position by a factor '''ε'''.
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| Expectation value of momentum in the translated state is calculated in the similar manner:
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| : <math>\begin{align} \langle\psi_{\varepsilon}|\mathbf{\hat P}|\psi_\varepsilon\rangle & = \int_{-\infty}^{\infty} \, \psi_{\varepsilon}^{*} (x) \, \left(-i\hbar\frac{d}{dx}\right) \, \psi_{\varepsilon}(x) \, dx \\
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| & = \int_{-\infty}^{\infty} \, \psi^{*}(x-\varepsilon) \, \left(-i\hbar\frac{d}{dx}\right) \, \psi(x-\varepsilon) \, dx \end{align} </math>
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| Substituting x-ε = x',
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| : <math> \begin{align} \langle\psi_{\varepsilon}|\mathbf{\hat P}|\psi_\varepsilon\rangle & = \int_{-\infty}^{\infty} \, \psi^{*}(x') \, \left(-i\hbar\frac{d}{dx'}\right) \, \psi(x') \, dx' \\
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| & = \langle\psi|\mathbf{\hat P}|\psi\rangle \end{align} </math>
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| Therefore, when a translation operator acts on a state of a particle,the expectation value of its momentum remains unchanged.
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| ==F. Translational Invariance==
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| [[Translational symmetry|Translational invariance]] in classical mechanics is defined as- Whenever p remains conserved for a system, the [[Hamiltonian]],('''H''') remains invariant under the transformations <math> x\rightarrow x+\varepsilon </math> and <math> p\rightarrow p </math>. In quantum mechanics, it is defined in the similar way, but the physical variables '''x''' and '''p''' are replaced by the expectation values of position and momentum. Therefore, whenever <math> \frac{d}{dt}\langle \mathbf{\hat P}\rangle = 0 </math> , the Hamiltonian remains invariant under the transformations:
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| : <math> \langle\mathbf{\hat X}\rangle \rightarrow \langle\mathbf{\hat X}\rangle + \varepsilon </math>
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| : <math> \langle\mathbf{\hat P}\rangle \rightarrow \langle\mathbf{\hat P}\rangle </math>
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| Since translation of the state of a particle is done by translation operator, invariance of Hamiltonian implies that:
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| :<math> \langle\psi|\mathbf{\hat T^ \dagger} (\varepsilon) \mathbf{\hat H} \mathbf{\hat T} (\varepsilon)|\psi\rangle = \langle\psi_{\varepsilon}|\mathbf{\hat H}|\psi_\varepsilon\rangle = \langle\psi|\mathbf{\hat H}|\psi\rangle </math>
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| where <math> \mathbf{\hat H} </math> is the [[Hamiltonian operator]].
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| Like in classical mechanics, conversely we can also say that whenever the Hamiltonian for a system remains invariant under the above transformations, the momentum remains conserved.It is proved below for an infinitesimal translation:
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| For an infinitesimal translation, Translation operator can be defined as,
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| : <math> \mathbf{\hat T} (\varepsilon) = I - \frac{i\varepsilon}{\hbar}\mathbf{\hat P} </math>
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| For translation invariance,
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| : <math> \begin{align} \langle\psi|\mathbf{\hat H}|\psi\rangle & = \langle\psi_{\varepsilon}|\mathbf{\hat H}|\psi_\varepsilon\rangle \\
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| & = \langle\psi|\mathbf{\hat T^ \dagger} (\varepsilon) \mathbf{\hat H} \mathbf{\hat T} (\varepsilon)|\psi\rangle \\
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| & = \langle\psi|\left( I + \frac{i\varepsilon}{\hbar}\mathbf{\hat P}\right)\mathbf{\hat H}\left( I - \frac{i\varepsilon}{\hbar}\mathbf{\hat P}\right)|\psi\rangle \\
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| & = \langle\psi|\mathbf{\hat H}|\psi\rangle + \frac{i\varepsilon}{\hbar}\langle\psi|\mathbf{[\hat P, \hat H]}|\psi\rangle + O(\varepsilon^2) \end{align} </math>
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| so that, we get, upon equating the coefficient of ε to zero,
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| : <math> \langle\psi|\mathbf{[\hat P, \hat H]}|\psi\rangle = 0 </math>
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| It now follows from [[Ehrenfest theorem]]<ref>Page no. 180, Chapter 6, R. Shankar, Principles of Quantum Mechanics</ref> that
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| : <math> \langle[\hat P, \hat H]\rangle = 0 \rightarrow \frac{d}{dt}\langle \mathbf{\hat P}\rangle = 0 </math>
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| ==G. Time Evolution and Translational Invariance ==
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| In the passive transformation picture, we define translational invariance by the requirement
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| :<math> \mathbf{\hat T^ \dagger} (\varepsilon) \mathbf{\hat H} \mathbf{\hat T}(\varepsilon) = \mathbf{\hat H} </math>
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| Pre-multiplying both sides with <math> \mathbf{\hat T} (\varepsilon) </math> and using the unitarity of Translation operator,we get
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| : <math> [\mathbf{\hat T} (\varepsilon),\mathbf{\hat H}] = 0 </math>
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| It follows that
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| : <math> [\mathbf{\hat T} (\varepsilon),\mathbf{\hat U} (t)] = 0 </math>
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| where <math> \mathbf{\hat U} (t) </math> is the Unitary Time Evolution operator.<ref>Page no.-308, Chapter-3,Volume-1, Claude Cohen-Tannoudji, Bernard Diu, Franck Laloë</ref><big>(</big> When the Hamiltonian is time independent, <math> \hat U(t)= \exp\left(\frac{-i\hat H t}{\hbar}\right) </math>. If the Hamiltonian is time dependent, the above commutation relation is satisfied if <math> \hat P </math> or <math> \hat T(\varepsilon) </math> commutes with <math> \hat H (t) </math> for all t.<big>)</big>
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| [[File:Translation Operator.png|frame|Translational Invariance: The states are represented by wave functions.]]
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| '''<big>Example:</big>''' At t=0 two observers A and B prepare identical systems at x=0 and x=a (fig. 1), respectively.Let <math> |\psi(0)> </math> be the state vector of the system prepared by A, then the state vector of the system prepared by B wil be given by <math> \mathbf{ \hat T} (a)|\psi(0)> </math>. Both the systems look identical to the observers who prepared them. After time t, the state vectors evolve into
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| <math> \mathbf{ \hat U} (t)|\psi(0)> </math> and <math> \mathbf{ \hat U} (t) \mathbf{ \hat T} (t)|\psi(0)> </math>. Using the above commutation relation, the later may be written as <math> \mathbf{ \hat T} (t) \mathbf{ \hat U} (t)|\psi(0)> </math>, which is just the translated version of A's system at time t. Therefore, the two systems, which differed only by a translation at t=0, differ only by the same translation at future times. The time evolution of both the systems appear the same to the observers who prepared them. We can conclude that the translational invariance of Hamiltonian implies that the same experiment repeated at two different places will give the same result (as seen by the local observers).
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| {{reflist}}
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| ==References==
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| * {{cite book
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| | author=Claude Cohen-Tannoudji, Bernard Diu, Franck Laloë
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| | title = Quantum Mechanics(Vol.1)
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| | publisher = John Wiley and Sons
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| }}
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| * {{cite book
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| | author = R. Shankar
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| | title = Principles of Quantum Mechanics
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| | publisher = Springer
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| }}
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| * {{cite book
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| | author = D. J. Griffiths
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| | title = Introduction to Quantum Mechanics
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| | publisher = Prentice Hall
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| }}
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| [[Category:Quantum mechanics]]
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